3004 lines
101 KiB
TeX
3004 lines
101 KiB
TeX
\documentclass{report}
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\usepackage{graphicx}
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\graphicspath{{./Apostol/images/}}
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\input{../preamble}
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\makeleancommands{..}
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\begin{document}
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\header
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{One-Variable Calculus, with an Introduction to Linear Algebra}
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{Tom M. Apostol}
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\tableofcontents
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\begingroup
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\renewcommand\thechapter{R}
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\setcounter{chapter}{0}
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\addtocounter{chapter}{-1}
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\chapter{Reference}%
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\label{chap:reference}
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\section{\defined{Characteristic Function}}%
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\label{ref:characteristic-function}
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Let $S$ be a set of points on the real line.
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The \textbf{characteristic function} of $S$ is the function $\mathcal{X}_S$ such
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that $\mathcal{X}_S(x) = 1$ for every $x$ in $S$, and $\mathcal{X}_S(x) = 0$
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for those $x$ not in $S$.
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\begin{definition}
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\lean*{Common/Set/Basic}{Set.characteristic}
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\end{definition}
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\section{\defined{Completeness Axiom}}%
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\label{ref:completeness-axiom}
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Every nonempty set $S$ of real numbers which is bounded above has a supremum;
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that is, there is a real number $B$ such that $B = \sup{S}$.
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\begin{axiom}
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\lean*{Mathlib/Data/Real/Basic}{Real.exists\_isLUB}
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\end{axiom}
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\section{\defined{Infimum}}%
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\label{ref:infimum}
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A number $B$ is called an \textbf{infimum} of a nonempty set $S$ if $B$ has
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the following two properties:
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\begin{enumerate}[(a)]
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\item $B$ is a lower bound for $S$.
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\item No number greater than $B$ is a lower bound for $S$.
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\end{enumerate}
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Such a number $B$ is also known as the \textbf{greatest lower bound}.
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\begin{definition}
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\lean*{Mathlib/Order/Bounds/Basic}{IsGLB}
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\end{definition}
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\section{\pending{Integrable}}%
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\label{ref:integrable}
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Let $f$ be a function defined and bounded on $[a, b]$.
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$f$ is said to be \textbf{integrable} if there exists one and only one number
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$I$ such that \eqref{ref:integral-bounded-function-eq2} holds.
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If $f$ is integrable on $[a, b]$, we say that the integral
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$\int_a^b f(x) \mathop{dx}$ \textbf{exists}.
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\section{\pending{Integral of a Bounded Function}}%
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\label{ref:integral-bounded-function}
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Let $f$ be a function defined and bounded on $[a, b]$.
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Let $s$ and $t$ denote arbitrary step functions defined on $[a, b]$ such that
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\begin{equation}
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\label{ref:integral-bounded-function-eq1}
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s(x) \leq f(x) \leq t(x)
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\end{equation}
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for every $x$ in $[a, b]$.
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If there is one and only one number $I$ such that
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\begin{equation}
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\label{ref:integral-bounded-function-eq2}
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\int_a^b s(x) \mathop{dx} \leq I \leq \int_a^b t(x) \mathop{dx}
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\end{equation}
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for every pair of step functions $s$ and $t$ satisfying
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\eqref{ref:integral-bounded-function-eq1}, then this number $I$ is called
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the \textbf{integral of $f$ from $a$ to $b$}, and is denoted by the symbol
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$\int_a^b f(x) \mathop{dx}$ or by $\int_a^b f$.
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If $a < b$, we define $\int_b^a f(x) \mathop{dx} = -\int_a^b f(x) \mathop{dx}$,
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provided $f$ is \nameref{ref:integrable} on $[a, b]$.
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We also define $\int_a^a f(x) \mathop{dx} = 0$.
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The function $f$ is called the \textbf{integrand}, the numbers $a$ and $b$ are
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called the \textbf{limits of integration}, and the interval $[a, b]$ the
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\textbf{interval of integration}.
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\section{\pending{Integral of a Step Function}}%
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\label{ref:integral-step-function}
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Let $s$ be a \nameref{ref:step-function} defined on $[a, b]$, and let
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$P = \{x_0, x_1, \ldots, x_n\}$ be a \nameref{ref:partition} of $[a, b]$
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such that $s$ is constant on the open subintervals of $P$.
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Denote by $s_k$ the constant value that $s$ takes in the $k$th open subinterval
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of $P$, so that
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$$s(x) = s_k \quad\text{if}\quad x_{k-1} < x < x_k, \quad k
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= 1, 2, \ldots, n.$$
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The \textbf{integral of $s$ from $a$ to $b$}, denoted by the symbol
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$\int_a^b s(x)\mathop{dx}$, is defined by the following formula:
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$$\int_a^b s(x) \mathop{dx} = \sum_{k=1}^n s_k \cdot (x_k - x_{k-1}).$$
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If $a < b$, we define $\int_b^a s(x) \mathop{dx} = -\int_a^b s(x) \mathop{dx}$.
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We also define $\int_a^a s(x) \mathop{dx} = 0$.
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\section{\pending{Lower Integral}}%
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\label{ref:lower-integral}
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Let $f$ be a function bounded on $[a, b]$ and $S$ denote the set of numbers
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$\int_a^b s(x) \mathop{dx}$ obtained as $s$ runs through all
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\nameref{ref:step-function}s below $f$.
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That is, let $$S = \left\{ \int_a^b s(x) \mathop{dx} : s \leq f \right\}.$$
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The number $\sup{S}$ is called the \textbf{lower integral of $f$}.
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It is denoted as $\ubar{I}(f)$.
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\section{\pending{Monotonic}}%
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\label{ref:monotonic}
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A function $f$ is called \textbf{monotonic} on set $S$ if it is increasing on
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$S$ or if it is decreasing on $S$.
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$f$ is said to be \textbf{strictly monotonic} if it is strictly increasing on
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$S$ or strictly decreasing on $S$.
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A function $f$ is said to be \textbf{piecewise monotonic} on an interval if its
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graph consists of a finite number of monotonic pieces.
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In other words, $f$ is piecewise monotonic on $[a, b]$ if there is a
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\nameref{ref:partition} of $[a, b]$ such that $f$ is monotonic on each of
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the open subintervals of $P$.
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\section{\defined{Partition}}%
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\label{ref:partition}
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Let $[a, b]$ be a closed interval decomposed into $n$ subintervals by inserting
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$n - 1$ points of subdivision, say $x_1$, $x_2$, $\ldots$, $x_{n-1}$, subject
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only to the restriction
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\begin{equation}
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\label{sec:partition-eq1}
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a < x_1 < x_2 < \cdots < x_{n-1} < b.
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\end{equation}
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It is convenient to denote the point $a$ itself by $x_0$ and the point $b$ by
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$x_n$.
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A collection of points satisfying \eqref{sec:partition-eq1} is called a
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\textbf{partition} $P$ of $[a, b]$, and we use the symbol
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$$P = \{x_0, x_1, \ldots, x_n\}$$ to designate this partition.
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\begin{definition}
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\lean*{Common/Set/Partition}{Set.Partition}
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\end{definition}
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\section{\pending{Refinement}}%
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\label{ref:refinement}
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Let $P$ be a \nameref{ref:partition} of closed interval $[a, b]$.
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A \textbf{refinement} $P'$ of $P$ is a partition formed by adjoining more
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subdivision points to those already in $P$.
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$P'$ is said to be \textbf{finer than} $P$. The union of two partitions $P_1$
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and $P_2$ is called the \textbf{common refinement} of $P_1$ and $P_2$.
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\section{\defined{Step Function}}%
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\label{ref:step-function}
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A function $s$, whose domain is a closed interval $[a, b]$, is called a
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\textbf{step function} if there is a \nameref{ref:partition}
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$P = \{x_0, x_1, \ldots, x_n\}$ of $[a, b]$ such that $s$ is constant on each
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open subinterval of $P$.
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That is to say, for each $k = 1, 2, \ldots, n$, there is a real number $s_k$
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such that $$s(x) = s_k \quad\text{if}\quad x_{k-1} < x < x_k.$$
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Step functions are sometimes called \textbf{piecewise constant functions}.
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\note{At each of the endpoints $x_{k-1}$ and $x_k$ the function must
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have some well-defined value, but this need not be the same as $s_k$.}
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\begin{definition}
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\lean*{Common/Geometry/StepFunction}{Geometry.StepFunction}
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\end{definition}
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\section{\defined{Supremum}}%
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\label{ref:supremum}
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A number $B$ is called a \textbf{supremum} of a nonempty set $S$ if $B$ has
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the following two properties:
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\begin{enumerate}[(a)]
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\item $B$ is an upper bound for $S$.
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\item No number less than $B$ is an upper bound for $S$.
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\end{enumerate}
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Such a number $B$ is also known as the \textbf{least upper bound}.
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\begin{definition}
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\lean*{Mathlib/Order/Bounds/Basic}{IsLUB}
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\end{definition}
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\section{\pending{Upper Integral}}%
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\label{ref:upper-integral}
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Let $f$ be a function bounded on $[a, b]$ and $T$ denote the set of numbers
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$\int_a^b t(x) \mathop{dx}$ obtained as $t$ runs through all
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\nameref{ref:step-function}s above $f$.
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That is, let $$T = \left\{ \int_a^b t(x) \mathop{dx} : f \leq t \right\}.$$
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The number $\inf{T}$ is called the \textbf{upper integral of $f$}.
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It is denoted as $\bar{I}(f)$.
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\endgroup
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\chapter{A Set of Axioms for the Real-Number System}%
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\label{chap:set-axioms-real-number-system}
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\section{\verified{Lemma 1}}%
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\label{sec:lemma-1}
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\begin{lemma}[1]
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Nonempty set $S$ has supremum $L$ if and only if set $-S$ has infimum $-L$.
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\end{lemma}
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\begin{proof}
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\lean{Bookshelf/Apostol/Chapter\_I\_03}
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{Apostol.Chapter\_I\_03.is\_lub\_neg\_set\_iff\_is\_glb\_set\_neg}
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Suppose $L = \sup{S}$ and fix $x \in S$.
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By definition of the \nameref{ref:supremum}, $x \leq L$ and $L$ is the
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smallest value satisfying this inequality.
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Negating both sides of the inequality yields $-x \geq -L$.
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Furthermore, $-L$ must be the largest value satisfying this inequality.
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Therefore $-L = \inf{-S}$.
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\end{proof}
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\section{\verified{Existence of a Greatest Lower Bound}}
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\label{sec:existence-greatest-lower-bound}
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\label{sec:theorem-i.27}
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\begin{theorem}[I.27]
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Every nonempty set $S$ that is bounded below has a greatest lower bound; that
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is, there is a real number $L$ such that $L = \inf{S}$.
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\end{theorem}
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\begin{proof}
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\lean{Bookshelf/Apostol/Chapter\_I\_03}
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{Apostol.Chapter\_I\_03.exists\_isGLB}
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Let $S$ be a nonempty set bounded below by $x$.
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Then $-S$ is nonempty and bounded above by $x$.
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By the \nameref{ref:completeness-axiom}, there exists a
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\nameref{ref:supremum} $L$ of $-S$.
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By \nameref{sec:lemma-1}, $L$ is a supremum of $-S$ if and only if $-L$ is an
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infimum of $S$.
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\end{proof}
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\section{\verified{Positive Integers Unbounded Above}}%
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\label{sec:positive-integers-unbounded-above}
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\label{sec:theorem-i.29}
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\begin{theorem}[I.29]
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For every real $x$ there exists a positive integer $n$ such that $n > x$.
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\end{theorem}
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\begin{proof}
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\lean{Bookshelf/Apostol/Chapter\_I\_03}
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{Apostol.Chapter\_I\_03.exists\_pnat\_geq\_self}
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Let $n = \abs{\ceil{x}} + 1$.
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It is trivial to see $n$ is a positive integer satisfying $n \geq 1$.
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Thus all that remains to be shown is that $n > x$.
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If $x$ is nonpositive, $n > x$ immediately follows from $n \geq 1$.
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If $x$ is positive,
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$$x = \abs{x} \leq \abs{\ceil{x}} < \abs{\ceil{x}} + 1 = n.$$
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\end{proof}
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\section{\verified{Archimedean Property of the Reals}}%
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\label{sec:archimedean-property-reals}
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\label{sec:theorem-i.30}
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\begin{theorem}[I.30]
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If $x > 0$ and if $y$ is an arbitrary real number, there exists a positive
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integer $n$ such that $nx > y$.
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\end{theorem}
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\begin{proof}
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\lean{Bookshelf/Apostol/Chapter\_I\_03}
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{Apostol.Chapter\_I\_03.exists\_pnat\_mul\_self\_geq\_of\_pos}
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Let $x > 0$ and $y$ be an arbitrary real number.
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By \nameref{sec:theorem-i.29}, there exists a positive integer $n$ such that
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$n > y / x$.
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Multiplying both sides of the inequality yields $nx > y$ as expected.
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\end{proof}
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\section{\verified{Theorem I.31}}%
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\label{sec:theorem-i.31}
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\begin{theorem}[I.31]
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If three real numbers $a$, $x$, and $y$ satisfy the inequalities
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$$a \leq x \leq a + \frac{y}{n}$$ for every integer $n \geq 1$, then
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$x = a$.
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\end{theorem}
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\begin{proof}
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\lean{Bookshelf/Apostol/Chapter\_I\_03}
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{Apostol.Chapter\_I\_03.forall\_pnat\_leq\_self\_leq\_frac\_imp\_eq}
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By the trichotomy of the reals, there are three cases to consider:
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\paragraph{Case 1}%
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Suppose $x = a$.
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Then we are immediately finished.
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\paragraph{Case 2}%
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Suppose $x < a$.
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But by hypothesis, $a \leq x$.
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Thus $a < a$, a contradiction.
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\paragraph{Case 3}%
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Suppose $x > a$.
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Then there exists some $c > 0$ such that $a + c = x$.
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By \nameref{sec:archimedean-property-reals}, there exists an integer $n > 0$
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such that $nc > y$.
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Rearranging terms, we see $y / n < c$.
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Therefore $a + y / n < a + c = x$.
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But by hypothesis, $x \leq a + y / n$.
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Thus $a + y / n < a + y / n$, a contradiction.
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\paragraph{Conclusion}%
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Since these cases are exhaustive and both case 2 and 3 lead to
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contradictions, $x = a$ is the only possibility.
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\end{proof}
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\section{\verified{Lemma 2}}%
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\label{sec:lemma-2}
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\begin{lemma}[2]
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If three real numbers $a$, $x$, and $y$ satisfy the inequalities
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$$a - y / n \leq x \leq a$$ for every integer $n \geq 1$, then $x = a$.
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\end{lemma}
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\begin{proof}
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\lean{Bookshelf/Apostol/Chapter\_I\_03}
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{Apostol.Chapter\_I\_03.forall\_pnat\_frac\_leq\_self\_leq\_imp\_eq}
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By the trichotomy of the reals, there are three cases to consider:
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\paragraph{Case 1}%
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Suppose $x = a$.
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Then we are immediately finished.
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\paragraph{Case 2}%
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Suppose $x < a$.
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Then there exists some $c > 0$ such that $x = a - c$.
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By \nameref{sec:archimedean-property-reals}, there exists an integer $n > 0$
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such that $nc > y$.
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Rearranging terms, we see that $y / n < c$.
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Therefore $a - y / n > a - c = x$.
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But by hypothesis, $x \geq a - y / n$.
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Thus $a - y / n < a - y / n$, a contradiction.
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\paragraph{Case 3}%
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Suppose $x > a$.
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But by hypothesis $x \leq a$.
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Thus $a < a$, a contradiction.
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\paragraph{Conclusion}%
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Since these cases are exhaustive and both case 2 and 3 lead to
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contradictions, $x = a$ is the only possibility.
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\end{proof}
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\section{\verified{Theorem I.32}}%
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\label{sec:theorem-i.32}
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Let $h$ be a given positive number and let $S$ be a set of real numbers.
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\subsection{\verified{Theorem I.32a}}%
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\label{sub:theorem-i.32a}
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\begin{theorem}[I.32a]
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If $S$ has a supremum, then for some $x$ in $S$ we have $x > \sup{S} - h$.
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\end{theorem}
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\begin{proof}
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\lean{Bookshelf/Apostol/Chapter\_I\_03}
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{Apostol.Chapter\_I\_03.sup\_imp\_exists\_gt\_sup\_sub\_delta}
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By definition of a \nameref{ref:supremum}, $\sup{S}$ is the least upper
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bound of $S$.
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For the sake of contradiction, suppose for all $x \in S$,
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$x \leq \sup{S} - h$.
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This immediately implies $\sup{S} - h$ is an upper bound of $S$.
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But $\sup{S} - h < \sup{S}$, contradicting $\sup{S}$ being the \textit{least}
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upper bound.
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Therefore our original hypothesis was wrong.
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That is, there exists some $x \in S$ such that $x > \sup{S} - h$.
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\end{proof}
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\subsection{\verified{Theorem I.32b}}%
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\label{sub:theorem-i.32b}
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\begin{theorem}[I.32b]
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If $S$ has an infimum, then for some $x$ in $S$ we have $x < \inf{S} + h$.
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\end{theorem}
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\begin{proof}
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\lean{Bookshelf/Apostol/Chapter\_I\_03}
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{Apostol.Chapter\_I\_03.inf\_imp\_exists\_lt\_inf\_add\_delta}
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By definition of an \nameref{ref:infimum}, $\inf{S}$ is the greatest lower
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bound of $S$.
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For the sake of contradiction, suppose for all $x \in S$,
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$x \geq \inf{S} + h$.
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This immediately implies $\inf{S} + h$ is a lower bound of $S$.
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But $\inf{S} + h > \inf{S}$, contradicting $\inf{S}$ being the
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\textit{greatest} lower bound.
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Therefore our original hypothesis was wrong.
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That is, there exists some $x \in S$ such that $x < \inf{S} + h$.
|
|
|
|
\end{proof}
|
|
|
|
\section{\verified{Additive Property of Supremums and Infimums}}%
|
|
\label{sec:additive-property-supremums-infimums}
|
|
\label{sec:theorem-i.33}
|
|
|
|
Given nonempty subsets $A$ and $B$ of $\mathbb{R}$, let $C$ denote the set
|
|
$$C = \{a + b : a \in A, b \in B\}.$$
|
|
|
|
\note{This is known as the "Additive Property."}
|
|
|
|
\subsection{\verified{Theorem I.33a}}%
|
|
\label{sub:theorem-i.33a}
|
|
|
|
\begin{theorem}[I.33a]
|
|
|
|
If each of $A$ and $B$ has a supremum, then $C$ has a supremum, and
|
|
$$\sup{C} = \sup{A} + \sup{B}.$$
|
|
|
|
\end{theorem}
|
|
|
|
\begin{proof}
|
|
|
|
\lean{Bookshelf/Apostol/Chapter\_I\_03}
|
|
{Apostol.Chapter\_I\_03.sup\_minkowski\_sum\_eq\_sup\_add\_sup}
|
|
|
|
We prove (i) $\sup{A} + \sup{B}$ is an upper bound of $C$ and (ii)
|
|
$\sup{A} + \sup{B}$ is the \textit{least} upper bound of $C$.
|
|
|
|
\paragraph{(i)}%
|
|
\label{par:theorem-i.33a-i}
|
|
|
|
Let $x \in C$.
|
|
By definition of $C$, there exist elements $a' \in A$ and $b' \in B$ such
|
|
that $x = a' + b'$.
|
|
By definition of a \nameref{ref:supremum}, $a' \leq \sup{A}$.
|
|
Likewise, $b' \leq \sup{B}$.
|
|
Therefore $a' + b' \leq \sup{A} + \sup{B}$.
|
|
Since $x = a' + b'$ was arbitrarily chosen, it follows $\sup{A} + \sup{B}$
|
|
is an upper bound of $C$.
|
|
|
|
\paragraph{(ii)}%
|
|
|
|
Since $A$ and $B$ have supremums, $C$ is nonempty.
|
|
By \nameref{par:theorem-i.33a-i}, $C$ is bounded above.
|
|
Therefore the completeness axiom tells us $C$ has a supremum.
|
|
Let $n > 0$ be an integer.
|
|
We now prove that
|
|
\begin{equation}
|
|
\label{par:theorem-i.33a-ii-eq1}
|
|
\sup{C} \leq \sup{A} + \sup{B} \leq \sup{C} + 1 / n.
|
|
\end{equation}
|
|
|
|
\subparagraph{Left-Hand Side}%
|
|
|
|
First consider the left-hand side of \eqref{par:theorem-i.33a-ii-eq1}.
|
|
By \nameref{par:theorem-i.33a-i}, $\sup{A} + \sup{B}$ is an upper bound of
|
|
$C$.
|
|
Since $\sup{C}$ is the \textit{least} upper bound of $C$, it follows
|
|
$\sup{C} \leq \sup{A} + \sup{B}$.
|
|
|
|
\subparagraph{Right-Hand Side}%
|
|
|
|
Next consider the right-hand side of \eqref{par:theorem-i.33a-ii-eq1}.
|
|
By \nameref{sub:theorem-i.32a}, there exists some $a' \in A$ such that
|
|
$\sup{A} < a' + 1 / (2n)$.
|
|
Likewise, there exists some $b' \in B$ such that
|
|
$\sup{B} < b' + 1 / (2n)$.
|
|
Adding these two inequalities together shows
|
|
\begin{align*}
|
|
\sup{A} + \sup{B}
|
|
& < a' + b' + 1 / n \\
|
|
& \leq \sup{C} + 1 / n.
|
|
\end{align*}
|
|
|
|
\subparagraph{Conclusion}%
|
|
|
|
Applying \nameref{sec:theorem-i.31} to \eqref{par:theorem-i.33a-ii-eq1}
|
|
proves $\sup{C} = \sup{A} + \sup{B}$ as expected.
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\verified{Theorem I.33b}}%
|
|
\label{sub:theorem-i.33b}
|
|
|
|
\begin{theorem}[I.33b]
|
|
|
|
If each of $A$ and $B$ has an infimum, then $C$ has an infimum, and
|
|
$$\inf{C} = \inf{A} + \inf{B}.$$
|
|
|
|
\end{theorem}
|
|
|
|
\begin{proof}
|
|
|
|
\lean{Bookshelf/Apostol/Chapter\_I\_03}
|
|
{Apostol.Chapter\_I\_03.inf\_minkowski\_sum\_eq\_inf\_add\_inf}
|
|
|
|
We prove (i) $\inf{A} + \inf{B}$ is a lower bound of $C$ and (ii)
|
|
$\inf{A} + \inf{B}$ is the \textit{greatest} lower bound of $C$.
|
|
|
|
\paragraph{(i)}%
|
|
\label{par:theorem-i.33b-i}
|
|
|
|
Let $x \in C$.
|
|
By definition of $C$, there exist elements $a' \in A$ and $b' \in B$ such
|
|
that $x = a' + b'$.
|
|
By definition of an \nameref{ref:infimum}, $a' \geq \inf{A}$.
|
|
Likewise, $b' \geq \inf{B}$.
|
|
Therefore $a' + b' \geq \inf{A} + \inf{B}$.
|
|
Since $x = a' + b'$ was arbitrarily chosen, it follows $\inf{A} + \inf{B}$
|
|
is a lower bound of $C$.
|
|
|
|
\paragraph{(ii)}%
|
|
|
|
Since $A$ and $B$ have infimums, $C$ is nonempty.
|
|
By \nameref{par:theorem-i.33b-i}, $C$ is bounded below.
|
|
Therefore \nameref{sec:theorem-i.27} tells us $C$ has an infimum.
|
|
Let $n > 0$ be an integer.
|
|
We now prove that
|
|
\begin{equation}
|
|
\label{par:theorem-i.33b-ii-eq1}
|
|
\inf{C} - 1 / n \leq \inf{A} + \inf{B} \leq \inf{C}.
|
|
\end{equation}
|
|
|
|
\subparagraph{Right-Hand Side}%
|
|
|
|
First consider the right-hand side of \eqref{par:theorem-i.33b-ii-eq1}.
|
|
By \nameref{par:theorem-i.33b-i}, $\inf{A} + \inf{B}$ is a lower bound of
|
|
$C$.
|
|
Since $\inf{C}$ is the \textit{greatest} upper bound of $C$, it follows
|
|
$\inf{C} \geq \inf{A} + \inf{B}$.
|
|
|
|
\subparagraph{Left-Hand Side}%
|
|
|
|
Next consider the left-hand side of \eqref{par:theorem-i.33b-ii-eq1}.
|
|
By \nameref{sub:theorem-i.32b}, there exists some $a' \in A$ such that
|
|
$\inf{A} > a' - 1 / (2n)$.
|
|
Likewise, there exists some $b' \in B$ such that
|
|
$\inf{B} > b' - 1 / (2n)$.
|
|
Adding these two inequalities together shows
|
|
\begin{align*}
|
|
\inf{A} + \inf{B}
|
|
& > a' + b' - 1 / n \\
|
|
& \geq \inf{C} - 1 / n.
|
|
\end{align*}
|
|
|
|
\subparagraph{Conclusion}%
|
|
|
|
Applying \nameref{sec:lemma-2} to \eqref{par:theorem-i.33b-ii-eq1}
|
|
proves $\inf{C} = \inf{A} + \inf{B}$ as expected.
|
|
|
|
\end{proof}
|
|
|
|
\section{\verified{Theorem I.34}}%
|
|
\label{sec:theorem-i.34}
|
|
|
|
\begin{theorem}[I.34]
|
|
|
|
Given two nonempty subsets $S$ and $T$ of $\mathbb{R}$ such that $$s \leq t$$
|
|
for every $s$ in $S$ and every $t$ in $T$. Then $S$ has a supremum, and $T$
|
|
has an infimum, and they satisfy the inequality $$\sup{S} \leq \inf{T}.$$
|
|
|
|
\end{theorem}
|
|
|
|
\begin{proof}
|
|
|
|
\lean{Bookshelf/Apostol/Chapter\_I\_03}
|
|
{Apostol.Chapter\_I\_03.forall\_mem\_le\_forall\_mem\_imp\_sup\_le\_inf}
|
|
|
|
By hypothesis, $S$ and $T$ are nonempty sets.
|
|
Let $s \in S$ and $t \in T$.
|
|
Then $t$ is an upper bound of $S$ and $s$ is a lower bound of $T$.
|
|
By the completeness axiom, $S$ has a supremum.
|
|
By \nameref{sec:theorem-i.27}, $T$ has an infimum.
|
|
All that remains is showing $\sup{S} \leq \inf{T}$.
|
|
|
|
For the sake of contradiction, suppose $\sup{S} > \inf{T}$.
|
|
Then there exists some $c > 0$ such that $\sup{S} = \inf{T} + c$.
|
|
Therefore $\inf{T} < \sup{S} - c / 2$.
|
|
By \nameref{sub:theorem-i.32a}, there exists some $x \in S$ such that
|
|
$\sup{S} - c / 2 < x$.
|
|
Thus $$\inf{T} < \sup{S} - c / 2 < x.$$
|
|
But by hypothesis, $x \in S$ is a lower bound of $T$ meaning $x \leq \inf{T}$.
|
|
Therefore $x < x$, a contradiction.
|
|
Out original assumption is incorrect; that is, $\sup{S} \leq \inf{T}$.
|
|
|
|
\end{proof}
|
|
|
|
\chapter{The Concepts of Integral Calculus}%
|
|
\label{chap:concepts-integral-calculus}
|
|
|
|
\section{The Concept of Area as a Set Function}%
|
|
\label{sec:concept-area-set-function}
|
|
|
|
We assume there exists a class $\mathscr{M}$ of measurable sets in the plane and
|
|
a set function $a$, whose domain is $\mathscr{M}$, with the following
|
|
properties:
|
|
|
|
\subsection{\defined{Nonnegative Property}}%
|
|
\label{sub:nonnegative-property}
|
|
|
|
For each set $S$ in $\mathscr{M}$, we have $a(S) \geq 0$.
|
|
|
|
\begin{axiom}
|
|
|
|
\leanp*{Common/Geometry/Area}{Nonnegative-Property}
|
|
{Nonnegative Property}
|
|
|
|
\end{axiom}
|
|
|
|
\subsection{\defined{Additive Property}}%
|
|
\label{sub:area-additive-property}
|
|
|
|
If $S$ and $T$ are in $\mathscr{M}$, then $S \cup T$ and $S \cap T$ are in
|
|
$\mathscr{M}$, and we have $a(S \cup T) = a(S) + a(T) - a(S \cap T)$.
|
|
|
|
\begin{axiom}
|
|
|
|
\leanp*{Common/Geometry/Area}{Additive-Property}
|
|
{Additive Property}
|
|
|
|
\end{axiom}
|
|
|
|
\subsection{\defined{Difference Property}}%
|
|
\label{sub:area-difference-property}
|
|
|
|
If $S$ and $T$ are in $\mathscr{M}$ with $S \subseteq T$, then $T - S$ is in
|
|
$\mathscr{M}$, and we have $a(T - S) = a(T) - a(S)$.
|
|
|
|
\begin{axiom}
|
|
|
|
\leanp*{Common/Geometry/Area}{Difference-Property}
|
|
{Difference Property}
|
|
|
|
\end{axiom}
|
|
|
|
\subsection{\defined{Invariance Under Congruence}}%
|
|
\label{sub:area-invariance-under-congruence}
|
|
|
|
If a set $S$ is in $\mathscr{M}$ and if $T$ is congruent to $S$, then $T$ is
|
|
also in $\mathscr{M}$ and we have $a(S) = a(T)$.
|
|
|
|
\begin{axiom}
|
|
|
|
\leanp*{Common/Geometry/Area}{Invariance-Under-Congruence}
|
|
{Invariance Under Congruence}
|
|
|
|
\end{axiom}
|
|
|
|
\subsection{\defined{Choice of Scale}}%
|
|
\label{sub:area-choice-scale}
|
|
|
|
Every rectangle $R$ is in $\mathscr{M}$.
|
|
If the edges of $R$ have lengths $h$ and $k$, then $a(R) = hk$.
|
|
|
|
\begin{axiom}
|
|
|
|
\leanp*{Common/Geometry/Area}{Choice-of-Scale}
|
|
{Choice of Scale}
|
|
|
|
\end{axiom}
|
|
|
|
\subsection{\pending{Exhaustion Property}}%
|
|
\label{sub:area-exhaustion-property}
|
|
|
|
Let $Q$ be a set that can be enclosed between two step regions $S$ and $T$, so
|
|
that
|
|
\begin{equation}
|
|
\label{sub:exhaustion-property-eq1}
|
|
S \subseteq Q \subseteq T.
|
|
\end{equation}
|
|
If there is one and only one number $c$ which satisfies the inequalities
|
|
$$a(S) \leq c \leq a(T)$$ for all step regions $S$ and $T$ satisfying
|
|
\eqref{sub:exhaustion-property-eq1}, then $Q$ is measurable and $a(Q) = c$.
|
|
|
|
\begin{axiom}
|
|
|
|
\leanp*{Common/Geometry/Area}{Exhaustion-Property}
|
|
{Exhaustion Property}
|
|
|
|
\end{axiom}
|
|
|
|
\section{Exercises 1.7}%
|
|
\label{sec:exercises-1.7}
|
|
|
|
\subsection{\pending{Exercise 1.7.1}}%
|
|
\label{sub:exercise-1.7.1}
|
|
|
|
Prove that each of the following sets is measurable and has zero area:
|
|
|
|
\subsubsection{\pending{Exercise 1.7.1a}}%
|
|
\label{ssub:exercise-1.7.1a}
|
|
|
|
A set consisting of a single point.
|
|
|
|
\begin{proof}
|
|
|
|
Let $S$ be a set consisting of a single point.
|
|
By definition of a point, $S$ is a rectangle in which all vertices coincide.
|
|
By \nameref{sub:area-choice-scale}, $S$ is measurable with area its width
|
|
times its height.
|
|
The width and height of $S$ is trivially zero.
|
|
Therefore $a(S) = (0)(0) = 0$.
|
|
|
|
\end{proof}
|
|
|
|
\subsubsection{\pending{Exercise 1.7.1b}}%
|
|
\label{ssub:exercise-1.7.1b}
|
|
|
|
A set consisting of a finite number of points in a plane.
|
|
|
|
\begin{proof}
|
|
|
|
Define predicate $P(n)$ as "A set consisting of $n$ points in a plane is
|
|
measurable with area $0$".
|
|
We use induction to prove $P(n)$ holds for all $n > 0$.
|
|
|
|
\paragraph{Base Case}%
|
|
|
|
Consider a set $S$ consisting of a single point in a plane.
|
|
By \nameref{ssub:exercise-1.7.1a}, $S$ is measurable with area $0$.
|
|
Thus $P(1)$ holds.
|
|
|
|
\paragraph{Induction Step}%
|
|
|
|
Assume induction hypothesis $P(k)$ holds for some $k > 0$.
|
|
Let $S_{k+1}$ be a set consisting of $k + 1$ points in a plane.
|
|
Pick an arbitrary point of $S_{k+1}$.
|
|
Denote the set containing just this point as $T$.
|
|
Denote the remaining set of points as $S_k$.
|
|
By construction, $S_{k+1} = S_k \cup T$.
|
|
By the induction hypothesis, $S_k$ is measurable with area $0$.
|
|
By \nameref{ssub:exercise-1.7.1a}, $T$ is measurable with area $0$.
|
|
By the \nameref{sub:area-additive-property}, $S_k \cup T$ is
|
|
measurable, $S_k \cap T$ is measurable, and
|
|
\begin{align}
|
|
a(S_{k+1})
|
|
& = a(S_k \cup T) \nonumber \\
|
|
& = a(S_k) + a(T) - a(S_k \cap T) \nonumber \\
|
|
& = 0 + 0 - a(S_k \cap T). \label{ssub:exercise-1.7.1b-eq1}
|
|
\end{align}
|
|
There are two cases to consider:
|
|
|
|
\subparagraph{Case 1}%
|
|
|
|
$S_k \cap T = \emptyset$.
|
|
Then it trivially follows that $a(S_k \cap T) = 0$.
|
|
|
|
\subparagraph{Case 2}%
|
|
|
|
$S_k \cap T \neq \emptyset$.
|
|
Since $T$ consists of a single point, $S_k \cap T = T$.
|
|
By \nameref{ssub:exercise-1.7.1a}, $a(S_k \cap T) = a(T) = 0$.
|
|
|
|
\vspace{8pt}
|
|
\noindent
|
|
In both cases, \eqref{ssub:exercise-1.7.1b-eq1} evaluates to $0$, implying
|
|
$P(k + 1)$ as expected.
|
|
|
|
\paragraph{Conclusion}%
|
|
|
|
By mathematical induction, it follows for all $n > 0$, $P(n)$ is true.
|
|
|
|
\end{proof}
|
|
|
|
\subsubsection{\pending{Exercise 1.7.1c}}%
|
|
\label{ssub:exercise-1.7.1c}
|
|
|
|
The union of a finite collection of line segments in a plane.
|
|
|
|
\begin{proof}
|
|
|
|
Define predicate $P(n)$ as "A set consisting of $n$ line segments in a plane
|
|
is measurable with area $0$".
|
|
We use induction to prove $P(n)$ holds for all $n > 0$.
|
|
|
|
\paragraph{Base Case}%
|
|
|
|
Consider a set $S$ consisting of a single line segment in a plane.
|
|
By definition of a line segment, $S$ is a rectangle in which one side has
|
|
dimension $0$.
|
|
By \nameref{sub:area-choice-scale}, $S$ is measurable with area its width
|
|
$w$ times its height $h$.
|
|
Therefore $a(S) = wh = 0$.
|
|
Thus $P(1)$ holds.
|
|
|
|
\paragraph{Induction Step}%
|
|
|
|
Assume induction hypothesis $P(k)$ holds for some $k > 0$.
|
|
Let $S_{k+1}$ be a set consisting of $k + 1$ line segments in a plane.
|
|
Pick an arbitrary line segment of $S_{k+1}$.
|
|
Denote the set containing just this line segment as $T$.
|
|
Denote the remaining set of line segments as $S_k$.
|
|
By construction, $S_{k+1} = S_k \cup T$.
|
|
By the induction hypothesis, $S_k$ is measurable with area $0$.
|
|
By the base case, $T$ is measurable with area $0$.
|
|
By the \nameref{sub:area-additive-property}, $S_k \cup T$ is measurable,
|
|
$S_k \cap T$ is measurable, and
|
|
\begin{align}
|
|
a(S_{k+1})
|
|
& = a(S_k \cup T) \nonumber \\
|
|
& = a(S_k) + a(T) - a(S_k \cap T) \nonumber \\
|
|
& = 0 + 0 - a(S_k \cap T). \label{ssub:exercise-1.7.1c-eq1}
|
|
\end{align}
|
|
There are two cases to consider:
|
|
|
|
\subparagraph{Case 1}%
|
|
|
|
$S_k \cap T = \emptyset$.
|
|
Then it trivially follows that $a(S_k \cap T) = 0$.
|
|
|
|
\subparagraph{Case 2}%
|
|
|
|
$S_k \cap T \neq \emptyset$.
|
|
Since $T$ consists of a single point, $S_k \cap T = T$.
|
|
By the base case, $a(S_k \cap T) = a(T) = 0$.
|
|
|
|
\vspace{8pt}
|
|
\noindent
|
|
In both cases, \eqref{ssub:exercise-1.7.1c-eq1} evaluates to $0$, implying
|
|
$P(k + 1)$ as expected.
|
|
|
|
\paragraph{Conclusion}%
|
|
|
|
By mathematical induction, it follows for all $n > 0$, $P(n)$ is true.
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\pending{Exercise 1.7.2}}%
|
|
\label{sub:exercise-1.7.2}
|
|
|
|
Every right triangular region is measurable because it can be obtained as the
|
|
intersection of two rectangles.
|
|
Prove that every triangular region is measurable and that its area is one half
|
|
the product of its base and altitude.
|
|
|
|
\begin{proof}
|
|
|
|
Let $T'$ be a triangular region with base of length $a$, height of length $b$,
|
|
and hypotenuse of length $c$.
|
|
Consider the translation and rotation of $T'$, say $T$, such that its
|
|
hypotenuse is entirely within quadrant I and the vertex opposite the
|
|
hypotenuse is situated at point $(0, 0)$.
|
|
|
|
Let $R$ be a rectangle of width $a$, height $b$, and bottom-left corner at
|
|
$(0, 0)$.
|
|
By construction, $R$ covers all of $T$.
|
|
Let $S$ be a rectangle of width $c$ and height $a\sin{\theta}$, where $\theta$
|
|
is the acute angle measured from the bottom-right corner of $T$ relative
|
|
to the $x$-axis.
|
|
As an example, consider the image below of triangle $T$ with width $4$ and
|
|
height $3$:
|
|
|
|
\begin{figure}[ht]
|
|
\includegraphics{right-triangle}
|
|
\centering
|
|
\end{figure}
|
|
|
|
By \nameref{sub:area-choice-scale}, both $R$ and $S$ are measurable.
|
|
By this same axiom, $a(R) = ab$ and $a(S) = ca\sin{\theta}$.
|
|
By the \nameref{sub:area-additive-property}, $R \cup S$ and $R \cap S$ are
|
|
both measurable.
|
|
$a(R \cap S) = a(T)$ and $a(R \cup S)$ can be determined by noting that
|
|
$R$'s construction implies identity $a(R) = 2a(T)$.
|
|
Therefore
|
|
\begin{align*}
|
|
a(T)
|
|
& = a(R \cap S) \\
|
|
& = a(R) + a(S) - a(R \cup S) \\
|
|
& = ab + ca\sin{\theta} - a(R \cup S) \\
|
|
& = ab + ca\sin{\theta} - (ca\sin{\theta} + \frac{1}{2}a(R)) \\
|
|
& = ab + ca\sin{\theta} - ca\sin{\theta} - a(T).
|
|
\end{align*}
|
|
Solving for $a(T)$ gives the desired identity: $$a(T) = \frac{1}{2}ab.$$
|
|
By \nameref{sub:area-invariance-under-congruence}, $a(T') = a(T)$, concluding
|
|
our proof.
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\pending{Exercise 1.7.3}}%
|
|
\label{sub:exercise-1.7.3}
|
|
|
|
Prove that every trapezoid and every parallelogram is measurable and derive the
|
|
usual formulas for their areas.
|
|
|
|
\begin{proof}
|
|
|
|
We begin by proving the formula for a trapezoid.
|
|
Let $S$ be a trapezoid with height $h$ and bases $b_1$ and $b_2$, $b_1 < b_2$.
|
|
There are three cases to consider:
|
|
|
|
\begin{figure}[ht]
|
|
\includegraphics[width=\textwidth]{trapezoid-cases}
|
|
\centering
|
|
\end{figure}
|
|
|
|
\paragraph{Case 1}%
|
|
|
|
Suppose $S$ is a right trapezoid.
|
|
Then $S$ is the union of non-overlapping rectangle $R$ of width $b_1$ and
|
|
height $h$ with right triangle $T$ of base $b_2 - b_1$ and height $h$.
|
|
By \nameref{sub:area-choice-scale}, $R$ is measurable.
|
|
By \nameref{sub:exercise-1.7.2}, $T$ is measurable.
|
|
By the \nameref{sub:area-additive-property}, $R \cup T$ and $R \cap T$ are
|
|
both measurable and
|
|
\begin{align*}
|
|
a(S)
|
|
& = a(R \cup T) \\
|
|
& = a(R) + a(T) - a(R \cap T) \\
|
|
& = a(R) + a(T) & \text{by construction} \\
|
|
& = b_1h + a(T) & \text{Choice of Scale} \\
|
|
& = b_1h + \frac{1}{2}(b_2 - b_1)h & \textref{sub:exercise-1.7.2} \\
|
|
& = \frac{b_1 + b_2}{2}h.
|
|
\end{align*}
|
|
|
|
\paragraph{Case 2}%
|
|
|
|
Suppose $S$ is an acute trapezoid.
|
|
Then $S$ is the union of non-overlapping triangle $T$ and right trapezoid $R$.
|
|
Let $c$ denote the length of base $T$.
|
|
Then $R$ has longer base edge of length $b_2 - c$.
|
|
By \nameref{sub:exercise-1.7.2}, $T$ is measurable.
|
|
By Case 1, $R$ is measurable.
|
|
By the \nameref{sub:area-additive-property}, $R \cup T$ and $R \cap T$ are
|
|
both measurable and
|
|
\begin{align*}
|
|
a(S)
|
|
& = a(T) + a(R) - a(R \cap T) \\
|
|
& = a(T) + a(R) & \text{by construction} \\
|
|
& = \frac{1}{2}ch + a(R) & \textref{sub:exercise-1.7.2} \\
|
|
& = \frac{1}{2}ch + \frac{b_1 + b_2 - c}{2}h & \text{Case 1} \\
|
|
& = \frac{b_1 + b_2}{2}h.
|
|
\end{align*}
|
|
|
|
\paragraph{Case 3}%
|
|
|
|
Suppose $S$ is an obtuse trapezoid.
|
|
Then $S$ is the union of non-overlapping triangle $T$ and right trapezoid $R$.
|
|
Let $c$ denote the length of base $T$.
|
|
Reflect $T$ vertically to form another right triangle, say $T'$.
|
|
Then $T' \cup R$ is an acute trapezoid.
|
|
By \nameref{sub:area-invariance-under-congruence},
|
|
\begin{equation}
|
|
\label{sub:exercise-1.7.3-eq1}
|
|
\tag{3.1}
|
|
a(T' \cup R) = a(T \cup R).
|
|
\end{equation}
|
|
By construction, $T' \cup R$ has height $h$ and bases $b_1 - c$ and $b_2 + c$
|
|
meaning
|
|
\begin{align*}
|
|
a(T \cup R)
|
|
& = a(T' \cup R) & \eqref{sub:exercise-1.7.3-eq1} \\
|
|
& = \frac{b_1 - c + b_2 + c}{2}h & \text{Case 2} \\
|
|
& = \frac{b_1 + b_2}{2}h.
|
|
\end{align*}
|
|
|
|
\paragraph{Conclusion}%
|
|
|
|
These cases are exhaustive and in agreement with one another.
|
|
Thus $S$ is measurable and $$a(S) = \frac{b_1 + b_2}{2}h.$$
|
|
|
|
\suitdivider
|
|
|
|
Let $P$ be a parallelogram with base $b$ and height $h$.
|
|
Then $P$ is the union of non-overlapping triangle $T$ and right trapezoid $R$.
|
|
Let $c$ denote the length of base $T$.
|
|
Reflect $T$ vertically to form another right triangle, say $T'$.
|
|
Then $T' \cup R$ is an acute trapezoid.
|
|
By \nameref{sub:area-invariance-under-congruence},
|
|
\begin{equation}
|
|
\label{sub:exercise-1.7.3-eq2}
|
|
a(T' \cup R) = a(T \cup R).
|
|
\end{equation}
|
|
By construction, $T' \cup R$ has height $h$ and bases $b - c$ and $b + c$
|
|
meaning
|
|
\begin{align*}
|
|
a(T \cup R)
|
|
& = a(T' \cup R) & \eqref{sub:exercise-1.7.3-eq2} \\
|
|
& = \frac{b - c + b + c}{2}h & \text{Area of Trapezoid} \\
|
|
& = bh.
|
|
\end{align*}
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\pending{Exercise 1.7.4}}%
|
|
\label{sub:exercise-1.7.4}
|
|
|
|
Let $P$ be a polygon whose vertices are lattice points.
|
|
The area of $P$ is $I + \frac{1}{2}B - 1$, where $I$ denotes the number of
|
|
lattice points inside the polygon and $B$ denotes the number on the boundary.
|
|
|
|
\subsubsection{\pending{Exercise 1.7.4a}}%
|
|
\label{ssub:exercise-1.7.4a}
|
|
|
|
Prove that the formula is valid for rectangles with sides parallel to the
|
|
coordinate axes.
|
|
|
|
\begin{proof}
|
|
|
|
Let $P$ be a rectangle with sides parallel to the coordinate axes, with width
|
|
$w$, height $h$, and lattice points for vertices.
|
|
We assume $P$ has three non-collinear points, ruling out any instances of
|
|
points or line segments.
|
|
|
|
By \nameref{sub:area-choice-scale}, $P$ is measurable with area $a(P) = wh$.
|
|
By construction, $P$ has $I = (w - 1)(h - 1)$ interior lattice points and
|
|
$B = 2(w + h)$ lattice points on its boundary.
|
|
The following shows the lattice point area formula is in agreement with
|
|
the expected result:
|
|
\begin{align*}
|
|
I + \frac{1}{2}B - 1
|
|
& = (w - 1)(h - 1) + \frac{1}{2}\left[ 2(w + h) \right] - 1 \\
|
|
& = (wh - w - h + 1) + \frac{1}{2}\left[ 2(w + h) \right] - 1 \\
|
|
& = (wh - w - h + 1) + (w + h) - 1 \\
|
|
& = wh.
|
|
\end{align*}
|
|
|
|
\end{proof}
|
|
|
|
\subsubsection{\pending{Exercise 1.7.4b}}%
|
|
\label{ssub:exercise-1.7.4b}
|
|
|
|
Prove that the formula is valid for right triangles and parallelograms.
|
|
|
|
\begin{proof}
|
|
|
|
Let $P$ be a right triangle with width $w > 0$, height $h > 0$, and lattice
|
|
points for vertices.
|
|
Let $T$ be the triangle $P$ translated, rotated, and reflected such that the
|
|
its vertices are $(0, 0)$, $(0, w)$, and $(w, h)$.
|
|
Let $I_T$ and $B_T$ be the number of interior and boundary points of $T$
|
|
respectively.
|
|
Let $H_L$ denote the number of lattice points on $T$'s hypotenuse.
|
|
|
|
Let $R$ be the overlapping rectangle of width $w$ and height $h$, situated
|
|
with bottom-left corner at $(0, 0)$.
|
|
Let $I_R$ and $B_R$ be the number of interior and boundary points
|
|
of $R$ respectively.
|
|
|
|
By construction, $T$ shares two sides with $R$.
|
|
Therefore
|
|
\begin{equation}
|
|
\label{ssub:exercise-1.7.4b-eq1}
|
|
B_T = \frac{1}{2}B_R - 1 + H_L.
|
|
\end{equation}
|
|
Likewise,
|
|
\begin{equation}
|
|
\label{ssub:exercise-1.7.4b-eq2}
|
|
I_T = \frac{1}{2}(I_R - (H_L - 2)).
|
|
\end{equation}
|
|
The following shows the lattice point area formula is in agreement with
|
|
the expected result:
|
|
\begin{align*}
|
|
I_T + \frac{1}{2}B_T - 1
|
|
& = \frac{1}{2}(I_R - (H_L - 2)) + \frac{1}{2}B_T - 1
|
|
& \eqref{ssub:exercise-1.7.4b-eq2} \\
|
|
& = \frac{1}{2}\left[ I_R - H_L + B_T \right] \\
|
|
& = \frac{1}{2}\left[ I_R - H_L + \frac{1}{2}B_R - 1 + H_L \right]
|
|
& \eqref{ssub:exercise-1.7.4b-eq1} \\
|
|
& = \frac{1}{2}\left[ I_R + \frac{1}{2}B_R - 1 \right] \\
|
|
& = \frac{1}{2}\left[ wh \right] & \textref{ssub:exercise-1.7.4a}.
|
|
\end{align*}
|
|
|
|
We do not prove this formula is valid for parallelograms here.
|
|
Instead, refer to \nameref{ssub:exercise-1.7.4c} below.
|
|
|
|
\end{proof}
|
|
|
|
\subsubsection{\pending{Exercise 1.7.4c}}%
|
|
\label{ssub:exercise-1.7.4c}
|
|
|
|
Use induction on the number of edges to construct a proof for general polygons.
|
|
|
|
\begin{proof}
|
|
|
|
Define predicate $P(n)$ as "An $n$-polygon with vertices on lattice points has
|
|
area $I + \frac{1}{2}B - 1$."
|
|
We use induction to prove $P(n)$ holds for all $n \geq 3$.
|
|
|
|
\paragraph{Base Case}%
|
|
|
|
A $3$-polygon is a triangle.
|
|
By \nameref{ssub:exercise-1.7.4b}, the lattice point area formula holds.
|
|
Thus $P(3)$ holds.
|
|
|
|
\paragraph{Induction Step}%
|
|
|
|
Assume induction hypothesis $P(k)$ holds for some $k \geq 3$.
|
|
Let $P$ be a $(k + 1)$-polygon with vertices on lattice points.
|
|
Such a polygon is equivalent to the union of a $k$-polygon $S$ with a
|
|
triangle $T$.
|
|
That is, $P = S \cup T$.
|
|
|
|
Let $I_P$ be the number of interior lattice points of $P$.
|
|
Let $B_P$ be the number of boundary lattice points of $P$.
|
|
Similarly, let $I_S$, $I_T$, $B_S$, and $B_T$ be the number of interior
|
|
and boundary lattice points of $S$ and $T$.
|
|
Let $c$ denote the number of boundary points shared between $S$ and $T$.
|
|
|
|
By our induction hypothesis, $a(S) = I_S + \frac{1}{2}B_S - 1$.
|
|
By our base case, $a(T) = I_T + \frac{1}{2}B_T - 1$.
|
|
By construction, it follows:
|
|
\begin{align*}
|
|
I_P & = I_S + I_T + c - 2 \\
|
|
B_P & = B_S + B_T - (c - 2) - c \\
|
|
& = B_S + B_T - 2c + 2.
|
|
\end{align*}
|
|
Applying the lattice point area formula to $P$ yields the following:
|
|
\begin{align*}
|
|
& I_P + \frac{1}{2}B_P - 1 \\
|
|
& = (I_S + I_T + c - 2) + \frac{1}{2}(B_S + B_T - 2c + 2) - 1 \\
|
|
& = I_S + I_T + c - 2 + \frac{1}{2}B_S + \frac{1}{2}B_T - c + 1 - 1 \\
|
|
& = (I_S + \frac{1}{2}B_S - 1) + (I_T + \frac{1}{2}B_T - 1) \\
|
|
& = a(S) + (I_T + \frac{1}{2}B_T - 1) & \text{induction hypothesis} \\
|
|
& = a(S) + a(T). & \text{base case}
|
|
\end{align*}
|
|
By the \nameref{sub:area-additive-property}, $S \cup T$ is measurable,
|
|
$S \cap T$ is measurable, and
|
|
\begin{align*}
|
|
a(P)
|
|
& = a(S \cup T) \\
|
|
& = a(S) + a(T) - a(S \cap T) \\
|
|
& = a(S) + a(T). & \text{by construction}
|
|
\end{align*}
|
|
This shows the lattice point area formula is in agreement with our axiomatic
|
|
definition of area.
|
|
Thus $P(k + 1)$ holds.
|
|
|
|
\paragraph{Conclusion}%
|
|
|
|
By mathematical induction, it follows for all $n \geq 3$, $P(n)$ is true.
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\pending{Exercise 1.7.5}}%
|
|
\label{sub:exercise-1.7.5}
|
|
|
|
Prove that a triangle whose vertices are lattice points cannot be equilateral.
|
|
|
|
[\textit{Hint:} Assume there is such a triangle and compute its area in two
|
|
ways, using Exercises 2 and 4.]
|
|
|
|
\begin{proof}
|
|
|
|
Proceed by contradiction.
|
|
Let $T$ be an equilateral triangle whose vertices are lattice points.
|
|
Assume each side of $T$ has length $a$.
|
|
Then $T$ has height $h = (a\sqrt{3}) / 2$.
|
|
By \nameref{sub:exercise-1.7.2},
|
|
\begin{equation}
|
|
\label{sub:exercise-1.7.5-eq1}
|
|
\tag{5.1}
|
|
a(T) = \frac{1}{2}ah = \frac{a^2\sqrt{3}}{4}.
|
|
\end{equation}
|
|
Let $I$ and $B$ denote the number of interior and boundary lattice points of
|
|
$T$ respectively.
|
|
By \nameref{sub:exercise-1.7.4},
|
|
\begin{equation}
|
|
\label{sub:exercise-1.7.5-eq2}
|
|
\tag{5.2}
|
|
a(T) = I + \frac{1}{2}B - 1.
|
|
\end{equation}
|
|
But \eqref{sub:exercise-1.7.5-eq1} is irrational whereas
|
|
\eqref{sub:exercise-1.7.5-eq2} is not.
|
|
This is a contradiction.
|
|
Thus, there is \textit{no} equilateral triangle whose vertices are lattice
|
|
points.
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\pending{Exercise 1.7.6}}%
|
|
\label{sub:exercise-1.7.6}
|
|
|
|
Let $A = \{1, 2, 3, 4, 5\}$, and let $\mathscr{M}$ denote the class of all
|
|
subsets of $A$.
|
|
(There are 32 altogether, counting $A$ itself and the empty set $\emptyset$.)
|
|
For each set $S$ in $\mathscr{M}$, let $n(S)$ denote the number of distinct
|
|
elements in $S$.
|
|
If $S = \{1, 2, 3, 4\}$ and $T = \{3, 4, 5\}$, compute $n(S \cup T)$,
|
|
$n(S \cap T)$, $n(S - T)$, and $n(T - S)$.
|
|
Prove that the set function $n$ satisfies the first three axioms for area.
|
|
|
|
\begin{proof}
|
|
|
|
Let $S = \{1, 2, 3, 4\}$ and $T = \{3, 4, 5\}$.
|
|
Then
|
|
\begin{align*}
|
|
n(S \cup T)
|
|
& = n(\{1, 2, 3, 4\} \cup \{3, 4, 5\}) \\
|
|
& = n(\{1, 2, 3, 4, 5\}) \\
|
|
& = 5. \\
|
|
n(S \cap T)
|
|
& = n(\{1, 2, 3, 4\} \cap \{3, 4, 5\}) \\
|
|
& = n(\{3, 4\}) \\
|
|
& = 2. \\
|
|
n(S - T)
|
|
& = n(\{1, 2, 3, 4\} - \{3, 4, 5\}) \\
|
|
& = n(\{1, 2\}) \\
|
|
& = 2. \\
|
|
n(T - S)
|
|
& = n(\{3, 4, 5\} - \{1, 2, 3, 4\}) \\
|
|
& = n(\{5\}) \\
|
|
& = 1.
|
|
\end{align*}
|
|
We now prove $n$ satisfies the first three axioms for area.
|
|
|
|
\paragraph{Nonnegative Property}%
|
|
|
|
$n$ returns the length of some member of $\mathscr{M}$.
|
|
By hypothesis, the smallest possible input to $n$ is $\emptyset$.
|
|
Since $n(\emptyset) = 0$, it follows $n(S) \geq 0$ for all $S \subset A$.
|
|
|
|
\paragraph{Additive Property}%
|
|
|
|
Let $S$ and $T$ be members of $\mathscr{M}$.
|
|
It trivially follows that both $S \cup T$ and $S \cap T$ are in
|
|
$\mathscr{M}$.
|
|
Consider the value of $n(S \cup T)$.
|
|
There are two cases to consider:
|
|
|
|
\subparagraph{Case 1}%
|
|
|
|
Suppose $S \cap T = \emptyset$.
|
|
That is, there is no common element shared between $S$ and $T$.
|
|
Thus
|
|
\begin{align*}
|
|
n(S \cup T)
|
|
& = n(S) + n(T) \\
|
|
& = n(S) + n(T) - 0 \\
|
|
& = n(S) + n(T) - n(S \cap T).
|
|
\end{align*}
|
|
|
|
\subparagraph{Case 2}%
|
|
|
|
Suppose $S \cap T \neq \emptyset$.
|
|
Then $n(S) + n(T)$ counts each element of $S \cap T$ twice.
|
|
Therefore $n(S \cup T) = n(S) + n(T) - n(S \cap T)$.
|
|
|
|
\subparagraph{Conclusion}%
|
|
|
|
These cases are exhaustive and in agreement with one another.
|
|
Thus $n(S \cup T) = n(S) + n(T) - n(S \cap T)$.
|
|
|
|
\paragraph{Difference Property}%
|
|
|
|
Suppose $S, T \in \mathscr{M}$ such that $S \subseteq T$.
|
|
That is, every member of $S$ is a member of $T$.
|
|
By definition, $T - S$ consists of members in $T$ but not in $S$.
|
|
Thus $n(T - S) = n(T) - n(S)$.
|
|
|
|
\end{proof}
|
|
|
|
\section{Exercises 1.11}%
|
|
\label{sec:exercises-1-11}
|
|
|
|
\subsection{\pending{Exercise 1.11.4}}%
|
|
\label{sub:exercise-1.11.4}
|
|
|
|
Prove that the greatest-integer function has the properties indicated:
|
|
|
|
\subsubsection{\verified{Exercise 1.11.4a}}%
|
|
\label{ssub:exercise-1.11.4a}
|
|
|
|
$\floor{x + n} = \floor{x} + n$ for every integer $n$.
|
|
|
|
\begin{proof}
|
|
|
|
\lean{Bookshelf/Apostol/Chapter\_1\_11}
|
|
{Apostol.Chapter\_1\_11.exercise\_4a}
|
|
|
|
Let $x$ be a real number and $n$ an integer.
|
|
Let $m = \floor{x + n}$.
|
|
By definition of the floor function, $m$ is the unique integer such that
|
|
$m \leq x + n < m + 1$.
|
|
Then $m - n \leq x < (m - n) + 1$.
|
|
That is, $m - n = \floor{x}$.
|
|
Rearranging terms we see that $m = \floor{x} + n$ as expected.
|
|
|
|
\end{proof}
|
|
|
|
\subsubsection{\verified{Exercise 1.11.4b}}%
|
|
\label{ssub:exercise-1.11.4b}
|
|
|
|
$\floor{-x} =
|
|
\begin{cases}
|
|
-\floor{x} & \text{if } x \text{ is an integer}, \\
|
|
-\floor{x} - 1 & \text{otherwise}.
|
|
\end{cases}$
|
|
|
|
\begin{proof}
|
|
|
|
\statementpadding
|
|
|
|
\lean*{Bookshelf/Apostol/Chapter\_1\_11}
|
|
{Apostol.Chapter\_1\_11.exercise\_4b\_1}
|
|
|
|
\lean{Bookshelf/Apostol/Chapter\_1\_11}
|
|
{Apostol.Chapter\_1\_11.exercise\_4b\_2}
|
|
|
|
There are two cases to consider:
|
|
|
|
\paragraph{Case 1}%
|
|
|
|
Suppose $x$ is an integer.
|
|
Then $x = \floor{x}$ and $-x = \floor{-x}$.
|
|
It immediately follows that $$\floor{-x} = -x = -\floor{x}.$$
|
|
|
|
\paragraph{Case 2}%
|
|
|
|
Suppose $x$ is not an integer.
|
|
Let $m = \floor{-x}$.
|
|
By definition of the floor function, $m$ is the unique integer such that
|
|
$m \leq -x < m + 1$.
|
|
Equivalently, $-m - 1 < x \leq -m$.
|
|
Since $x$ is not an integer, it follows $-m - 1 \leq x < -m$.
|
|
Then, by definition of the floor function, $\floor{x} = -m - 1$.
|
|
Solving for $m$ yields $$\floor{-x} = m = -\floor{x} - 1.$$
|
|
|
|
\paragraph{Conclusion}%
|
|
|
|
The above two cases are exhaustive. Thus
|
|
$$\floor{-x} =
|
|
\begin{cases}
|
|
-\floor{x} & \text{if } x \text{ is an integer}, \\
|
|
-\floor{x} - 1 & \text{otherwise}.
|
|
\end{cases}$$
|
|
|
|
\end{proof}
|
|
|
|
\subsubsection{\verified{Exercise 1.11.4c}}%
|
|
\label{ssub:exercise-1.11.4c}
|
|
|
|
$\floor{x + y} = \floor{x} + \floor{y}$ or $\floor{x} + \floor{y} + 1$.
|
|
|
|
\begin{proof}
|
|
|
|
\lean{Bookshelf/Apostol/Chapter\_1\_11}
|
|
{Apostol.Chapter\_1\_11.exercise\_4c}
|
|
|
|
Rewrite $x$ and $y$ as the sum of their floor and fractional components:
|
|
$x = \floor{x} + \{x\}$ and $y = \floor{y} + \{y\}$.
|
|
Now
|
|
\begin{align}
|
|
\floor{x + y}
|
|
& = \floor{\floor{x} + \{x\} + \floor{y} + \{y\}} \nonumber \\
|
|
& = \floor{\floor{x} + \floor{y} + \{x\} + \{y\}} \nonumber \\
|
|
& = \floor{x} + \floor{y} + \floor{\{x\} + \{y\}}
|
|
& \textref{ssub:exercise-1.11.4a} \label{ssub:exercise-1.11.4c-eq1}
|
|
\end{align}
|
|
There are two cases to consider:
|
|
|
|
\paragraph{Case 1}%
|
|
|
|
Suppose $\{x\} + \{y\} < 1$.
|
|
Then $\floor{\{x\} + \{y\}} = 0$.
|
|
Substituting this value into \eqref{ssub:exercise-1.11.4c-eq1} yields
|
|
$$\floor{x + y} = \floor{x} + \floor{y}.$$
|
|
|
|
\paragraph{Case 2}%
|
|
|
|
Suppose $\{x\} + \{y\} \geq 1$.
|
|
Because $\{x\}$ and $\{y\}$ are both less than $1$, $\{x\} + \{y\} < 2$.
|
|
Thus $\floor{\{x\} + \{y\}} = 1$.
|
|
Substituting this value into \eqref{ssub:exercise-1.11.4c-eq1} yields
|
|
$$\floor{x + y} = \floor{x} + \floor{y} + 1.$$
|
|
|
|
\paragraph{Conclusion}%
|
|
|
|
Since the above two cases are exhaustive, it follows
|
|
$\floor{x + y} = \floor{x} + \floor{y}$ or $\floor{x} + \floor{y} + 1$.
|
|
|
|
\end{proof}
|
|
|
|
\subsubsection{\pending{Exercise 1.11.4d}}%
|
|
\label{ssub:exercise-1.11.4d}
|
|
|
|
$\floor{2x} = \floor{x} + \floor{x + \frac{1}{2}}.$
|
|
|
|
\begin{proof}
|
|
|
|
\lean{Bookshelf/Apostol/Chapter\_1\_11}
|
|
{Apostol.Chapter\_1\_11.exercise\_4d}
|
|
|
|
This is immediately proven by applying \nameref{sub:hermites-identity}.
|
|
|
|
\end{proof}
|
|
|
|
\subsubsection{\pending{Exercise 1.11.4e}}%
|
|
\label{ssub:exercise-1.11.4e}
|
|
|
|
$\floor{3x} = \floor{x} + \floor{x + \frac{1}{3}} + \floor{x + \frac{2}{3}}.$
|
|
|
|
\begin{proof}
|
|
|
|
\lean{Bookshelf/Apostol/Chapter\_1\_11}
|
|
{Apostol.Chapter\_1\_11.exercise\_4e}
|
|
|
|
This is immediately proven by applying \nameref{sub:hermites-identity}.
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\pending{Hermite's Identity}}%
|
|
\label{sub:hermites-identity}
|
|
\label{sub:exercise-1.11.5}
|
|
|
|
The formulas in Exercises 4(d) and 4(e) suggest a generalization for
|
|
$\floor{nx}$.
|
|
State and prove such a generalization.
|
|
|
|
\begin{proof}
|
|
|
|
\lean{Bookshelf/Apostol/Chapter\_1\_11}
|
|
{Apostol.Chapter\_1\_11.exercise\_5}
|
|
|
|
We prove that for all natural numbers $n$ and real numbers $x$, the following
|
|
identity holds:
|
|
\begin{equation}
|
|
\label{sub:exercise-1.11.5-eq1}
|
|
\floor{nx} = \sum_{i=0}^{n-1} \floor{x + \frac{i}{n}}
|
|
\end{equation}
|
|
By definition of the floor function, $x = \floor{x} + r$ for some
|
|
$r \in \ico{0}{1}$.
|
|
Define $S$ as the partition of non-overlapping subintervals
|
|
$$\ico{0}{\frac{1}{n}}, \ico{\frac{1}{n}}{\frac{2}{n}}, \ldots,
|
|
\ico{\frac{n-1}{n}}{1}.$$
|
|
By construction, $\cup\; S = \ico{0}{1}$.
|
|
Therefore there exists some $j \in \mathbb{N}$ such that
|
|
\begin{equation}
|
|
\label{sub:exercise-1.11.5-eq2}
|
|
r \in \ico{\frac{j}{n}}{\frac{j+1}{n}}.
|
|
\end{equation}
|
|
With these definitions established, we now show the left- and right-hand sides
|
|
of \eqref{sub:exercise-1.11.5-eq1} evaluate to the same number.
|
|
|
|
\paragraph{Left-Hand Side}%
|
|
|
|
Consider the left-hand side of identity \eqref{sub:exercise-1.11.5-eq1}.
|
|
By \eqref{sub:exercise-1.11.5-eq2}, $nr \in \ico{j}{j + 1}$.
|
|
Therefore $\floor{nr} = j$.
|
|
Thus
|
|
\begin{align}
|
|
\floor{nx}
|
|
& = \floor{n(\floor{x} + r)} \nonumber \\
|
|
& = \floor{n\floor{x} + nr} \nonumber \\
|
|
& = \floor{n\floor{x}} + \floor{nr}. \nonumber
|
|
& \textref{ssub:exercise-1.11.4a} \\
|
|
& = \floor{n\floor{x}} + j \nonumber \\
|
|
& = n\floor{x} + j. \label{sub:exercise-1.11.5-eq3}
|
|
\end{align}
|
|
|
|
\paragraph{Right-Hand Side}%
|
|
|
|
Now consider the right-hand side of identity
|
|
\eqref{sub:exercise-1.11.5-eq1}.
|
|
We note each summand, by construction, is the floor of $x$ added to a
|
|
nonnegative number less than one.
|
|
Therefore each summand contributes either $\floor{x}$ or $\floor{x} + 1$ to
|
|
the total.
|
|
Letting $z$ denote the number of summands that contribute $\floor{x} + 1$,
|
|
we have
|
|
\begin{equation}
|
|
\label{sub:exercise-1.11.5-eq4}
|
|
\sum_{i=0}^{n-1} \floor{x + \frac{i}{n}} = n\floor{x} + z.
|
|
\end{equation}
|
|
The value of $z$ corresponds to the number of indices $i$ that satisfy
|
|
$$\frac{i}{n} \geq 1 - r.$$
|
|
By \eqref{sub:exercise-1.11.5-eq2}, it follows
|
|
\begin{align*}
|
|
1 - r
|
|
& \in \ioc{1 - \frac{j+1}{n}}{1-\frac{j}{n}} \\
|
|
& = \ioc{\frac{n - j - 1}{n}}{\frac{n - j}{n}}.
|
|
\end{align*}
|
|
Thus we can determine the value of $z$ by instead counting the number of
|
|
indices $i$ that satisfy $$\frac{i}{n} \geq \frac{n - j}{n}.$$
|
|
Rearranging terms, we see that $i \geq n - j$ holds for
|
|
$z = (n - 1) - (n - j) + 1 = j$ of the $n$ summands.
|
|
Substituting the value of $z$ into \eqref{sub:exercise-1.11.5-eq4} yields
|
|
\begin{equation}
|
|
\label{sub:exercise-1.11.5-eq5}
|
|
\sum_{i=0}^{n-1} \floor{x + \frac{i}{n}} = n\floor{x} + j.
|
|
\end{equation}
|
|
|
|
\paragraph{Conclusion}%
|
|
|
|
Since \eqref{sub:exercise-1.11.5-eq3} and \eqref{sub:exercise-1.11.5-eq5}
|
|
agree with one another, it follows identity
|
|
\eqref{sub:exercise-1.11.5-eq1} holds.
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\pending{Exercise 1.11.6}}%
|
|
\label{sub:exercise-1.11.6}
|
|
|
|
Recall that a lattice point $(x, y)$ in the plane is one whose coordinates are
|
|
integers.
|
|
Let $f$ be a nonnegative function whose domain is the interval $[a, b]$, where
|
|
$a$ and $b$ are integers, $a < b$.
|
|
Let $S$ denote the set of points $(x, y)$ satisfying $a \leq x \leq b$,
|
|
$0 < y \leq f(x)$.
|
|
Prove that the number of lattice points in $S$ is equal to the sum
|
|
$$\sum_{n=a}^b \floor{f(n)}.$$
|
|
|
|
\begin{proof}
|
|
|
|
Let $i = a, \ldots, b$ and define $S_i = \mathbb{N} \cap \ioc{0}{f(i)}$.
|
|
By construction, the number of lattice points in $S$ is
|
|
\begin{equation}
|
|
\label{sub:exercise-1.11.6-eq1}
|
|
\sum_{n = a}^b \abs{S_n}.
|
|
\end{equation}
|
|
All that remains is to show $\abs{S_i} = \floor{f(i)}$.
|
|
There are two cases to consider:
|
|
|
|
\paragraph{Case 1}%
|
|
|
|
Suppose $f(i)$ is an integer.
|
|
Then the number of integers in $\ioc{0}{f(i)}$ is $f(i) = \floor{f(i)}$.
|
|
|
|
\paragraph{Case 2}%
|
|
|
|
Suppose $f(i)$ is not an integer.
|
|
Then the number of integers in $\ioc{0}{f(i)}$ is the same as that of
|
|
$\ioc{0}{\floor{f(i)}}$.
|
|
Once again, that number is $\floor{f(i)}$.
|
|
|
|
\paragraph{Conclusion}%
|
|
|
|
By cases 1 and 2, $\abs{S_i} = \floor{f(i)}$.
|
|
Substituting this identity into \eqref{sub:exercise-1.11.6-eq1} finishes the
|
|
proof.
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\pending{Exercise 1.11.7}}%
|
|
\label{sub:exercise-1.11.7}
|
|
|
|
If $a$ and $b$ are positive integers with no common factor, we have the formula
|
|
$$\sum_{n=1}^{b-1} \floor{\frac{na}{b}} = \frac{(a - 1)(b - 1)}{2}.$$
|
|
When $b = 1$, the sum on the left is understood to be $0$.
|
|
|
|
\note{When $b = 1$, the proofs of (a) and (b) are trivial. We continue under the
|
|
assumption $b > 1$.}
|
|
|
|
\subsubsection{\pending{Exercise 1.11.7a}}%
|
|
\label{ssub:exercise-1.11.7a}
|
|
|
|
Derive this result by a geometric argument, counting lattice points in a right
|
|
triangle.
|
|
|
|
\begin{proof}
|
|
|
|
Let $f \colon [1, b - 1] \rightarrow \mathbb{R}$ be given by $f(x) = ax / b$.
|
|
Let $S$ denote the set of points $(x, y)$ satisfying $1 \leq x \leq b - 1$,
|
|
$0 < y \leq f(x)$.
|
|
By \nameref{sub:exercise-1.11.6}, the number of lattice points of $S$ is equal
|
|
to the sum
|
|
\begin{equation}
|
|
\label{ssub:exercise-1.11.7a-eq1}
|
|
\sum_{n=1}^{b-1} \floor{f(n)} = \sum_{n=1}^{b-1} \floor{\frac{na}{b}}.
|
|
\end{equation}
|
|
Define $T$ to be the triangle of width $w = b$ and height $h = f(b) = a$
|
|
as $$T = \{ (x, y) : 0 < x < b, 0 < y \leq f(x) \}.$$
|
|
By construction, $T$ does not introduce any additional lattice points.
|
|
Thus $S$ and $T$ have the same number of lattice points.
|
|
Let $H_L$ denote the number of boundary points on $T$'s hypotenuse.
|
|
We prove that (i) $H_L = 2$ and (ii) that $T$ has $\frac{(a - 1)(b - 1)}{2}$
|
|
lattice points.
|
|
|
|
\paragraph{(i)}%
|
|
\label{par:exercise-1.11.7a-i}
|
|
|
|
Consider the line $L$ overlapping the hypotenuse of $T$.
|
|
By construction, $T$'s hypotenuse has endpoints $(0, 0)$ and $(b, a)$.
|
|
By hypothesis, $a$ and $b$ are positive, excluding the possibility of $L$
|
|
being vertical.
|
|
Define the slope of $L$ as $$m = \frac{a}{b}.$$
|
|
$H_L$ coincides with the number of indices $i = 0, \ldots, b$ such that
|
|
$(i, i * m)$ is a lattice point.
|
|
But $a$ and $b$ are coprime by hypothesis and $i \leq b$.
|
|
Thus $i * m$ is an integer if and only if $i = 0$ or $i = b$.
|
|
Thus $H_L = 2$.
|
|
|
|
\paragraph{(ii)}%
|
|
|
|
Next we count the number of lattice points in $T$.
|
|
Let $R$ be the overlapping retangle of width $w$ and height $h$, situated
|
|
with bottom-left corner at $(0, 0)$.
|
|
Let $I_R$ denote the number of interior lattice points of $R$.
|
|
Let $I_T$ and $B_T$ denote the interior and boundary lattice points of $T$
|
|
respectively.
|
|
By \nameref{ssub:exercise-1.7.4b-eq2},
|
|
\begin{align}
|
|
I_T
|
|
& = \frac{1}{2}(I_R - (H_L - 2)) \nonumber \\
|
|
& = \frac{1}{2}(I_R - (2 - 2))
|
|
& \textref{par:exercise-1.11.7a-i} \nonumber \\
|
|
& = \frac{1}{2}I_R. & \label{ssub:exercise-1.11.7a-eq2}
|
|
\end{align}
|
|
Furthermore, since both the adjacent and opposite side of $T$ are not
|
|
included in $T$ and there exist no lattice points on $T$'s hypotenuse
|
|
besides the endpoints, it follows
|
|
\begin{equation}
|
|
\label{ssub:exercise-1.11.7a-eq3}
|
|
B_T = 0.
|
|
\end{equation}
|
|
Thus the number of lattice points of $T$ equals
|
|
\begin{align}
|
|
I_T + B_T
|
|
& = I_T & \eqref{ssub:exercise-1.11.7a-eq3} \nonumber \\
|
|
& = \frac{1}{2}I_R & \eqref{ssub:exercise-1.11.7a-eq2} \nonumber \\
|
|
& = \frac{(b - 1)(a - 1)}{2}.
|
|
& \textref{ssub:exercise-1.7.4a} \label{ssub:exercise-1.11.7a-eq4}
|
|
\end{align}
|
|
|
|
\paragraph{Conclusion}%
|
|
|
|
By \eqref{ssub:exercise-1.11.7a-eq1} the number of lattice points of $S$ is
|
|
equal to the sum $$\sum_{n=1}^{b-1} \floor{\frac{na}{b}}.$$
|
|
But the number of lattice points of $S$ is the same as that of $T$.
|
|
By \eqref{ssub:exercise-1.11.7a-eq4}, the number of lattice points in $T$ is
|
|
equal to $$\frac{(b - 1)(a - 1)}{2}.$$
|
|
Thus $$\sum_{n=1}^{b-1} \floor{\frac{na}{b}} = \frac{(a - 1)(b - 1)}{2}.$$
|
|
|
|
\end{proof}
|
|
|
|
\subsubsection{\pending{Exercise 1.11.7b}}%
|
|
\label{ssub:exercise-1.11.7b}
|
|
|
|
Derive the result analytically as follows:
|
|
By changing the index of summation, note that
|
|
$\sum_{n=1}^{b-1} \floor{na / b} = \sum_{n=1}^{b-1} \floor{a(b - n) / b}$.
|
|
Now apply Exercises 4(a) and (b) to the bracket on the right.
|
|
|
|
\begin{proof}
|
|
|
|
\lean{Bookshelf/Apostol/Chapter\_1\_11}
|
|
{Apostol.Chapter\_1\_11.exercise\_7b}
|
|
|
|
Let $n = 1, \ldots, b - 1$.
|
|
By hypothesis, $a$ and $b$ are coprime.
|
|
Furthermore, $n < b$ for all values of $n$.
|
|
Thus $an / b$ is not an integer.
|
|
By \nameref{ssub:exercise-1.11.4b},
|
|
\begin{equation}
|
|
\label{ssub:exercise-1.11.7b-eq1}
|
|
\floor{-\frac{an}{b}} = -\floor{\frac{an}{b}} - 1.
|
|
\end{equation}
|
|
Consider the following:
|
|
\begin{align*}
|
|
\sum_{n=1}^{b-1} \floor{\frac{na}{b}}
|
|
& = \sum_{n=1}^{b-1} \floor{\frac{a(b - n)}{b}} \\
|
|
& = \sum_{n=1}^{b-1} \floor{\frac{ab - an}{b}} \\
|
|
& = \sum_{n=1}^{b-1} \floor{-\frac{an}{b} + a} \\
|
|
& = \sum_{n=1}^{b-1} \floor{-\frac{an}{b}} + a.
|
|
& \textref{ssub:exercise-1.11.4a} \\
|
|
& = \sum_{n=1}^{b-1} -\floor{\frac{an}{b}} - 1 + a
|
|
& \eqref{ssub:exercise-1.11.7b-eq1} \\
|
|
& = -\sum_{n=1}^{b-1} \floor{\frac{an}{b}} - \sum_{n=1}^{b-1} 1 +
|
|
\sum_{n=1}^{b-1} a \\
|
|
& = -\sum_{n=1}^{b-1} \floor{\frac{an}{b}} - (b - 1) + a(b - 1).
|
|
\end{align*}
|
|
Rearranging the above yields
|
|
$$2\sum_{n=1}^{b-1} \floor{\frac{an}{b}} = (a - 1)(b - 1).$$
|
|
Dividing both sides of the above identity concludes the proof.
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\pending{Exercise 1.11.8}}%
|
|
\label{sub:exercise-1.11.8}
|
|
|
|
Let $S$ be a set of points on the real line.
|
|
Let $\mathcal{X}_S$ denote the \nameref{ref:characteristic-function} of $S$.
|
|
Let $f$ be a \nameref{ref:step-function} which takes the constant value
|
|
$c_k$ on the $k$th open subinterval $I_k$ of some partition of an interval
|
|
$[a, b]$.
|
|
Prove that for each $x$ in the union $I_1 \cup I_2 \cup \cdots \cup I_n$ we have
|
|
$$f(x) = \sum_{k=1}^n c_k\mathcal{X}_{I_k}(x).$$
|
|
This property is described by saying that every step function is a linear
|
|
combination of characteristic functions of intervals.
|
|
|
|
\begin{proof}
|
|
|
|
Let $x \in I_1 \cup I_2 \cup \cdots \cup I_n$ and $N = \{1, \ldots, n\}$.
|
|
Let $k \in N$ such that $x \in I_k$.
|
|
Consider an arbitrary $j \in N - \{k\}$.
|
|
By definition of a nameref{ref:partition}, $I_j \cap I_k = \emptyset$.
|
|
That is, $I_j$ and $I_k$ are disjoint for all $j \in N - \{k\}$.
|
|
Therefore, by definition of the characteristic function,
|
|
$\mathcal{X}_{I_k}(x) = 1$ and $\mathcal{X}_{I_j}(x) = 0$ for all
|
|
$j \in N - \{k\}$.
|
|
Thus
|
|
\begin{align*}
|
|
f(x)
|
|
& = c_k \\
|
|
& = (c_k)(1) + \sum\nolimits_{j \in N - \{k\}} (c_j)(0) \\
|
|
& = c_k\mathcal{X}_{I_k}(x) +
|
|
\sum\nolimits_{j \in N - \{k\}} c_j\mathcal{X}_{I_j}(x) \\
|
|
& = \sum_{k=1}^n c_k\mathcal{X}_{I_k}(x).
|
|
\end{align*}
|
|
|
|
\end{proof}
|
|
|
|
\section{Properties of the Integral of a Step Function}%
|
|
\label{sec:properties-integral-step-function}
|
|
|
|
\subsection{\pending{Additive Property}}%
|
|
\label{sub:step-additive-property}
|
|
\label{sub:theorem-1.2}
|
|
|
|
\begin{theorem}[1.2]
|
|
|
|
Let $s$ and $t$ be \nameref{ref:step-function}s on closed interval
|
|
$[a, b]$.
|
|
Then
|
|
$$\int_a^b \left[ s(x) + t(x) \right] \mathop{dx} =
|
|
\int_a^b s(x) \mathop{dx} + \int_a^b t(x) \mathop{dx}.$$
|
|
|
|
\end{theorem}
|
|
|
|
\begin{proof}
|
|
|
|
Let $s$ and $t$ be step functions on closed interval $[a, b]$.
|
|
By definition of a step function, there exists a \nameref{ref:partition}
|
|
$P_s$ such that $s$ is constant on each open subinterval of $P_s$.
|
|
Likewise, there exists a partition $P_t$ such that $t$ is constant on each
|
|
open subinterval of $P_t$.
|
|
Therefore $s + t$ is a step function with step partition
|
|
$$P = P_s \cup P_t = \{x_0, x_1, \ldots, x_n\},$$ the common refinement of
|
|
$P_s$ and $P_t$ with subdivision points $x_0$, $x_1$, $\ldots$, $x_n$.
|
|
|
|
$s$ and $t$ remain constant on every open subinterval of $P$.
|
|
Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of
|
|
$P$.
|
|
Let $t_k$ denote the constant value of $t$ on the $k$th open subinterval of
|
|
$P$.
|
|
By definition of the \nameref{ref:integral-step-function},
|
|
\begin{align*}
|
|
\int_a^b \left[ s(x) + t(x) \right] \mathop{dx}
|
|
& = \sum_{k=1}^n (s_k + t_k) \cdot (x_k - x_{k-1}) \\
|
|
& = \sum_{k=1}^n \left[ s_k \cdot (x_k - x_{k-1}) +
|
|
t_k \cdot (x_k - x_{k-1}) \right] \\
|
|
& = \sum_{k=1}^n s_k \cdot (x_k - x_{k-1}) +
|
|
\sum_{k=1}^n t_k \cdot (x_k - x_{k-1}) \\
|
|
& = \int_a^b s(x) \mathop{dx} + \int_a^b t(x) \mathop{dx}.
|
|
\end{align*}
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\pending{Homogeneous Property}}%
|
|
\label{sub:step-homogeneous-property}
|
|
\label{sub:theorem-1.3}
|
|
|
|
\begin{theorem}[1.3]
|
|
|
|
Let $s$ be a \nameref{ref:step-function} on closed interval $[a, b]$.
|
|
For every real number $c$, we have
|
|
$$\int_a^b c \cdot s(x) \mathop{dx} = c\int_a^b s(x) \mathop{dx}.$$
|
|
|
|
\end{theorem}
|
|
|
|
\begin{proof}
|
|
|
|
Let $s$ be a step function on closed interval $[a, b]$.
|
|
By definition of a step function, there exists a \nameref{ref:partition}
|
|
$P = \{x_0, x_1, \ldots, x_n\}$ such that $s$ is constant on each open
|
|
subinterval of $P$.
|
|
Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of
|
|
$P$.
|
|
Then $c \cdot s$ is a step function with step partition $P$.
|
|
By definition of the \nameref{ref:integral-step-function},
|
|
\begin{align*}
|
|
\int_a^b c \cdot s(x) \mathop{dx}
|
|
& = \sum_{k=1}^n c \cdot s_k \cdot (x_k - x_{k-1}) \\
|
|
& = c \sum_{k=1}^n s_k \cdot (x_k - x_{k-1}) \\
|
|
& = c \int_a^b s(x) \mathop{dx}.
|
|
\end{align*}
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\pending{Linearity Property}}%
|
|
\label{sub:step-linearity-property}
|
|
\label{sub:theorem-1.4}
|
|
|
|
\begin{theorem}[1.4]
|
|
|
|
Let $s$ and $t$ be \nameref{ref:step-function}s on closed interval
|
|
$[a, b]$.
|
|
For every real $c_1$ and $c_2$, we have
|
|
$$\int_a^b \left[ c_1s(x) + c_2t(x) \right] \mathop{dx} =
|
|
c_1\int_a^b s(x) \mathop{dx} + c_2\int_a^b t(x) \mathop{dx}.$$
|
|
|
|
\end{theorem}
|
|
|
|
\begin{proof}
|
|
|
|
Let $s$ and $t$ be step functions on closed interval $[a, b]$.
|
|
Let $c_1$ and $c_2$ be real numbers.
|
|
Then $c_1 \cdot s$ and $c_2 \cdot t$ are step functions.
|
|
Then
|
|
\begin{align*}
|
|
& \int_a^b \left[ c_1s(x) + c_2t(x) \right] \mathop{dx} \\
|
|
& = \int_a^b c_1s(x) \mathop{dx} + \int_a^b c_2t(x) \mathop{dx}
|
|
& \textref{sub:step-additive-property} \\
|
|
& = c_1\int_a^b s(x) \mathop{dx} + c_2\int_a^b t(x). \mathop{dx}
|
|
& \textref{sub:step-homogeneous-property}
|
|
\end{align*}
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\pending{Comparison Theorem}}%
|
|
\label{sub:step-comparison-theorem}
|
|
\label{sub:theorem-1.5}
|
|
|
|
\begin{theorem}[1.5]
|
|
|
|
Let $s$ and $t$ be \nameref{ref:step-function}s on closed interval
|
|
$[a, b]$.
|
|
If $s(x) < t(x)$ for every $x$ in $[a, b]$, then
|
|
$$\int_a^b s(x) \mathop{dx} < \int_a^b t(x) \mathop{dx}.$$
|
|
|
|
\end{theorem}
|
|
|
|
\begin{proof}
|
|
|
|
Let $s$ and $t$ be step functions on closed interval $[a, b]$.
|
|
By definition of a step function, there exists a \nameref{ref:partition}
|
|
$P_s$ such that $s$ is constant on each open subinterval of $P_s$.
|
|
Likewise, there exists a partition $P_t$ such that $t$ is constant on each
|
|
open subinterval of $P_t$.
|
|
Let $$P = P_s \cup P_t = \{x_0, x_1, \ldots, x_n\}$$ be the common refinement
|
|
of $P_s$ and $P_t$ with subdivision points $x_0$, $x_1$, $\ldots$, $x_n$.
|
|
|
|
By construction, $P$ is a step partition for both $s$ and $t$.
|
|
Thus $s$ and $t$ remain constant on every open subinterval of $P$.
|
|
Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of
|
|
$P$.
|
|
Let $t_k$ denote the constant value of $t$ on the $k$th open subinterval of
|
|
$P$.
|
|
By definition of the \nameref{ref:integral-step-function},
|
|
\begin{align*}
|
|
\int_a^b s(x) \mathop{dx}
|
|
& = \sum_{k=1}^n s_k \cdot (x_k - x_{k-1}) \\
|
|
& < \sum_{k=1}^n t_k \cdot (x_k - x_{k-1}) & \text{by hypothesis} \\
|
|
& = \int_a^b t(x) \mathop{dx}.
|
|
\end{align*}
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\pending{Additivity With Respect to the Interval of Integration}}%
|
|
\label{sub:step-additivity-with-respect-interval-integration}
|
|
\label{sub:theorem-1.6}
|
|
|
|
\begin{theorem}[1.6]
|
|
|
|
Let $a, b, c \in \mathbb{R}$ and $s$ a \nameref{ref:step-function} on the
|
|
smallest closed interval containing them.
|
|
Then
|
|
$$\int_a^c s(x) \mathop{dx} + \int_c^b s(x) \mathop{dx} +
|
|
\int_b^a s(x) \mathop{dx} = 0.$$
|
|
|
|
\end{theorem}
|
|
|
|
\begin{proof}
|
|
|
|
WLOG, suppose $a < c < b$ and $s$ be a step function on closed interval
|
|
$[a, b]$.
|
|
By definition of a step function, there exists a \nameref{ref:partition}
|
|
$P$ such that $s$ is constant on each open subinterval of $P$.
|
|
|
|
Let $Q = \{x_0, x_1, \ldots, x_n\}$ be a refinement of $P$ that includes $c$
|
|
as a subdivision point.
|
|
Then $Q$ is a step partition of $s$ and there exists some $0 < i < n$ such
|
|
that $x_i = c$.
|
|
Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of
|
|
$Q$.
|
|
By definition of the \nameref{ref:integral-step-function},
|
|
\begin{align*}
|
|
\int_a^b s(x) \mathop{dx}
|
|
& = \sum_{k=1}^n s_k \cdot (x_k - x_{k - 1}) \\
|
|
& = \sum_{k=1}^i s_k \cdot (x_k - x_{k - 1}) +
|
|
\sum_{k=i+1}^n s_k \cdot (x_k - x_{k - 1}) \\
|
|
& = \int_a^c s(x) \mathop{dx} + \int_c^b s(x) \mathop{dx}.
|
|
\end{align*}
|
|
Rearranging terms shows
|
|
\begin{align*}
|
|
0
|
|
& = \int_a^c s(x) \mathop{dx} + \int_c^b s(x) \mathop{dx} -
|
|
\int_a^b s(x) \mathop{dx} \\
|
|
& = \int_a^c s(x) \mathop{dx} + \int_c^b s(x) \mathop{dx} +
|
|
\int_b^a s(x) \mathop{dx}.
|
|
\end{align*}
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\pending{Invariance Under Translation}}%
|
|
\label{sub:step-invariance-under-translation}
|
|
\label{sub:theorem-1.7}
|
|
|
|
\begin{theorem}[1.7]
|
|
|
|
Let $s$ be a step function on closed interval $[a, b]$.
|
|
Then
|
|
$$\int_a^b s(x) \mathop{dx} =
|
|
\int_{a+c}^{b+x} s(x - c) \mathop{dx} \quad\text{for every real } c.$$
|
|
|
|
\end{theorem}
|
|
|
|
\begin{proof}
|
|
|
|
Let $s$ be a step function on closed interval $[a, b]$.
|
|
By definition of a step function, there exists a \nameref{ref:partition}
|
|
$P = \{x_0, x_1, \ldots, x_n\}$ such that $s$ is constant on each open
|
|
subinterval of $P$.
|
|
Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of
|
|
$P$.
|
|
|
|
Let $c$ be a real number.
|
|
Then $t(x) = s(x - c)$ is a step function on closed interval $[a + c, b + c]$
|
|
with partition $Q = \{x_0 + c, x_1 + c, \ldots, x_n + c\}$.
|
|
Furthermore, $t$ is constant on each open subinterval of $Q$.
|
|
Let $t_k$ denote the value of $t$ on the $k$th open subinterval of $Q$.
|
|
By construction, $t_k = s_k$.
|
|
|
|
By definition of the \nameref{ref:integral-step-function},
|
|
\begin{align*}
|
|
\int_{a+c}^{b+c} s(x - c) \mathop{dx}
|
|
& = \int_{a+c}^{b+c} t(x) \mathop{dx} \\
|
|
& = \sum_{k=1}^n t_k \cdot ((x_k + c) - (x_{k - 1} + c)) \\
|
|
& = \sum_{k=1}^n t_k \cdot (x_k - x_{k - 1}) \\
|
|
& = \sum_{k=1}^n s_k \cdot (x_k - x_{k - 1}) \\
|
|
& = \int_a^b s(x) \mathop{dx}.
|
|
\end{align*}
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\pending{Expansion or Contraction of the Interval of Integration}}%
|
|
\label{sub:step-expansion-contraction-interval-integration}
|
|
\label{sub:theorem-1.8}
|
|
|
|
\begin{theorem}[1.8]
|
|
|
|
Let $s$ be a step function on closed interval $[a, b]$.
|
|
Then
|
|
$$\int_{ka}^{kb} s \left( \frac{x}{k} \right) \mathop{dx} =
|
|
k \int_a^b s(x) \mathop{dx} \quad\text{for every } k \neq 0.$$
|
|
|
|
\end{theorem}
|
|
|
|
\begin{proof}
|
|
|
|
Let $s$ be a step function on closed interval $[a, b]$.
|
|
By definition of a step function, there exists a \nameref{ref:partition}
|
|
$P = \{x_0, x_1, \ldots, x_n\}$ such that $s$ is constant on each open
|
|
subinterval of $P$.
|
|
Let $s_i$ denote the value of $s$ on the $i$th open subinterval of $P$.
|
|
|
|
Let $k \neq 0$ be a real number.
|
|
There are two cases to consider:
|
|
|
|
\paragraph{Case 1}%
|
|
|
|
Suppose $k > 0$.
|
|
Then $t(x) = s(x / k)$ is a step function on closed interval $[ka, kb]$
|
|
with partition $Q = \{kx_0, kx_1, \ldots, kx_n\}$.
|
|
Furthermore $t_i = s_i$.
|
|
By definition of the \nameref{ref:integral-step-function},
|
|
\begin{align*}
|
|
\int_{ka}^{kb} s(x / k) \mathop{dx}
|
|
& = \int_{ka}^{kb} t(x) \mathop{dx} \\
|
|
& = \sum_{i=1}^n t_i \cdot (kx_i - kx_{i-1}) \\
|
|
& = k \sum_{i=1}^n t_i \cdot (x_i - x_{i-1}) \\
|
|
& = k \sum_{i=1}^n s_i \cdot(x_i - x_{i-1}) \\
|
|
& = k \int_a^b s(x) \mathop{dx}.
|
|
\end{align*}
|
|
|
|
\paragraph{Case 2}%
|
|
|
|
Let $k < 0$ be a real number.
|
|
Then $t(x) = s(x / k)$ is a step function on closed interval $[kb, ka]$
|
|
with partition $Q = \{kx_n, kx_{n-1}, \ldots, kx_0\}$.
|
|
Furthermore $t_i = s_i$.
|
|
By definition of the \nameref{ref:integral-step-function},
|
|
\begin{align*}
|
|
\int_{ka}^{kb} s(x / k) \mathop{dx}
|
|
& = -\int_{kb}^{ka} s(x / k) \mathop{dx} \\
|
|
& = -\int_{kb}^{ka} t(x) \mathop{dx} \\
|
|
& = -\sum_{i=1}^n t_i \cdot (kx_{i-1} - kx_i) \\
|
|
& = -\sum_{i=1}^n s_i \cdot (kx_{i-1} - kx_i) \\
|
|
& = -\sum_{i=1}^n s_i \cdot (-k) \cdot (x_i - x_{i-1}) \\
|
|
& = k \sum_{i=1}^n s_i \cdot (x_i - x_{i-1}) \\
|
|
& = k \int_a^b s(x) \mathop{dx}.
|
|
\end{align*}
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\pending{Reflection Property}}%
|
|
\label{sub:step-reflection-property}
|
|
|
|
Let $s$ be a step function on closed interval $[a, b]$.
|
|
Then
|
|
$$\int_a^b s(x) \mathop{dx} = -\int_{-b}^{-a} s(-x) \mathop{dx}.$$
|
|
|
|
\begin{proof}
|
|
|
|
Let $k = -1$.
|
|
By \nameref{sub:step-expansion-contraction-interval-integration},
|
|
$$\int_{-a}^{-b} s \left( \frac{x}{-1} \right) =
|
|
-\int_a^b s(x) \mathop{dx}.$$
|
|
Simplifying the left-hand side of the above identity, and multiplying both
|
|
sides by $-1$ yields the desired result.
|
|
|
|
\end{proof}
|
|
|
|
\section{Exercises 1.15}%
|
|
\label{sec:exercises-1.15}
|
|
|
|
\subsection{\pending{Exercise 1.15.1}}%
|
|
\label{sub:exercise-1.15.1}
|
|
|
|
Compute the value of each of the following integrals.
|
|
|
|
\subsubsection{\pending{Exercise 1.15.1a}}%
|
|
\label{ssub:exercise-1.15.1a}
|
|
|
|
$\int_{-1}^3 \floor{x} \mathop{dx}$.
|
|
|
|
\begin{proof}
|
|
|
|
Let $s(x) = \floor{x}$ with domain $[-1, 3]$.
|
|
By construction, $s$ is a step function with partition
|
|
$P = \{-1, 0, 1, 2, 3\} = \{x_0, x_1, x_2, x_3, x_4\}$.
|
|
Let $s_k$ denote the constant value $s$ takes on the $k$th open subinterval of
|
|
$P$.
|
|
By definition of the \nameref{ref:integral-step-function},
|
|
\begin{align*}
|
|
\int_{-1}^3 \floor{x} \mathop{dx}
|
|
& = \sum_{k=1}^4 s_k \cdot (x_k - x_{k-1}) \\
|
|
& = -1 + 0 + 1 + 2 \\
|
|
& = 2.
|
|
\end{align*}
|
|
|
|
\end{proof}
|
|
|
|
\subsubsection{\pending{Exercise 1.15.1c}}%
|
|
\label{ssub:exercise-1.15.1c}
|
|
|
|
$\int_{-1}^3 \left(\floor{x} + \floor{x + \frac{1}{2}}\right) \mathop{dx}$.
|
|
|
|
\begin{proof}
|
|
|
|
Let $s(x) = \floor{x} + \floor{x + \frac{1}{2}}$ with domain $[-1, 3]$.
|
|
By construction, $s$ is a step function with partition
|
|
\begin{align*}
|
|
P
|
|
& = \{-1, -\frac{1}{2}, 0, \frac{1}{2}, 1, \frac{3}{2}, 2, \frac{5}{2},
|
|
3\} \\
|
|
& = \{x_0, x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8\}.
|
|
\end{align*}
|
|
Let $s_k$ denote the constant value $s$ takes on the $k$th open subinterval of
|
|
$P$.
|
|
By definition of the \nameref{ref:integral-step-function},
|
|
\begin{align*}
|
|
\int_{-1}^3 \floor{x} + \floor{x + \frac{1}{2}}
|
|
& = \sum_{k=1}^8 s_k \cdot (x_k - x_{k-1}) \\
|
|
& = \frac{1}{2} \sum_{k=1}^8 s_k \\
|
|
& = \frac{1}{2} \left( -2 - 1 + 0 + 1 + 2 + 3 + 4 + 5 \right) \\
|
|
& = 6.
|
|
\end{align*}
|
|
|
|
\end{proof}
|
|
|
|
\subsubsection{\pending{Exericse 1.15.1e}}%
|
|
\label{ssub:exercise-1.15.1e}
|
|
|
|
$\int_{-1}^3 \floor{2x} \mathop{dx}$.
|
|
|
|
\begin{proof}
|
|
|
|
Let $s(x) = \floor{2x}$.
|
|
By \nameref{sub:hermites-identity},
|
|
$s(x) = \floor{x} + \floor{x + \frac{1}{2}}$.
|
|
Thus, by \nameref{ssub:exercise-1.15.1c},
|
|
$$\int_{-1}^3 \floor{2x} \mathop{dx} = 6.$$
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\pending{Exercise 1.15.3}}%
|
|
\label{sub:exercise-1.15.3}
|
|
|
|
Show that
|
|
$\int_a^b \floor{x} \mathop{dx} + \int_a^b \floor{-x} \mathop{dx} = a - b$.
|
|
|
|
\begin{proof}
|
|
|
|
Let $s(x) = \floor{x}$ and $t(x) = \floor{-x}$, both with domain $[a, b]$.
|
|
Let $x_1$, $\ldots$, $x_{n-1}$ denote the integers found in interval $(a, b)$.
|
|
Then $P = \{x_0, x_1, \ldots, x_n\}$, $x_0 = a$ and $x_n = b$, is a step
|
|
\nameref{ref:partition} of both $s$ and $t$.
|
|
Let $s_k$ and $t_k$ denote the constant values $s$ and $t$ take on the $k$th
|
|
open subinterval of $P$ respectively.
|
|
By \nameref{ssub:exercise-1.11.4b}, $\floor{-x} = -\floor{x} - 1$ for all $x$
|
|
in every open subinterval of $P$.
|
|
That is, $s_k = -t_k - 1$.
|
|
By definition of the \nameref{ref:integral-step-function},
|
|
\begin{align*}
|
|
\int_a^b \floor{x} \mathop{dx} + \int_a^b \floor{-x} \mathop{dx}
|
|
& = \sum_{k=1}^n s_k (x_k - x_{k-1}) +
|
|
\sum_{k=1}^n t_k (x_k - x_{k-1}) \\
|
|
& = \sum_{k=1}^n (x_k - x_{k-1}) \cdot (s_k + t_k) \\
|
|
& = \sum_{k=1}^n (x_k - x_{k-1}) \cdot (-t_k - 1 + t_k) \\
|
|
& = \sum_{k=1}^n (x_{k-1} - x_k) \\
|
|
& = x_0 - x_n \\
|
|
& = a - b.
|
|
\end{align*}
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\pending{Exercise 1.15.5}}%
|
|
\label{sub:exercise-1.15.5}
|
|
|
|
\subsubsection{\pending{Exercise 1.15.5a}}%
|
|
\label{ssub:exercise-1.15.5a}
|
|
|
|
Prove that $\int_0^2 \floor{t^2} \mathop{dt} = 5 - \sqrt{2} - \sqrt{3}$.
|
|
|
|
\begin{proof}
|
|
|
|
Let $s(t) = \floor{t^2}$ with domain $[0, 2]$.
|
|
Then $s$ is a \nameref{ref:step-function} with partition
|
|
$P = \{0, 1, \sqrt{2}, \sqrt{3}, 2\} = \{x_0, x_1, \ldots, x_4\}$.
|
|
Let $s_k$ denote the constant value that $s$ takes in the $k$th open
|
|
subinterval of $P$.
|
|
By the \nameref{ref:integral-step-function},
|
|
\begin{align*}
|
|
\int_0^2 \floor{t^2} \mathop{dt}
|
|
& = \sum_{k=1}^4 s_k \cdot (x_k - x_{k-1}) \\
|
|
& = 0 \cdot (1 - 0) + 1 \cdot (\sqrt{2} - 1) +
|
|
2 \cdot (\sqrt{3} - \sqrt{2}) + 3 \cdot (2 - \sqrt{3}) \\
|
|
& = 5 - \sqrt{2} - \sqrt{3}.
|
|
\end{align*}
|
|
|
|
\end{proof}
|
|
|
|
\subsubsection{\pending{Exercise 1.15.5b}}%
|
|
\label{ssub:exercise-1.15.5b}
|
|
|
|
Compute $\int_{-3}^3 \floor{t^2} \mathop{dt}$.
|
|
|
|
\begin{proof}
|
|
|
|
Let $s(t) = \floor{t^2}$ with domain $[0, 3]$.
|
|
Then $s$ is a \nameref{ref:step-function} with \nameref{ref:partition}
|
|
\begin{align*}
|
|
P
|
|
& = \{\sqrt{0}, \sqrt{1}, \sqrt{2}, \sqrt{3}, \sqrt{4}, \sqrt{5},
|
|
\sqrt{6}, \sqrt{7}, \sqrt{8}, \sqrt{9}\} \\
|
|
& = \{x_0, x_1, \ldots, x_9\}.
|
|
\end{align*}
|
|
Let $s_k$ denote the constant value that $s$ takes in the $k$th open
|
|
subinterval of $P$.
|
|
By the \nameref{ref:integral-step-function},
|
|
\begin{align}
|
|
\int_0^3 \floor{t^2} \mathop{dt}
|
|
& = \sum_{k=1}^9 s_k \cdot (x_k - x_{k-1})
|
|
\nonumber \\
|
|
& = \sum_{k=0}^8 k \cdot (\sqrt{k + 1} - \sqrt{k}).
|
|
\label{sub:exercise-1.15.5b-eq1}
|
|
\end{align}
|
|
We notice $\floor{t^2}$ is symmetric about the $y$-axis.
|
|
Thus
|
|
\begin{equation}
|
|
\label{sub:exercise-1.15.5b-eq2}
|
|
\int_{-3}^0 \floor{t^2} \mathop{dt} = \int_0^3 \floor{t^2}.
|
|
\end{equation}
|
|
By \nameref{sub:step-additivity-with-respect-interval-integration},
|
|
\begin{align*}
|
|
\int_{-3}^3 \floor{t^2} \mathop{dt}
|
|
& = \int_{-3}^0 \floor{t^2} \mathop{dt} +
|
|
\int_0^3 \floor{t^2} \mathop{dt} \\
|
|
& = 2\int_0^3 \floor{t^2} \mathop{dt}
|
|
& \eqref{sub:exercise-1.15.5b-eq2} \\
|
|
& = 2 \left[\sum_{k=0}^8 k \cdot (\sqrt{k + 1} - \sqrt{k})\right].
|
|
& \eqref{sub:exercise-1.15.5b-eq1}
|
|
\end{align*}
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\pending{Exercise 1.15.7}}%
|
|
\label{sub:exercise-1.15.7}
|
|
|
|
\subsubsection{\pending{Exercise 1.15.7a}}%
|
|
\label{ssub:exercise-1.15.7a}
|
|
|
|
Compute $\int_0^9 \floor{\sqrt{t}} \mathop{dt}$.
|
|
|
|
\begin{proof}
|
|
|
|
Let $s(t) = \floor{\sqrt{t}}$ with domain $[0, 9]$.
|
|
Then $s$ is a \nameref{ref:step-function} with \nameref{ref:partition}
|
|
$P = \{0, 1, 4, 9\} = \{x_0, x_1, x_2, x_3\}$.
|
|
Let $s_k$ denote the constant value that $s$ takes in the $k$th open
|
|
subinterval of $P$.
|
|
By the \nameref{ref:integral-step-function},
|
|
\begin{align*}
|
|
\int_0^9 \floor{\sqrt{t}} \mathop{dt}
|
|
& = \sum_{k=1}^3 s_k \cdot (x_k - x_{k-1}) \\
|
|
& = 0 \cdot (1 - 0) + 1 \cdot (4 - 1) + 2 \cdot (9 - 4) \\
|
|
& = 13.
|
|
\end{align*}
|
|
|
|
\end{proof}
|
|
|
|
\subsubsection{\pending{Exercise 1.15.7b}}%
|
|
\label{ssub:exercise-1.15.7b}
|
|
|
|
If $n$ is a positive integer, prove that
|
|
$$\int_0^{n^2} \floor{\sqrt{t}} \mathop{dt} = n(n - 1)(4n + 1) / 6.$$
|
|
|
|
\begin{proof}
|
|
|
|
Define predicate $P(n)$ as
|
|
\begin{equation}
|
|
\label{sub:exercise-1.15.7b-eq1}
|
|
\int_0^{n^2} \floor{\sqrt{t}} \mathop{dt} = \frac{n(n - 1)(4n + 1)}{6}.
|
|
\end{equation}
|
|
We use induction to prove $P(n)$ holds for all integers satisfying $n > 0$.
|
|
|
|
\paragraph{Base Case}%
|
|
|
|
Let $n = 1$.
|
|
Define $s(t) = \floor{\sqrt{t}}$ with domain $[0, 1]$.
|
|
Then $s$ is a \nameref{ref:step-function} with \nameref{ref:partition}
|
|
$P = \{0, 1\} = \{x_0, x_1\}$.
|
|
Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of
|
|
$P$.
|
|
By definition of the \nameref{ref:integral-step-function}, the left-hand
|
|
side of \eqref{sub:exercise-1.15.7b-eq1} evaluates to
|
|
\begin{align*}
|
|
\int_0^{n^2} \floor{\sqrt{t}} \mathop{dt}
|
|
& = \int_0^1 \floor{\sqrt{t}} \mathop{dt} \\
|
|
& = \sum_{k=1}^1 s_k \cdot (x_k - x_{k-1}) \\
|
|
& = 0.
|
|
\end{align*}
|
|
The right-hand side of \eqref{sub:exercise-1.15.7b-eq1} likewise evaluates
|
|
to $0$.
|
|
Thus $P(1)$ holds.
|
|
|
|
\paragraph{Induction Step}%
|
|
|
|
Let $n > 0$ be a positive integer and suppose $P(n)$ is true.
|
|
Define $s(t) = \floor{\sqrt{t}}$ with domain $[0, (n + 1)^2]$.
|
|
Then $s$ is a \nameref{ref:step-function} with \nameref{ref:partition}
|
|
\begin{align*}
|
|
P
|
|
& = \{0, 1, 4, \ldots, n^2, (n + 1)^2\} \\
|
|
& = \{x_0, x_1, \ldots, x_n, x_{n + 1}\}.
|
|
\end{align*}
|
|
Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of
|
|
$P$.
|
|
By definition of the \nameref{ref:integral-step-function}, it follows
|
|
that
|
|
\begin{align*}
|
|
& \int_0^{(n + 1)^2} s(x) \mathop{dx} \\
|
|
& = \sum_{k=1}^{n + 1} s_k \cdot (x_k - x_{k-1}) \\
|
|
& = \sum_{k=1}^n s_k \cdot (x_k - x_{k-1}) +
|
|
\left[s_{n+1} \cdot (x_{n + 1} - x_n)\right] \\
|
|
& = \int_0^{n^2} s(x) \mathop{dx} +
|
|
\left[s_{n+1} \cdot (x_{n + 1} - x_n)\right] \\
|
|
& = \int_0^{n^2} s(x) \mathop{dx} +
|
|
\left[ n \cdot ((n + 1)^2 - n^2) \right] \\
|
|
& = \int_0^{n^2} s(x) \mathop{dx} + \left[ 2n^2 + n \right] \\
|
|
& = \frac{n(n - 1)(4n + 1)}{6} + 2n^2 + n
|
|
& \text{induction hypothesis} \\
|
|
& = \frac{n(n - 1)(4n + 1) + 12n^2 + 6n}{6} \\
|
|
& = \frac{4n^3 + 9n^2 + 5n}{6} \\
|
|
& = \frac{(n^2 + n)(4n + 5)}{6} \\
|
|
& = \frac{(n + 1)((n + 1) - 1)(4(n + 1) + 1)}{6}.
|
|
\end{align*}
|
|
Thus $P(n + 1)$ holds.
|
|
|
|
\paragraph{Conclusion}%
|
|
|
|
By mathematical induction, it follows for all positive integers $n$, $P(n)$
|
|
is true.
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\pending{Exercise 1.15.9}}%
|
|
\label{sub:exercise-1.15.9}
|
|
|
|
Show that the following property is equivalent to
|
|
\nameref{sub:step-expansion-contraction-interval-integration}:
|
|
\begin{equation}
|
|
\label{sub:exercise-1.15.9-eq1}
|
|
\int_{ka}^{kb} f(x) \mathop{dx} = k \int_a^b f(kx) \mathop{dx}.
|
|
\end{equation}
|
|
|
|
\begin{proof}
|
|
|
|
Let $f$ be a step function on closed interval $[a, b]$ and $k \neq 0$.
|
|
Applying \nameref{sub:step-expansion-contraction-interval-integration} to the
|
|
right-hand side of \eqref{sub:exercise-1.15.9-eq1} yields
|
|
$$k\int_{ka}^{kb} f(kx / k) \mathop{dx} =
|
|
k\left[k\int_a^b f(kx) \mathop{dx}\right].$$
|
|
Simplifying the left-hand side and dividing both sides by $k$ immediately
|
|
yields the desired result.
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\pending{Exercise 1.15.11}}%
|
|
\label{sub:exercise-1.15.11}
|
|
|
|
If we instead defined the integral of step functions as
|
|
\begin{equation*}
|
|
\label{sub:exercise-1.15.11-eq1}
|
|
\int_a^b s(x) \mathop{dx} = \sum_{k=1}^n s_k^3 \cdot (x_k - x_{k-1}),
|
|
\end{equation*}
|
|
a new and different theory of integration would result.
|
|
Which of the following properties would remain valid in this new theory?
|
|
|
|
\subsubsection{\pending{Exercise 1.15.11a}}%
|
|
\label{ssub:exercise-1.15.11a}
|
|
|
|
$\int_a^b s + \int_b^c s = \int_a^c s$.
|
|
|
|
\note{This property mirrors
|
|
\nameref{sub:step-additivity-with-respect-interval-integration}.}
|
|
|
|
\begin{proof}
|
|
|
|
The above property is \textbf{valid}.
|
|
|
|
\vspace{6pt}
|
|
|
|
WLOG, suppose $a < b < c$.
|
|
Let $s$ be a step function defined on closed interval $[a, c]$.
|
|
By definition of a \nameref{ref:step-function}, there exists a
|
|
\nameref{ref:partition} such that $s$ is constant on each open
|
|
subinterval of $P$.
|
|
Let $Q = \{x_0, x_1, \ldots, x_n\}$ be a refinement of $P$ that includes $b$
|
|
as a subdivision point.
|
|
Then $Q$ is a step partition of $s$ and there exists some $0 < i < n$ such
|
|
that $x_i = c$.
|
|
Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of
|
|
$Q$.
|
|
By \eqref{sub:exercise-1.15.11-eq1},
|
|
\begin{align*}
|
|
\int_a^c s
|
|
& = \sum_{k=1}^n s_k^3 \cdot (x_k - x_{k-1}) \\
|
|
& = \sum_{k=1}^i s_k^3 \cdot (x_k - x_{k-1}) +
|
|
\sum_{k=i+1}^n s_k^3 \cdot (x_k - x_{k-1}) \\
|
|
& = \int_a^b s + \int_b^c s.
|
|
\end{align*}
|
|
|
|
\end{proof}
|
|
|
|
\subsubsection{\pending{Exercise 1.15.11b}}%
|
|
\label{ssub:exercise-1.15.11b}
|
|
|
|
$\int_a^b (s + t) = \int_a^b s + \int_a^b t$.
|
|
|
|
\note{This property mirrors the \nameref{sub:step-additive-property}.}
|
|
|
|
\begin{proof}
|
|
|
|
The above property is \textbf{invalid}.
|
|
|
|
\vspace{6pt}
|
|
|
|
Let $s$ and $t$ be step functions on closed interval $[a, b]$.
|
|
By definition of a step function, there exists a \nameref{ref:partition}
|
|
$P_s$ such that $s$ is constant on each open subinterval of $P_s$.
|
|
Likewise, there exists a partition $P_t$ such that $t$ is constant on each
|
|
open subinterval of $P_t$.
|
|
Therefore $s + t$ is a step function with step partition
|
|
$$P = P_s \cup P_t = \{x_0, x_1, \ldots, x_n\},$$
|
|
the common refinement of $P_s$ and $P_t$ with subdivision points
|
|
$x_0$, $x_1$, $\ldots$, $x_n$.
|
|
|
|
$s$ and $t$ remain constant on every open subinterval of $P$.
|
|
Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of
|
|
$P_s$.
|
|
Let $t_k$ denote the constant value of $t$ on the $k$th open subinterval of
|
|
$P_t$.
|
|
By \eqref{sub:exercise-1.15.11-eq1},
|
|
\begin{align*}
|
|
\int_a^b s + t
|
|
& = \sum_{k=1}^n (s_k + t_k)^3 \cdot (x_k - x_{k-1}) \\
|
|
& = \sum_{k=1}^n
|
|
\left[ s_k^3 + 3s_k^2t_k + 3s_kt_k^2 + t_k^3 \right] \\
|
|
& = \sum_{k=1}^n s_k^3 \cdot (x_k - x_{k-1}) \;+ \\
|
|
& \quad\qquad
|
|
\sum_{k=1}^n t_k^3 \cdot (x_k - x_{k-1}) \;+ \\
|
|
& \quad\qquad
|
|
\sum_{k=1}^n (3s_k^2t_k + 3s_kt_k^2) \cdot (x_k - x_{k - 1}) \\
|
|
& = \int_a^b s + \int_a^b t +
|
|
\sum_{k=1}^n (3s_k^2t_k + 3s_kt_k^2) \cdot (x_k - x_{k - 1}).
|
|
\end{align*}
|
|
Since this last addend does not necessarily equal $0$, the desired property is
|
|
invalid.
|
|
|
|
\end{proof}
|
|
|
|
\subsubsection{\pending{Exercise 1.15.11c}}%
|
|
\label{ssub:exercise-1.15.11c}
|
|
|
|
$\int_a^b c \cdot s = c \int_a^b s$.
|
|
|
|
\note{This property mirrors the \nameref{sub:step-homogeneous-property}.}
|
|
|
|
\begin{proof}
|
|
|
|
The above property is \textbf{invalid}.
|
|
|
|
\vspace{6pt}
|
|
|
|
Let $s$ be a step function on closed interval $[a, b]$.
|
|
By definition of a step function, there exists a \nameref{ref:partition}
|
|
$P = \{x_0, x_1, \ldots, x_n\}$ such that $s$ is constant on each open
|
|
subinterval of $P$.
|
|
Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of
|
|
$P$.
|
|
Then $c \cdot s$ is a step function with step partition $P$.
|
|
By \eqref{sub:exercise-1.15.11-eq1},
|
|
\begin{align*}
|
|
\int_a^b c \cdot s
|
|
& = \sum_{k=1}^n (c \cdot s_k)^3 \cdot (x_k - x_{k-1}) \\
|
|
& = \sum_{k=1}^n c^3 \cdot s_k^3 \cdot (x_k - x_{k-1}) \\
|
|
& = c^3 \sum_{k=1}^n s_k^3 \cdot (x_k - x_{k-1}) \\
|
|
& = c^3 \int_a^b s.
|
|
\end{align*}
|
|
Since $c^3$ does not necessarily equal $c$, the desired property is invalid.
|
|
|
|
\end{proof}
|
|
|
|
\subsubsection{\pending{Exercise 1.15.11d}}%
|
|
\label{ssub:exercise-1.15.11d}
|
|
|
|
$\int_{a+c}^{b+c} s(x) \mathop{dx} = \int_a^b s(x + c) \mathop{dx}$.
|
|
|
|
\note{This property mirrors \nameref{sub:step-invariance-under-translation}.}
|
|
|
|
\begin{proof}
|
|
|
|
The above property is \textbf{valid}.
|
|
|
|
\vspace{6pt}
|
|
|
|
Let $s$ be a step function on closed interval $[a + c, b + c]$.
|
|
By definition of a \nameref{ref:step-function}, there exists a
|
|
\nameref{ref:partition} $P = \{x_0, x_1, \ldots, x_n\}$ such that $s$ is
|
|
constant on each open subinterval of $P$.
|
|
Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of
|
|
$P$.
|
|
|
|
Let $c$ be a real number.
|
|
Then $t(x) = s(x + c)$ is a step function on closed interval $[a, b]$ with
|
|
partition $Q = \{x_0 - c, x_1 - c, \ldots, x_n - c\}$.
|
|
Furthermore, $t$ is constant on each open subinterval of $Q$.
|
|
Let $t_k$ denote the value of $t$ on the $k$th open subinterval of $Q$.
|
|
By construction, $t_k = s_k$.
|
|
By \eqref{sub:exercise-1.15.11-eq1},
|
|
\begin{align*}
|
|
\int_{a+c}^{b+c} s(x) \mathop{dx}
|
|
& = \sum_{k=1}^n s_k^3 \cdot (x_k - x_{k-1}) \\
|
|
& = \sum_{k=1}^n s_k^3 \cdot ((x_k - c) - (x_{k-1} - c)) \\
|
|
& = \sum_{k=1}^n t_k^3 \cdot ((x_k - c) - (x_{k-1} - c)) \\
|
|
& = \int_a^b t(x) \mathop{dx} \\
|
|
& = \int_a^b s(x + c) \mathop{dx}.
|
|
\end{align*}
|
|
|
|
\end{proof}
|
|
|
|
\subsubsection{\pending{Exercise 1.15.11e}}%
|
|
\label{ssub:exercise-1.15.11e}
|
|
|
|
If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
|
|
|
|
\note{This property mirrors the \nameref{sub:step-comparison-theorem}.}
|
|
|
|
\begin{proof}
|
|
|
|
The above property is \textbf{valid}.
|
|
|
|
\vspace{6pt}
|
|
|
|
Let $s$ and $t$ be step functions on closed interval $[a, b]$.
|
|
By definition of a \nameref{ref:step-function}, there exists a
|
|
\nameref{ref:partition} $P_s$ such that $s$ is constant on each open
|
|
subinterval of $P_s$.
|
|
Likewise, there exists a partition $P_t$ such that $t$ is constant on each
|
|
open subinterval of $P_t$.
|
|
Let $$P = P_s \cup P_t = \{x_0, x_1, \ldots, x_n\}$$ be the common refinement
|
|
of $P_s$ and $P_t$ with subdivision points $x_0$, $x_1$, $\ldots$, $x_n$.
|
|
|
|
By construction, $P$ is a step partition for both $s$ and $t$.
|
|
Thus $s$ and $t$ remain constant on every open subinterval of $P$.
|
|
Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of
|
|
$P$.
|
|
Let $t_k$ denote the constant value of $t$ on the $k$th open subinterval of
|
|
$P$.
|
|
By \eqref{sub:exercise-1.15.11-eq1},
|
|
\begin{align*}
|
|
\int_a^b s
|
|
& = \sum_{k=1}^n s_k^3 \cdot (x_k - x_{k-1}) \\
|
|
& < \sum_{k=1}^n t_k^3 \cdot (x_k - x_{k-1}) \\
|
|
& = \int_a^b t.
|
|
\end{align*}
|
|
|
|
\end{proof}
|
|
|
|
\section{Upper and Lower Integrals}%
|
|
\label{sec:upper-lower-integrals}
|
|
|
|
\subsection{\pending{Theorem 1.9}}%
|
|
\label{sub:theorem-1.9}
|
|
|
|
\begin{theorem}[1.9]
|
|
|
|
Every function $f$ which is bounded on $[a, b]$ has a lower integral
|
|
$\ubar{I}(f)$ and an upper integral $\overline{I}(f)$ satisfying the
|
|
inequalities
|
|
\begin{equation}
|
|
\label{sub:theorem-1.9-eq1}
|
|
\int_a^b s(x) \mathop{dx} \leq \ubar{I}(f) \leq
|
|
\bar{I}(f) \leq \int_a^b t(x) \mathop{dx}
|
|
\end{equation}
|
|
for all \nameref{ref:step-function}s $s$ and $t$ with $s \leq f \leq t$.
|
|
The function $f$ is \nameref{ref:integrable} on $[a, b]$ if and only if
|
|
its upper and lower integrals are equal, in which case we have
|
|
$$\int_a^b f(x) \mathop{dx} = \ubar{I}(f) = \bar{I}(f).$$
|
|
|
|
\end{theorem}
|
|
|
|
\begin{proof}
|
|
|
|
Let $f$ be a function bounded on $[a, b]$.
|
|
We prove that (i) $f$ has a lower and upper integral satisfying
|
|
\eqref{sub:theorem-1.9-eq1} and (ii) that $f$ is integrable on $[a, b]$ if
|
|
and only if its lower and upper integrals are equal.
|
|
|
|
\paragraph{(i)}%
|
|
|
|
Because $f$ is bounded, there exists some $M > 0$ such that
|
|
$\abs{f(x)} \leq M$ for all $x \in [a, b]$.
|
|
|
|
Let $S$ denote the set of numbers $\int_a^b s(x) \mathop{dx}$ obtained as
|
|
$s$ runs through all step functions below $f$.
|
|
That is, let $$S = \left\{ \int_a^b s(x) \mathop{dx} : s \leq f \right\}.$$
|
|
Note $S$ is nonempty since, e.g. constant function $c(x) = -M$ is a member.
|
|
|
|
Likewise, let $T$ denote the set of numbers $\int_a^b t(x) \mathop{dx}$
|
|
obtained as $t$ runs through all step functions above $f$.
|
|
That is, let $$T = \left\{ \int_a^b t(x) \mathop{dx} : f \leq t \right\}.$$
|
|
Note $T$ is nonempty since e.g. constant function $c(x) = M$ is a member.
|
|
|
|
By construction, $s \leq t$ for every $s$ in $S$ and $t$ in $T$.
|
|
Therefore \nameref{sec:theorem-i.34} tells us $S$ has a
|
|
\nameref{ref:supremum}, $T$ has an \nameref{ref:infimum}, and
|
|
$\sup{S} \leq \inf{T}$.
|
|
By definition of the \nameref{ref:lower-integral}, $\ubar{I}(f) = \sup{S}$.
|
|
By definition of the \nameref{ref:upper-integral}, $\bar{I}(f) = \inf{S}$.
|
|
Thus \eqref{sub:theorem-1.9-eq1} holds.
|
|
|
|
\paragraph{(ii)}%
|
|
|
|
By definition of integrability, $f$ is integrable on $[a, b]$ if and only if
|
|
there exists one and only one number $I$ such that
|
|
$$\int_a^b s(x) \mathop{dx} \leq I \leq \int_a^b t(x) \mathop{dx}$$
|
|
for every pair of step functions $s$ and $t$ satisfying
|
|
\eqref{ref:integral-bounded-function-eq1}.
|
|
By \eqref{sub:theorem-1.9-eq1} and the definition of the supremum/infimum,
|
|
this holds if and only if $\ubar{I}(f) = \bar{I}(f)$, concluding the
|
|
proof.
|
|
|
|
\end{proof}
|
|
|
|
\section{The Area of an Ordinate Set Expressed as an Integral}%
|
|
\label{sec:area-ordinate-set-expressed-integral}
|
|
|
|
\subsection{\pending{Theorem 1.10}}%
|
|
\label{sub:theorem-1.10}
|
|
|
|
\begin{theorem}[1.10]
|
|
|
|
Let $f$ be a nonnegative function, \nameref{ref:integrable} on an interval
|
|
$[a, b]$, and let $Q$ denote the ordinate set of $f$ over $[a, b]$.
|
|
Then $Q$ is measurable and its area is equal to the integral
|
|
$\int_a^b f(x) \mathop{dx}$.
|
|
|
|
\end{theorem}
|
|
|
|
\begin{proof}
|
|
|
|
Let $f$ be a nonnegative function, \nameref{ref:integrable} on $[a, b]$.
|
|
By definition of integrability, there exists one and only one number $I$ such
|
|
that $$\int_a^b s(x) \mathop{dx} \leq I \leq \int_a^b t(x) \mathop{dx}$$ for
|
|
every pair of step functions $s$ and $t$ satisfying
|
|
\eqref{ref:integral-bounded-function-eq1}.
|
|
In other words, $I$ is the one and only number that satisfies
|
|
$$a(S) \leq I \leq a(T)$$ for every pair of step regions
|
|
$S \subseteq Q \subseteq T$.
|
|
By the \nameref{sub:area-exhaustion-property}, $Q$ is measurable and its area
|
|
is equal to $I = \int_a^b f(x) \mathop{dx}$.
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\pending{Theorem 1.11}}%
|
|
\label{sub:theorem-1.11}
|
|
|
|
\begin{theorem}[1.11]
|
|
|
|
Let $f$ be a nonnegative function, integrable on an interval $[a, b]$.
|
|
Then the graph of $f$, that is, the set
|
|
\begin{equation}
|
|
\label{sub:theorem-1.11-eq1}
|
|
\{(x, y) \mid a \leq x \leq b, y = f(x)\},
|
|
\end{equation}
|
|
is measurable and has area equal to $0$.
|
|
|
|
\end{theorem}
|
|
|
|
\begin{proof}
|
|
|
|
Let $f$ be a nonnegative function, integrable on an interval $[a, b]$.
|
|
Let $$Q' = \{(x, y) \mid a \leq x \leq b, 0 \leq y < f(x)\}.$$
|
|
We show that (i) $Q'$ is measurable with area equal to
|
|
$\int_a^b f(x) \mathop{dx}$ and (ii) the graph of $f$ is meaurable with area
|
|
equal to $0$.
|
|
|
|
\paragraph{(i)}%
|
|
\label{par:theorem-1.11-i}
|
|
|
|
By definition of integrability, there exists one and only one number $I$
|
|
such that
|
|
$$\int_a^b s(x) \mathop{dx} \leq I \leq \int_a^b t(x) \mathop{dx}$$
|
|
for every pair of step functions $s$ and $t$ satisfying
|
|
\eqref{ref:integral-bounded-function-eq1}.
|
|
In other words, $I$ is the one and only number that satisfies
|
|
$$a(S) \leq I \leq a(T)$$ for every pair of step regions
|
|
$S \subseteq Q' \subseteq T$.
|
|
By the \nameref{sub:area-exhaustion-property}, $Q'$ is measurable and its
|
|
area is equal to $I = \int_a^b f(x) \mathop{dx}$.
|
|
|
|
\paragraph{(ii)}%
|
|
|
|
Let $Q$ denote the ordinate set of $f$.
|
|
By \nameref{sub:theorem-1.10}, $Q$ is measurable with area equal to the
|
|
integral $I = \int_a^b f(x) \mathop{dx}$.
|
|
By \nameref{par:theorem-1.11-i}, $Q'$ is
|
|
measurable with area also equal to $I$.
|
|
We note the graph of $f$, \eqref{sub:theorem-1.11-eq1}, is equal to set
|
|
$Q - Q'$.
|
|
By the \nameref{sub:area-difference-property}, $Q - Q'$ is measurable and
|
|
$$a(Q - Q') = a(Q) - a(Q') = I - I = 0.$$
|
|
Thus the graph of $f$ is measurable and has area equal to $0$.
|
|
|
|
\end{proof}
|
|
|
|
\section
|
|
[Integrability of Bounded Monotonic Functions]
|
|
{Integrability of Bounded Monotonic \texorpdfstring{\\}{}Functions}
|
|
\label{sec:integrability-bounded-monotonic-functions}
|
|
|
|
\subsection{\pending{Theorem 1.12}}%
|
|
\label{sub:theorem-1.12}
|
|
|
|
\begin{theorem}[1.12]
|
|
|
|
If $f$ is \nameref{ref:monotonic} on a closed interval $[a, b]$, then $f$
|
|
is \nameref{ref:integrable} on $[a, b]$.
|
|
|
|
\end{theorem}
|
|
|
|
\begin{proof}
|
|
|
|
Let $f$ be a monotonic function on closed interval $[a, b]$.
|
|
That is to say, either $f$ is increasing on $[a, b]$ or $f$ is decreasing on
|
|
$[a, b]$.
|
|
Because $f$ is on a closed interval, it is bounded.
|
|
By \nameref{sub:theorem-1.9}, $f$ has a \nameref{ref:lower-integral}
|
|
$\ubar{I}(f)$, $f$ has an \nameref{ref:upper-integral} $\bar{I}(f)$,
|
|
and $f$ is integrable if and only if $\ubar{I}(f) = \bar{I}(f)$.
|
|
|
|
Consider a partition $P = \{x_0, x_1, \ldots, x_n\}$ of $[a, b]$ in which
|
|
$x_k - x_{k-1} = (b - a) / n$ for each $k = 1, \ldots, n$.
|
|
There are two cases to consider:
|
|
|
|
\paragraph{Case 1}%
|
|
|
|
Suppose $f$ is increasing.
|
|
Let $s$ be the step function below $f$ with constant value $f(x_{k-1})$
|
|
on every $k$th open subinterval of $P$.
|
|
Let $t$ be the step function above $f$ with constant value $f(x_k)$
|
|
on every $k$th open subinterval of $P$.
|
|
Then, by \eqref{sub:theorem-1.9-eq1}, it follows
|
|
\begin{equation}
|
|
\label{sub:theorem-1.12-eq1}
|
|
\int_a^b s(x) \mathop{dx} \leq \ubar{I}(f)
|
|
\leq \bar{I}(f) \leq \int_a^b t(x) \mathop{dx}.
|
|
\end{equation}
|
|
By definition of the \nameref{ref:integral-step-function},
|
|
\begin{align*}
|
|
\int_a^b s(x) \mathop{dx}
|
|
& = \sum_{k=1}^n f(x_{k-1})\left[\frac{b - a}{n}\right] \\
|
|
\int_a^b t(x) \mathop{dx}
|
|
& = \sum_{k=1}^n f(x_k)\left[\frac{b - a}{n}\right].
|
|
\end{align*}
|
|
Thus
|
|
\begin{align*}
|
|
\int_a^b t(x) \mathop{dx} - \int_a^b s(x) \mathop{dx}
|
|
& = \sum_{k=1}^n f(x_k)\left[\frac{b - a}{n}\right] -
|
|
\sum_{k=1}^n f(x_{k-1})\left[\frac{b - a}{n}\right] \\
|
|
& = \left[\frac{b - a}{n}\right] \sum_{k=1}^n f(x_k) - f(x_{k-1}) \\
|
|
& = \frac{(b - a)(f(b) - f(a))}{n}.
|
|
\end{align*}
|
|
By \eqref{sub:theorem-1.12-eq1},
|
|
\begin{align*}
|
|
\ubar{I}(f)
|
|
& \leq \bar{I}(f) \\
|
|
& \leq \int_a^b t(x) \mathop{dx} \\
|
|
& = \int_a^b s(x) \mathop{dx} + \frac{(b - a)(f(b) - f(a))}{n} \\
|
|
& \leq \ubar{I}(f) + \frac{(b - a)(f(b) - f(a))}{n}.
|
|
\end{align*}
|
|
Since the above holds for all positive integers $n$,
|
|
\nameref{sec:theorem-i.31} indicates $\ubar{I}(f) = \bar{I}(f)$.
|
|
|
|
\paragraph{Case 2}%
|
|
|
|
Suppose $f$ is decreasing.
|
|
Let $s$ be the step function below $f$ with constant value $f(x_k)$
|
|
on every $k$th open subinterval of $P$.
|
|
Let $t$ be the step function above $f$ with constant value $f(x_{k-1})$
|
|
on every $k$th open subinterval of $P$.
|
|
Then, by \eqref{sub:theorem-1.9-eq1}, it follows
|
|
\begin{equation}
|
|
\label{sub:theorem-1.12-eq2}
|
|
\int_a^b s(x) \mathop{dx} \leq \ubar{I}(f)
|
|
\leq \bar{I}(f) \leq \int_a^b t(x) \mathop{dx}.
|
|
\end{equation}
|
|
By definition of the \nameref{ref:integral-step-function},
|
|
\begin{align*}
|
|
\int_a^b s(x) \mathop{dx}
|
|
& = \sum_{k=1}^n f(x_k)\left[\frac{b - a}{n}\right] \\
|
|
\int_a^b t(x) \mathop{dx}
|
|
& = \sum_{k=1}^n f(x_{k-1})\left[\frac{b - a}{n}\right].
|
|
\end{align*}
|
|
Thus
|
|
\begin{align*}
|
|
\int_a^b t(x) \mathop{dx} - \int_a^b s(x) \mathop{dx}
|
|
& = \sum_{k=1}^n f(x_{k-1})\left[\frac{b - a}{n}\right] -
|
|
\sum_{k=1}^n f(x_k)\left[\frac{b - a}{n}\right] \\
|
|
& = \left[\frac{b - a}{n}\right] \sum_{k=1}^n f(x_{k-1}) - f(x_k) \\
|
|
& = \frac{(b - a)(f(a) - f(b))}{n}.
|
|
\end{align*}
|
|
By \eqref{sub:theorem-1.12-eq2},
|
|
\begin{align*}
|
|
\ubar{I}(f)
|
|
& \leq \bar{I}(f) \\
|
|
& \leq \int_a^b t(x) \mathop{dx} \\
|
|
& = \int_a^b s(x) \mathop{dx} + \frac{(b - a)(f(a) - f(b))}{n} \\
|
|
& \leq \ubar{I}(f) + \frac{(b - a)(f(a) - f(b))}{n}.
|
|
\end{align*}
|
|
Since the above holds for all positive integers $n$,
|
|
\nameref{sec:theorem-i.31} indicates $\ubar{I}(f) = \bar{I}(f)$.
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\pending{Theorem 1.13}}%
|
|
\label{sub:theorem-1.13}
|
|
|
|
\begin{theorem}[1.13]
|
|
|
|
Assume $f$ is increasing on a closed interval $[a, b]$.
|
|
Let $x_k = a + k(b - a) / n$ for $k = 0, 1, \ldots, n$.
|
|
If $I$ is any number which satisfies the inequalities
|
|
\begin{equation}
|
|
\label{sub:theorem-1.13-eq1}
|
|
\frac{b - a}{n} \sum_{k=0}^{n-1} f(x_k)
|
|
\leq I \leq
|
|
\frac{b - a}{n} \sum_{k=1}^n f(x_k)
|
|
\end{equation}
|
|
for every integer $n \geq 1$, then $I = \int_a^b f(x) \mathop{dx}$.
|
|
|
|
\end{theorem}
|
|
|
|
\begin{proof}
|
|
|
|
Let $f$ be increasing on a closed interval $[a, b]$ and $I$ be a number
|
|
satisfying \eqref{sub:theorem-1.13-eq1}.
|
|
Let $s$ be the step function below $f$ with constant value $f(x_{k-1})$
|
|
on every $k$th open subinterval of $P$.
|
|
Let $t$ be the step function above $f$ with constant value $f(x_k)$
|
|
on every $k$th open subinterval of $P$.
|
|
By definition of the \nameref{ref:integral-step-function},
|
|
\begin{align*}
|
|
\int_a^b s(x) \mathop{dx}
|
|
& = \sum_{k=1}^n f(x_{k-1})\left[\frac{b - a}{n}\right] \\
|
|
& = \sum_{k=0}^{n-1} f(x_k)\left[\frac{b - a}{n}\right] \\
|
|
\int_a^b t(x) \mathop{dx}
|
|
& = \sum_{k=1}^n f(x_k)\left[\frac{b - a}{n}\right].
|
|
\end{align*}
|
|
Therefore \eqref{sub:theorem-1.13-eq1} can alternatively be written as
|
|
\begin{equation}
|
|
\label{sub:theorem-1.13-eq2}
|
|
\int_a^b s(x) \mathop{dx} \leq I \leq \int_a^b t(x) \mathop{dx}.
|
|
\end{equation}
|
|
By \nameref{sub:theorem-1.12}, $f$ is integrable.
|
|
Therefore \nameref{sub:theorem-1.9} indicates $f$ satisfies
|
|
\begin{equation}
|
|
\label{sub:theorem-1.13-eq3}
|
|
\int_a^b s(x) \mathop{dx}
|
|
\leq \int_a^b f(x) \mathop{dx}
|
|
\leq \int_a^b t(x) \mathop{dx}.
|
|
\end{equation}
|
|
Manipulating \eqref{sub:theorem-1.13-eq2} and \eqref{sub:theorem-1.13-eq3}
|
|
together yields
|
|
\begin{align*}
|
|
I - \int_a^b f(x) \mathop{dx}
|
|
& \leq \int_a^b t(x) \mathop{dx} - \int_a^b s(x) \mathop{dx}, \\
|
|
\int_a^b f(x) \mathop{dx} - I
|
|
& \leq \int_a^b t(x) \mathop{dx} - \int_a^b s(x) \mathop{dx}.
|
|
\end{align*}
|
|
Combining the above inequalities in turn yields
|
|
\begin{align*}
|
|
0
|
|
& \leq \abs{\int_a^b f(x) \mathop{dx} - I} \\
|
|
& \leq \int_a^b t(x) \mathop{dx} - \int_a^b s(x) \mathop{dx} \\
|
|
& = \sum_{k=1}^n f(x_k) \left[ \frac{b - a}{n} \right] -
|
|
\sum_{k=1}^n f(x_{k-1}) \left[ \frac{b - a}{n} \right] \\
|
|
& = \frac{b - a}{n} \sum_{k=1}^n f(x_k) - f(x_{k-1}) \\
|
|
& = \frac{(b - a)(f(b) - f(a))}{n}.
|
|
\end{align*}
|
|
The above chain of inequalities holds for all positive integers $n \geq 1$,
|
|
meaning \nameref{sec:theorem-i.31} applies.
|
|
Thus $$\abs{\int_a^b f(x) \mathop{dx} - I} = 0,$$ which immediately implies
|
|
the desired result.
|
|
|
|
\end{proof}
|
|
|
|
\subsection{\pending{Theorem 1.14}}%
|
|
\label{sub:theorem-1.14}
|
|
|
|
\begin{theorem}[1.14]
|
|
|
|
Assume $f$ is descreasing on $[a, b]$.
|
|
Let $x_k = a + k(b - a) / n$ for $k = 0, 1, \ldots, n$.
|
|
If $I$ is any number which satisfies the inequalities
|
|
\begin{equation}
|
|
\label{sub:theorem-1.14-eq1}
|
|
\frac{b - a}{n} \sum_{k=1}^n f(x_k)
|
|
\leq I \leq
|
|
\frac{b - a}{n} \sum_{k=0}^{n-1} f(x_k)
|
|
\end{equation}
|
|
for every integer $n \geq 1$, then $I = \int_a^b f(x) \mathop{dx}$.
|
|
|
|
\end{theorem}
|
|
|
|
\begin{proof}
|
|
|
|
Let $f$ be decreasing on a closed interval $[a, b]$ and $I$ be a number
|
|
satisfying \eqref{sub:theorem-1.14-eq1}.
|
|
Let $s$ be the step function below $f$ with constant value $f(x_k)$
|
|
on every $k$th open subinterval of $P$.
|
|
Let $t$ be the step function above $f$ with constant value $f(x_{k-1})$
|
|
on every $k$th open subinterval of $P$.
|
|
By definition of the \nameref{ref:integral-step-function},
|
|
\begin{align*}
|
|
\int_a^b s(x) \mathop{dx}
|
|
& = \sum_{k=1}^n f(x_k) \left[\frac{b - a}{n}\right] \\
|
|
\int_a^b t(x) \mathop{dx}
|
|
& = \sum_{k=1}^n f(x_{k-1}) \left[\frac{b - a}{n}\right] \\
|
|
& = \sum_{k=0}^{n-1} f(x) \left[\frac{b - a}{n}\right].
|
|
\end{align*}
|
|
Therefore \eqref{sub:theorem-1.14-eq1} can alternatively be written as
|
|
\begin{equation}
|
|
\label{sub:theorem-1.14-eq2}
|
|
\int_a^b s(x) \mathop{dx} \leq I \leq \int_a^b t(x) \mathop{dx}.
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\end{equation}
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By \nameref{sub:theorem-1.12}, $f$ is integrable.
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Therefore \nameref{sub:theorem-1.9} indicates $f$ satisfies
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\begin{equation}
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\label{sub:theorem-1.14-eq3}
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\int_a^b s(x) \mathop{dx}
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\leq \int_a^b f(x) \mathop{dx}
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\leq \int_a^b t(x) \mathop{dx}.
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|
\end{equation}
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|
Manipulating \eqref{sub:theorem-1.14-eq2} and \eqref{sub:theorem-1.14-eq3}
|
|
together yields
|
|
\begin{align*}
|
|
I - \int_a^b f(x) \mathop{dx}
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|
& \leq \int_a^b t(x) \mathop{dx} - \int_a^b s(x) \mathop{dx}, \\
|
|
\int_a^b f(x) \mathop{dx} - I
|
|
& \leq \int_a^b t(x) \mathop{dx} - \int_a^b s(x) \mathop{dx}.
|
|
\end{align*}
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|
Combining the above inequalities in turn yields
|
|
\begin{align*}
|
|
0
|
|
& \leq \abs{\int_a^b f(x) \mathop{dx} - I} \\
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& \leq \int_a^b t(x) \mathop{dx} - \int_a^b s(x) \mathop{dx} \\
|
|
& = \sum_{k=1}^n f(x_k) \left[ \frac{b - a}{n} \right] -
|
|
\sum_{k=1}^n f(x_{k-1}) \left[ \frac{b - a}{n} \right] \\
|
|
& = \frac{b - a}{n} \sum_{k=1}^n f(x_k) - f(x_{k-1}) \\
|
|
& = \frac{(b - a)(f(b) - f(a))}{n}.
|
|
\end{align*}
|
|
The above chain of inequalities holds for all positive integers $n \geq 1$,
|
|
meaning \nameref{sec:theorem-i.31} applies.
|
|
Thus $$\abs{\int_a^b f(x) \mathop{dx} - I} = 0,$$ which immediately implies
|
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the desired result.
|
|
|
|
\end{proof}
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|
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\subsection{\sorry{%
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Integral of \texorpdfstring{$\int_0^b x^p \mathop{dx}$}{int-x-p} when
|
|
\texorpdfstring{$p$}{p} is a Positive Integer}}%
|
|
\label{sub:calculation-integral-int-x-p-p-positive-integer}
|
|
\label{sub:theorem-1.15}
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|
|
|
\begin{theorem}[1.15]
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|
|
|
If $p$ is a positive integer and $b > 0$, we have
|
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$$\int_0^b x^p \mathop{dx} = \frac{b^{p+1}}{p+1}.$$
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|
|
\end{theorem}
|
|
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\begin{proof}
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TODO
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\end{proof}
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\end{document}
|