Apostol 1.20, 21.

finite-set-exercises
Joshua Potter 2023-05-17 10:32:49 -06:00
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\newcommand{\lean}[2]{\leanref{../#1.html\##2}{#2}} \newcommand{\lean}[2]{\leanref{../#1.html\##2}{#2}}
\newcommand{\leanPretty}[3]{\leanref{../#1.html\##2}{#3}} \newcommand{\leanPretty}[3]{\leanref{../#1.html\##2}{#3}}
\newcommand{\ubar}[1]{\text{\b{$#1$}}} \newcommand{\ubar}[1]{\text{\b{$#1$}}}
\newcommand{\aliasref}[2]{\hyperref[#1]{#2}}
\begin{document} \begin{document}
@ -107,8 +105,8 @@ The \textbf{integral of $s$ from $a$ to $b$}, denoted by the symbol
If $a < b$, we define $\int_b^a s(x) \mathop{dx} = -\int_a^b s(x) \mathop{dx}$. If $a < b$, we define $\int_b^a s(x) \mathop{dx} = -\int_a^b s(x) \mathop{dx}$.
We also define $\int_a^a s(x) \mathop{dx} = 0$. We also define $\int_a^a s(x) \mathop{dx} = 0$.
\section{\partial{Lower Integral of \texorpdfstring{$f$}{f}}}% \section{\partial{Lower Integral}}%
\label{sec:def-lower-integral-f} \label{sec:def-lower-integral}
Let $f$ be a function bounded on $[a, b]$ and $S$ denote the set of numbers Let $f$ be a function bounded on $[a, b]$ and $S$ denote the set of numbers
$\int_a^b s(x) \mathop{dx}$ obtained as $s$ runs through all $\int_a^b s(x) \mathop{dx}$ obtained as $s$ runs through all
@ -117,6 +115,20 @@ That is, let $$S = \left\{ \int_a^b s(x) \mathop{dx} : s \leq f \right\}.$$
The number $\sup{S}$ is called the \textbf{lower integral of $f$}. The number $\sup{S}$ is called the \textbf{lower integral of $f$}.
It is denoted as $\ubar{I}(f)$. It is denoted as $\ubar{I}(f)$.
\section{\partial{Monotonic}}%
\label{sec:def-monotonic}
A function $f$ is called \textbf{monotonic} on set $S$ if it is increasing on
$S$ or if it is decreasing on $S$.
$f$ is said to be \textbf{strictly monotonic} if it is strictly increasing on
$S$ or strictly decreasing on $S$.
A function $f$ is said to be \textbf{piecewise monotonic} on an interval if its
graph consists of a finite number of monotonic pieces.
In other words, $f$ is piecewise monotonic on $[a, b]$ if there is a
\nameref{sec:def-partition} of $[a, b]$ such that $f$ is monotonic on each of
the open subintervals of $P$.
\section{\defined{Partition}}% \section{\defined{Partition}}%
\label{sec:def-partition} \label{sec:def-partition}
@ -188,8 +200,8 @@ Such a number $B$ is also known as the \textbf{least upper bound}.
\end{definition} \end{definition}
\section{\partial{Upper Integral of \texorpdfstring{$f$}{f}}}% \section{\partial{Upper Integral}}%
\label{sec:def-upper-integral-f} \label{sec:def-upper-integral}
Let $f$ be a function bounded on $[a, b]$ and $T$ denote the set of numbers Let $f$ be a function bounded on $[a, b]$ and $T$ denote the set of numbers
$\int_a^b t(x) \mathop{dx}$ obtained as $t$ runs through all $\int_a^b t(x) \mathop{dx}$ obtained as $t$ runs through all
@ -216,7 +228,11 @@ Every nonempty set $S$ of real numbers which is bounded above has a supremum;
\section{\verified{Lemma 1}}% \section{\verified{Lemma 1}}%
\label{sec:lemma-1} \label{sec:lemma-1}
Nonempty set $S$ has supremum $L$ if and only if set $-S$ has infimum $-L$. \begin{lemma}{1}
Nonempty set $S$ has supremum $L$ if and only if set $-S$ has infimum $-L$.
\end{lemma}
\begin{proof} \begin{proof}
@ -237,9 +253,13 @@ Nonempty set $S$ has supremum $L$ if and only if set $-S$ has infimum $-L$.
\section{\verified{Theorem I.27}}% \section{\verified{Theorem I.27}}%
\label{sec:theorem-i.27} \label{sec:theorem-i.27}
Every nonempty set $S$ that is bounded below has a greatest lower bound; that \begin{theorem}{I.27}
Every nonempty set $S$ that is bounded below has a greatest lower bound; that
is, there is a real number $L$ such that $L = \inf{S}$. is, there is a real number $L$ such that $L = \inf{S}$.
\end{theorem}
\begin{proof} \begin{proof}
\lean{Bookshelf/Apostol/Chapter\_I\_03} \lean{Bookshelf/Apostol/Chapter\_I\_03}
@ -259,7 +279,11 @@ Every nonempty set $S$ that is bounded below has a greatest lower bound; that
\section{\verified{Theorem I.29}}% \section{\verified{Theorem I.29}}%
\label{sec:theorem-i.29} \label{sec:theorem-i.29}
For every real $x$ there exists a positive integer $n$ such that $n > x$. \begin{theorem}{I.29}
For every real $x$ there exists a positive integer $n$ such that $n > x$.
\end{theorem}
\begin{proof} \begin{proof}
@ -279,11 +303,14 @@ For every real $x$ there exists a positive integer $n$ such that $n > x$.
\section{\verified{Archimedean Property of the Reals}}% \section{\verified{Archimedean Property of the Reals}}%
\label{sec:archimedean-property-reals} \label{sec:archimedean-property-reals}
\label{sec:theorem-i.30}
If $x > 0$ and if $y$ is an arbitrary real number, there exists a positive \begin{theorem}{I.30}
If $x > 0$ and if $y$ is an arbitrary real number, there exists a positive
integer $n$ such that $nx > y$. integer $n$ such that $nx > y$.
\note{This is known as the "Archimedean Property of the Reals."} \end{theorem}
\begin{proof} \begin{proof}
@ -302,8 +329,13 @@ If $x > 0$ and if $y$ is an arbitrary real number, there exists a positive
\section{\verified{Theorem I.31}}% \section{\verified{Theorem I.31}}%
\label{sec:theorem-i.31} \label{sec:theorem-i.31}
If three real numbers $a$, $x$, and $y$ satisfy the inequalities \begin{theorem}{I.31}
$$a \leq x \leq a + \frac{y}{n}$$ for every integer $n \geq 1$, then $x = a$.
If three real numbers $a$, $x$, and $y$ satisfy the inequalities
$$a \leq x \leq a + \frac{y}{n}$$ for every integer $n \geq 1$, then
$x = a$.
\end{theorem}
\begin{proof} \begin{proof}
@ -346,9 +378,13 @@ If three real numbers $a$, $x$, and $y$ satisfy the inequalities
\section{\verified{Lemma 2}}% \section{\verified{Lemma 2}}%
\label{sec:lemma-2} \label{sec:lemma-2}
If three real numbers $a$, $x$, and $y$ satisfy the inequalities \begin{lemma}{2}
If three real numbers $a$, $x$, and $y$ satisfy the inequalities
$$a - y / n \leq x \leq a$$ for every integer $n \geq 1$, then $x = a$. $$a - y / n \leq x \leq a$$ for every integer $n \geq 1$, then $x = a$.
\end{lemma}
\begin{proof} \begin{proof}
\lean{Bookshelf/Apostol/Chapter\_I\_03} \lean{Bookshelf/Apostol/Chapter\_I\_03}
@ -395,7 +431,11 @@ Let $h$ be a given positive number and let $S$ be a set of real numbers.
\subsection{\verified{Theorem I.32a}}% \subsection{\verified{Theorem I.32a}}%
\label{sub:theorem-i.32a} \label{sub:theorem-i.32a}
If $S$ has a supremum, then for some $x$ in $S$ we have $x > \sup{S} - h$. \begin{theorem}{I.32a}
If $S$ has a supremum, then for some $x$ in $S$ we have $x > \sup{S} - h$.
\end{theorem}
\begin{proof} \begin{proof}
@ -419,7 +459,11 @@ If $S$ has a supremum, then for some $x$ in $S$ we have $x > \sup{S} - h$.
\subsection{\verified{Theorem I.32b}}% \subsection{\verified{Theorem I.32b}}%
\label{sub:theorem-i.32b} \label{sub:theorem-i.32b}
If $S$ has an infimum, then for some $x$ in $S$ we have $x < \inf{S} + h$. \begin{theorem}{I.32b}
If $S$ has an infimum, then for some $x$ in $S$ we have $x < \inf{S} + h$.
\end{theorem}
\begin{proof} \begin{proof}
@ -451,9 +495,13 @@ Given nonempty subsets $A$ and $B$ of $\mathbb{R}$, let $C$ denote the set
\subsection{\verified{Theorem I.33a}}% \subsection{\verified{Theorem I.33a}}%
\label{sub:theorem-i.33a} \label{sub:theorem-i.33a}
If each of $A$ and $B$ has a supremum, then $C$ has a supremum, and \begin{theorem}{I.33a}
If each of $A$ and $B$ has a supremum, then $C$ has a supremum, and
$$\sup{C} = \sup{A} + \sup{B}.$$ $$\sup{C} = \sup{A} + \sup{B}.$$
\end{theorem}
\begin{proof} \begin{proof}
\lean{Bookshelf/Apostol/Chapter\_I\_03} \lean{Bookshelf/Apostol/Chapter\_I\_03}
@ -520,9 +568,13 @@ If each of $A$ and $B$ has a supremum, then $C$ has a supremum, and
\subsection{\verified{Theorem I.33b}}% \subsection{\verified{Theorem I.33b}}%
\label{sub:theorem-i.33b} \label{sub:theorem-i.33b}
If each of $A$ and $B$ has an infimum, then $C$ has an infimum, and \begin{theorem}{I.33b}
If each of $A$ and $B$ has an infimum, then $C$ has an infimum, and
$$\inf{C} = \inf{A} + \inf{B}.$$ $$\inf{C} = \inf{A} + \inf{B}.$$
\end{theorem}
\begin{proof} \begin{proof}
\lean{Bookshelf/Apostol/Chapter\_I\_03} \lean{Bookshelf/Apostol/Chapter\_I\_03}
@ -589,10 +641,14 @@ If each of $A$ and $B$ has an infimum, then $C$ has an infimum, and
\section{\verified{Theorem I.34}}% \section{\verified{Theorem I.34}}%
\label{sec:theorem-i.34} \label{sec:theorem-i.34}
Given two nonempty subsets $S$ and $T$ of $\mathbb{R}$ such that $$s \leq t$$ \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 $T$ 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 $$\sup{S} \leq \inf{T}.$$ has an infimum, and they satisfy the inequality $$\sup{S} \leq \inf{T}.$$
\end{theorem}
\begin{proof} \begin{proof}
\lean{Bookshelf/Apostol/Chapter\_I\_03} \lean{Bookshelf/Apostol/Chapter\_I\_03}
@ -1749,12 +1805,18 @@ This property is described by saying that every step function is a linear
\section{\partial{Additive Property}}% \section{\partial{Additive Property}}%
\label{sec:step-additive-property} \label{sec:step-additive-property}
\label{sec:theorem-1.2}
Let $s$ and $t$ be \nameref{sec:def-step-function}s on closed interval $[a, b]$. \begin{theorem}{1.2}
Then
Let $s$ and $t$ be \nameref{sec:def-step-function}s on closed interval
$[a, b]$.
Then
$$\int_a^b \left[ s(x) + t(x) \right] \mathop{dx} = $$\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}.$$ \int_a^b s(x) \mathop{dx} + \int_a^b t(x) \mathop{dx}.$$
\end{theorem}
\begin{proof} \begin{proof}
Let $s$ and $t$ be step functions on closed interval $[a, b]$. Let $s$ and $t$ be step functions on closed interval $[a, b]$.
@ -1786,11 +1848,16 @@ Then
\section{\partial{Homogeneous Property}}% \section{\partial{Homogeneous Property}}%
\label{sec:step-homogeneous-property} \label{sec:step-homogeneous-property}
\label{sec:theorem-1.3}
Let $s$ be a \nameref{sec:def-step-function} on closed interval $[a, b]$. \begin{theorem}{1.3}
For every real number $c$, we have
Let $s$ be a \nameref{sec:def-step-function} on closed interval $[a, b]$.
For every real number $c$, we have
$$\int_a^b c \cdot s(x) \mathop{dx} = c\int_a^b s(x) \mathop{dx}.$$ $$\int_a^b c \cdot s(x) \mathop{dx} = c\int_a^b s(x) \mathop{dx}.$$
\end{theorem}
\begin{proof} \begin{proof}
Let $s$ be a step function on closed interval $[a, b]$. Let $s$ be a step function on closed interval $[a, b]$.
@ -1812,12 +1879,18 @@ For every real number $c$, we have
\section{\partial{Linearity Property}}% \section{\partial{Linearity Property}}%
\label{sec:step-linearity-property} \label{sec:step-linearity-property}
\label{sec:theorem-1.4}
Let $s$ and $t$ be \nameref{sec:def-step-function}s on closed interval $[a, b]$. \begin{theorem}{1.4}
For every real $c_1$ and $c_2$, we have
Let $s$ and $t$ be \nameref{sec:def-step-function}s on closed interval
$[a, b]$.
For every real $c_1$ and $c_2$, we have
$$\int_a^b \left[ c_1s(x) + c_2t(x) \right] \mathop{dx} = $$\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}.$$ c_1\int_a^b s(x) \mathop{dx} + c_2\int_a^b t(x) \mathop{dx}.$$
\end{theorem}
\begin{proof} \begin{proof}
Let $s$ and $t$ be step functions on closed interval $[a, b]$. Let $s$ and $t$ be step functions on closed interval $[a, b]$.
@ -1836,11 +1909,17 @@ For every real $c_1$ and $c_2$, we have
\section{\partial{Comparison Theorem}}% \section{\partial{Comparison Theorem}}%
\label{sec:step-comparison-theorem} \label{sec:step-comparison-theorem}
\label{sec:theorem-1.5}
Let $s$ and $t$ be \nameref{sec:def-step-function}s on closed interval $[a, b]$. \begin{theorem}{1.5}
If $s(x) < t(x)$ for every $x$ in $[a, b]$, then
Let $s$ and $t$ be \nameref{sec:def-step-function}s on closed interval
$[a, b]$.
If $s(x) < t(x)$ for every $x$ in $[a, b]$, then
$$\int_a^b s(x) \mathop{dx} < \int_a^b t(x) \mathop{dx}.$$ $$\int_a^b s(x) \mathop{dx} < \int_a^b t(x) \mathop{dx}.$$
\end{theorem}
\begin{proof} \begin{proof}
Let $s$ and $t$ be step functions on closed interval $[a, b]$. Let $s$ and $t$ be step functions on closed interval $[a, b]$.
@ -1869,13 +1948,18 @@ If $s(x) < t(x)$ for every $x$ in $[a, b]$, then
\section{\partial{Additivity With Respect to the Interval of Integration}}% \section{\partial{Additivity With Respect to the Interval of Integration}}%
\label{sec:step-additivity-with-respect-interval-integration} \label{sec:step-additivity-with-respect-interval-integration}
\label{sec:theorem-1.6}
Let $a, b, c \in \mathbb{R}$ and $s$ a \nameref{sec:def-step-function} on the \begin{theorem}{1.6}
Let $a, b, c \in \mathbb{R}$ and $s$ a \nameref{sec:def-step-function} on the
smallest closed interval containing them. smallest closed interval containing them.
Then Then
$$\int_a^c s(x) \mathop{dx} + \int_c^b s(x) \mathop{dx} + $$\int_a^c s(x) \mathop{dx} + \int_c^b s(x) \mathop{dx} +
\int_b^a s(x) \mathop{dx} = 0.$$ \int_b^a s(x) \mathop{dx} = 0.$$
\end{theorem}
\begin{proof} \begin{proof}
WLOG, suppose $a < c < b$ and $s$ be a step function on closed interval WLOG, suppose $a < c < b$ and $s$ be a step function on closed interval
@ -1910,12 +1994,17 @@ Then
\section{\partial{Invariance Under Translation}}% \section{\partial{Invariance Under Translation}}%
\label{sec:step-invariance-under-translation} \label{sec:step-invariance-under-translation}
\label{sec:theorem-1.7}
Let $s$ be a step function on closed interval $[a, b]$. \begin{theorem}{1.7}
Then
Let $s$ be a step function on closed interval $[a, b]$.
Then
$$\int_a^b s(x) \mathop{dx} = $$\int_a^b s(x) \mathop{dx} =
\int_{a+c}^{b+x} s(x - c) \mathop{dx} \quad\text{for every real } c.$$ \int_{a+c}^{b+x} s(x - c) \mathop{dx} \quad\text{for every real } c.$$
\end{theorem}
\begin{proof} \begin{proof}
Let $s$ be a step function on closed interval $[a, b]$. Let $s$ be a step function on closed interval $[a, b]$.
@ -1946,12 +2035,17 @@ Then
\section{\partial{Expansion or Contraction of the Interval of Integration}}% \section{\partial{Expansion or Contraction of the Interval of Integration}}%
\label{sec:step-expansion-contraction-interval-integration} \label{sec:step-expansion-contraction-interval-integration}
\label{sec:theorem-1.8}
Let $s$ be a step function on closed interval $[a, b]$. \begin{theorem}{1.8}
Then
Let $s$ be a step function on closed interval $[a, b]$.
Then
$$\int_{ka}^{kb} s \left( \frac{x}{k} \right) \mathop{dx} = $$\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.$$ k \int_a^b s(x) \mathop{dx} \quad\text{for every } k \neq 0.$$
\end{theorem}
\begin{proof} \begin{proof}
Let $s$ be a step function on closed interval $[a, b]$. Let $s$ be a step function on closed interval $[a, b]$.
@ -2511,13 +2605,15 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
\end{proof} \end{proof}
\chapter{Upper and Lower Integrals}% \chapter{Theorey of Integrability}%
\label{chap:upper-lower-integrals} \label{chap:theory-integrability}
\section{\partial{Theorem 1.9}}% \section{\partial{Theorem 1.9}}%
\label{sec:theorem-1.9} \label{sec:theorem-1.9}
Every function $f$ which is bounded on $[a, b]$ has a lower integral \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 $\ubar{I}(f)$ and an upper integral $\overline{I}(f)$ satisfying the
inequalities inequalities
\begin{equation} \begin{equation}
@ -2526,10 +2622,12 @@ Every function $f$ which is bounded on $[a, b]$ has a lower integral
\bar{I}(f) \leq \int_a^b t(x) \mathop{dx} \bar{I}(f) \leq \int_a^b t(x) \mathop{dx}
\end{equation} \end{equation}
for all \nameref{sec:def-step-function}s $s$ and $t$ with $s \leq f \leq t$. for all \nameref{sec:def-step-function}s $s$ and $t$ with $s \leq f \leq t$.
The function $f$ is \nameref{sec:def-integrable} on $[a, b]$ if and only if The function $f$ is \nameref{sec:def-integrable} on $[a, b]$ if and only if
its upper and lower integrals are equal, in which case we have its upper and lower integrals are equal, in which case we have
$$\int_a^b f(x) \mathop{dx} = \ubar{I}(f) = \bar{I}(f).$$ $$\int_a^b f(x) \mathop{dx} = \ubar{I}(f) = \bar{I}(f).$$
\end{theorem}
\begin{proof} \begin{proof}
Let $f$ be a function bounded on $[a, b]$. Let $f$ be a function bounded on $[a, b]$.
@ -2556,9 +2654,9 @@ The function $f$ is \nameref{sec:def-integrable} on $[a, b]$ if and only if
Therefore \nameref{sec:theorem-i.34} tells us $S$ has a Therefore \nameref{sec:theorem-i.34} tells us $S$ has a
\nameref{sec:def-supremum}, $T$ has an \nameref{sec:def-infimum}, and \nameref{sec:def-supremum}, $T$ has an \nameref{sec:def-infimum}, and
$\sup{S} \leq \inf{T}$. $\sup{S} \leq \inf{T}$.
By definition of the \nameref{sec:def-lower-integral-f}, By definition of the \nameref{sec:def-lower-integral},
$\ubar{I}(f) = \sup{S}$. $\ubar{I}(f) = \sup{S}$.
By definition of the \nameref{sec:def-upper-integral-f}, By definition of the \nameref{sec:def-upper-integral},
$\bar{I}(f) = \inf{S}$. $\bar{I}(f) = \inf{S}$.
Thus \eqref{sec:theorem-1.9-eq1} holds. Thus \eqref{sec:theorem-1.9-eq1} holds.
@ -2575,17 +2673,19 @@ The function $f$ is \nameref{sec:def-integrable} on $[a, b]$ if and only if
\end{proof} \end{proof}
\chapter{The Area of an Ordinate Set Expressed as an Interval}% \section{\partial{Measurability of Ordinate Sets}}%
\label{chap:area-ordinate-set-expressed-interval} \label{sec:measurability-ordinate-sets}
\section{\partial{Theorem 1.10}}%
\label{sec:theorem-1.10} \label{sec:theorem-1.10}
Let $f$ be a nonnegative function, \nameref{sec:def-integrable} on an interval \begin{theorem}{1.10}
Let $f$ be a nonnegative function, \nameref{sec:def-integrable} on an interval
$[a, b]$, and let $Q$ denote the ordinate set of $f$ over $[a, b]$. $[a, b]$, and let $Q$ denote the ordinate set of $f$ over $[a, b]$.
Then $Q$ is measurable and its area is equal to the integral Then $Q$ is measurable and its area is equal to the integral
$\int_a^b f(x) \mathop{dx}$. $\int_a^b f(x) \mathop{dx}$.
\end{theorem}
\begin{proof} \begin{proof}
Let $f$ be a nonnegative function, \nameref{sec:def-integrable} on $[a, b]$. Let $f$ be a nonnegative function, \nameref{sec:def-integrable} on $[a, b]$.
@ -2601,17 +2701,22 @@ Then $Q$ is measurable and its area is equal to the integral
\end{proof} \end{proof}
\section{\partial{Theorem 1.11}}% \section{\partial{Measurability of the Graph of a Nonnegative Function}}%
\label{sec:measurability-graph-nonnegative-function}
\label{sec:theorem-1.11} \label{sec:theorem-1.11}
Let $f$ be a nonnegative function, integrable on an interval $[a, b]$. \begin{theorem}{1.11}
Then the graph of $f$, that is, the set
Let $f$ be a nonnegative function, integrable on an interval $[a, b]$.
Then the graph of $f$, that is, the set
\begin{equation} \begin{equation}
\label{sec:theorem-1.11-eq1} \label{sec:measurability-graph-nonnegative-function-eq1}
\{(x, y) | a \leq x \leq b, y = f(x)\}, \{(x, y) \mid a \leq x \leq b, y = f(x)\},
\end{equation} \end{equation}
is measurable and has area equal to $0$. is measurable and has area equal to $0$.
\end{theorem}
\begin{proof} \begin{proof}
Let $f$ be a nonnegative function, integrable on an interval $[a, b]$. Let $f$ be a nonnegative function, integrable on an interval $[a, b]$.
@ -2621,7 +2726,7 @@ Then the graph of $f$, that is, the set
equal to $0$. equal to $0$.
\paragraph{(i)}% \paragraph{(i)}%
\label{par:theorem-1.11-i} \label{par:measurability-graph-nonnegative-function-i}
By definition of integrability, there exists one and only one number $I$ By definition of integrability, there exists one and only one number $I$
such that such that
@ -2637,11 +2742,12 @@ Then the graph of $f$, that is, the set
\paragraph{(ii)}% \paragraph{(ii)}%
Let $Q$ denote the ordinate set of $f$. Let $Q$ denote the ordinate set of $f$.
By \nameref{sec:theorem-1.11}, $Q$ is measurable with area equal to the By \nameref{sec:measurability-ordinate-sets}, $Q$ is measurable with area
integral $I = \int_a^b f(x) \mathop{dx}$. 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 By \nameref{par:measurability-graph-nonnegative-function-i}, $Q'$ is
$I$. measurable with area also equal to $I$.
We note the graph of $f$, \eqref{sec:theorem-1.11-eq1}, is equal to set We note the graph of $f$,
\eqref{sec:measurability-graph-nonnegative-function-eq1}, is equal to set
$Q - Q'$. $Q - Q'$.
By the \nameref{sec:area-difference-property}, $Q - Q'$ is measurable and By the \nameref{sec:area-difference-property}, $Q - Q'$ is measurable and
$$a(Q - Q') = a(Q) - a(Q') = I - I = 0.$$ $$a(Q - Q') = a(Q) - a(Q') = I - I = 0.$$
@ -2649,4 +2755,134 @@ Then the graph of $f$, that is, the set
\end{proof} \end{proof}
\section
[\partial{Integrability of Bounded Monotonic Functions}]
{\partial{Integrability of Bounded Monotonic \texorpdfstring{\\}{}Functions}}
\label{sec:integrability-bounded-monotonic-functions}
\label{sec:theorem-1.12}
\begin{theorem}{1.12}
If $f$ is \nameref{sec:def-monotonic} on a closed interval $[a, b]$, then $f$
is \nameref{sec:def-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{sec:theorem-1.9}, $f$ has a \nameref{sec:def-lower-integral}
$\ubar{I}(f)$, $f$ has an \nameref{sec:def-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{sec:theorem-1.9-eq1}, it follows
\begin{equation}
\label{sec:integrability-bounded-monotonic-functions-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{sec:def-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{sec:integrability-bounded-monotonic-functions-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{sec:theorem-1.9-eq1}, it follows
\begin{equation}
\label{sec:integrability-bounded-monotonic-functions-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{sec:def-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{sec:integrability-bounded-monotonic-functions-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}
\section{\unverified{%
Calculation of the Integral of a Bounded Monotonic Function}}%
\label{sec:calculation-integral-bounded-monotonic-function}
\label{sec: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
$$\frac{b - a}{n} \sum_{k=0}^{n-1} f(x_k)
\leq I \leq
\frac{b - a}{n} \sum_{k=1}^n f(x_k)$$
for every integer $n \geq 1$, then $I = \int_a^b f(x) \mathop{dx}$.
\end{theorem}
\begin{proof}
TODO
\end{proof}
\end{document} \end{document}

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@ -26,13 +26,7 @@
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