From 43dd9c2997a4a05bb3303ef327b8af3014808ec7 Mon Sep 17 00:00:00 2001 From: Joshua Potter Date: Mon, 15 May 2023 15:54:46 -0600 Subject: [PATCH] Add properties of the integral of a step function. --- Bookshelf/Apostol.tex | 279 +++++++++++++++++++++++++++++++++++++----- preamble.tex | 19 ++- 2 files changed, 259 insertions(+), 39 deletions(-) diff --git a/Bookshelf/Apostol.tex b/Bookshelf/Apostol.tex index 49cf68b..4225e87 100644 --- a/Bookshelf/Apostol.tex +++ b/Bookshelf/Apostol.tex @@ -31,7 +31,7 @@ The \textbf{characteristic function} of $S$ is the function $\mathcal{X}_S$ such \begin{definition} \lean{Common/Set/Basic}{Set.characteristic} - + \end{definition} \section{\defined{Infimum}}% @@ -51,10 +51,10 @@ Such a number $B$ is also known as the \textbf{greatest lower bound}. \end{definition} -\section{\partial{Integral of Step Function}}% +\section{\partial{Integral of a Step Function}}% \label{sec:def-integral-step-function} -Lset $s$ be a \nameref{sec:def-step-function} defined on $[a, b]$, and let +Let $s$ be a \nameref{sec:def-step-function} defined on $[a, b]$, and let $P = \{x_0, x_1, \ldots, x_n\}$ be a \nameref{sec:def-partition} of $[a, b]$ such that $s$ is constant on the open subintervals of $P$. Denote by $s_k$ the constant value that $s$ takes in the $k$th open subinterval, @@ -576,7 +576,7 @@ For each set $S$ in $\mathscr{M}$, we have $a(S) \geq 0$. \end{axiom} \section{\defined{Additive Property}}% -\label{sec:additive-property} +\label{sec: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 $a(S \cup T) = a(S) + a(T) - a(S \cap T)$. @@ -589,7 +589,7 @@ If $S$ and $T$ are in $\mathscr{M}$, then $S \cup T$ and $S \cap T$ are in \end{axiom} \section{\defined{Difference Property}}% -\label{sec:difference-property} +\label{sec: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 $a(T - S) = a(T) - a(S)$. @@ -602,7 +602,7 @@ If $S$ and $T$ are in $\mathscr{M}$ with $S \subseteq T$, then $T - S$ is in \end{axiom} \section{\defined{Invariance Under Congruence}}% -\label{sec:invariance-under-congruence} +\label{sec: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 $a(S) = a(T)$. @@ -615,7 +615,7 @@ If a set $S$ is in $\mathscr{M}$ and if $T$ is congruent to $S$, then $T$ is \end{axiom} \section{\defined{Choice of Scale}}% -\label{sec:choice-scale} +\label{sec:area-choice-scale} Every rectangle $R$ is in $\mathscr{M}$. If the edges of $R$ have lengths $h$ and $k$, then $a(R) = hk$. @@ -628,7 +628,7 @@ If the edges of $R$ have lengths $h$ and $k$, then $a(R) = hk$. \end{axiom} \section{\partial{Exhaustion Property}}% -\label{sec:exhaustion-property} +\label{sec:area-exhaustion-property} Let $Q$ be a set that can be enclosed between two step regions $S$ and $T$, so that @@ -664,8 +664,8 @@ A set consisting of a single point. 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{sec:choice-scale}, $S$ is measurable with area its width times - its height. + By \nameref{sec:area-choice-scale}, $S$ is measurable with area its width + times its height. The width and height of $S$ is trivially zero. Therefore $a(S) = (0)(0) = 0$. @@ -698,7 +698,7 @@ A set consisting of a finite number of points in a plane. By construction, $S_{k+1} = S_k \cup T$. By the induction hypothesis, $S_k$ is measurable with area $0$. By \nameref{sub:exercise-1.7.1a}, $T$ is measurable with area $0$. - By the \nameref{sec:additive-property}, $S_k \cup T$ is + By the \nameref{sec:area-additive-property}, $S_k \cup T$ is measurable, $S_k \cap T$ is measurable, and \begin{align} a(S_{k+1}) @@ -746,8 +746,8 @@ The union of a finite collection of line segments in a plane. 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{sec:choice-scale}, $S$ is measurable with area its width $w$ - times its height $h$. + By \nameref{sec:area-choice-scale}, $S$ is measurable with area its width + $w$ times its height $h$. Therefore $a(S) = wh = 0$. Thus $P(1)$ holds. @@ -761,7 +761,7 @@ The union of a finite collection of line segments in a plane. 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{sec:additive-property}, $S_k \cup T$ is measurable, + By the \nameref{sec:area-additive-property}, $S_k \cup T$ is measurable, $S_k \cap T$ is measurable, and \begin{align} a(S_{k+1}) @@ -823,10 +823,10 @@ Prove that every triangular region is measurable and that its area is one half \centering \end{figure} - By \nameref{sec:choice-scale}, both $R$ and $S$ are measurable. + By \nameref{sec: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{sec:additive-property}, $R \cup S$ and $R \cap S$ are both - measurable. + By the \nameref{sec: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 @@ -839,8 +839,8 @@ Prove that every triangular region is measurable and that its area is one half & = 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{sec:invariance-under-congruence}, $a(T') = a(T)$, concluding our - proof. + By \nameref{sec:area-invariance-under-congruence}, $a(T') = a(T)$, concluding + our proof. \end{proof} @@ -866,10 +866,10 @@ Prove that every trapezoid and every parallelogram is measurable and derive the 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{sec:choice-scale}, $R$ is measurable. + By \nameref{sec:area-choice-scale}, $R$ is measurable. By \nameref{sec:exercise-1.7.2}, $T$ is measurable. - By the \nameref{sec:additive-property}, $R \cup T$ and $R \cap T$ are both - measurable and + By the \nameref{sec:area-additive-property}, $R \cup T$ and $R \cap T$ are + both measurable and \begin{align*} a(S) & = a(R \cup T) \\ @@ -888,8 +888,8 @@ Prove that every trapezoid and every parallelogram is measurable and derive the Then $R$ has longer base edge of length $b_2 - c$. By \nameref{sec:exercise-1.7.2}, $T$ is measurable. By Case 1, $R$ is measurable. - By the \nameref{sec:additive-property}, $R \cup T$ and $R \cap T$ are both - measurable and + By the \nameref{sec: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) \\ @@ -906,7 +906,7 @@ Prove that every trapezoid and every parallelogram is measurable and derive the 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{sec:invariance-under-congruence}, + By \nameref{sec:area-invariance-under-congruence}, \begin{equation} \label{sec:exercise-1.7.3-eq1} \tag{3.1} @@ -933,7 +933,7 @@ Prove that every trapezoid and every parallelogram is measurable and derive the 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{sec:invariance-under-congruence}, + By \nameref{sec:area-invariance-under-congruence}, \begin{equation} \label{sec:exercise-1.7.3-eq2} \tag{3.2} @@ -970,7 +970,7 @@ Prove that the formula is valid for rectangles with sides parallel to the We assume $P$ has three non-collinear points, ruling out any instances of points or line segments. - By \nameref{sec:choice-scale}, $P$ is measurable with area $a(P) = wh$. + By \nameref{sec: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 @@ -1082,7 +1082,7 @@ Use induction on the number of edges to construct a proof for general polygons. & = 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{sec:additive-property}, $S \cup T$ is measurable, + By the \nameref{sec:area-additive-property}, $S \cup T$ is measurable, $S \cap T$ is measurable, and \begin{align*} a(P) @@ -1681,4 +1681,227 @@ This property is described by saying that every step function is a linear \end{proof} +\chapter{Properties of the Integral of a Step Function}% +\label{chap:properties-integral-step-function} + +\section{\partial{Additive Property}}% +\label{sec:step-additive-property} + +Let $s$ and $t$ be \nameref{sec:def-step-function}s on closed interval $[a, b]$. +Then + $$\int_a^b \left[ s(x) + t(x) \right] \mathop{dx} = + \int_a^b s(x) \mathop{dx} + \int_a^b t(x) \mathop{dx}.$$ + +\begin{proof} + + Let $s$ and $t$ be step functions on closed interval $[a, b]$. + By definition of a step function, there exists a \nameref{sec:def-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 value of $s$ on the $k$th open subinterval of $P$. + Let $t_k$ denote the value of $t$ on the $k$th open subinterval of $P$. + By definition of the \nameref{sec:def-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} + +\section{\partial{Homogeneous Property}}% +\label{sec:step-homogeneous-property} + +Let $s$ be a \nameref{sec:def-step-function} on closed interval $[a, b]$. +For every real number $c$, we have + $$\int_a^b c \cdot s(x) \mathop{dx} = c\int_a^b s(x) \mathop{dx}.$$ + +\begin{proof} + + Let $s$ be a step function on closed interval $[a, b]$. + By definition of a step function, there exists a \nameref{sec:def-partition} + $P = \{x_0, x_1, \ldots, x_n\}$ such that $s$ is constant on each open + subinterval of $P$. + Then $c \cdot s$ is a step function with step partition $P$. + + By definition of the \nameref{sec:def-integral-step-function}, + \begin{align*} + \int_a^b c \cdot s(x) \mathop{dx} + & = \sum_{k=1}^n (c \cdot s(x)) \cdot (x_k - x_{k-1}) \\ + & = \sum_{k=1}^n c \cdot (s(x) \cdot (x_k - x_{k-1})) \\ + & = c \sum_{k=1}^n s(x) \cdot (x_k - x_{k-1}) \\ + & = c \int_a^b s(x) \mathop{dx}. + \end{align*} + +\end{proof} + +\section{\partial{Linearity Property}}% +\label{sec:step-linearity-property} + +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 + $$\int_a^b \left[ c_1s(x) + c_2t(x) \right] \mathop{dx} = + c_1\int_a^b s(x) \mathop{dx} + c_2\int_a^b t(x) \mathop{dx}.$$ + +\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{sec:step-additive-property} \\ + & = c_1\int_a^b s(x) \mathop{dx} + c_2\int_a^b t(x). \mathop{dx} + & \textref{sec:step-homogeneous-property} + \end{align*} + +\end{proof} + +\section{\partial{Comparison Theorem}}% +\label{sec:step-comparison-theorem} + +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 + $$\int_a^b s(x) \mathop{dx} < \int_a^b t(x) \mathop{dx}.$$ + +\begin{proof} + + Let $s$ and $t$ be step functions on closed interval $[a, b]$. + By definition of a step function, there exists a \nameref{sec:def-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 value of $s$ on the $k$th open subinterval of $P$. + Let $t_k$ denote the value of $t$ on the $k$th open subinterval of $P$. + By definition of the \nameref{sec:def-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} + +\section{\partial{Additivity With Respect to the Interval of Integration}}% +\label{sec:step-additivity-with-respect-interval-integration} + +Let $s$ be a \nameref{sec:def-step-function} on closed interval $[a, b]$. +Then + $$\int_a^c s(x) \mathop{dx} + \int_c^b s(x) \mathop{dx} = + \int_a^b s(x) \mathop{dx} \quad\text{if}\quad a < c < b.$$ + +\todo{This holds for any arrangement of values $a$, $b$, and $c$.} + +\begin{proof} + + Let $s$ be a step function on closed interval $[a, b]$. + By definition of a step function, there exists a \nameref{sec:def-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 value of $s$ on the $k$th open subinterval of $Q$. + By definition of the \nameref{sec:def-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*} + +\end{proof} + +\section{\partial{Invariance Under Translation}}% +\label{sec:step-invariance-under-translation} + +Let $s$ be a step function on closed interval $[a, b]$. +Then + $$\int_a^b s(x) \mathop{dx} = + \int_{a+c}^{b+x} s(x - c) \mathop{dx} \quad\text{for every real } c.$$ + +\begin{proof} + + Let $s$ be a step function on closed interval $[a, b]$. + By definition of a step function, there exists a \nameref{sec:def-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 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{sec:def-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} + +\section{\partial{Expansion or Contraction of the Interval of Integration}}% +\label{sec:step-expansion-contraction-interval-integration} + +Let $s$ be a step function on closed interval $[a, b]$. +Then + $$\int_{ka}^{kb} s \left( \frac{x}{k} \right) \mathop{dx} = + k \int_a^b s(x) \mathop{dx} \quad\text{for every } k > 0.$$ + +\todo{This also holds for negative values of $k$.} + +\begin{proof} + + Let $s$ be a step function on closed interval $[a, b]$. + By definition of a step function, there exists a \nameref{sec:def-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 $k > 0$ be a real number. + 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, $t$ is constant on each open subinterval of $Q$. + Let $t_i$ denote the value of $t$ on the $i$th open subinterval of $Q$. + By construction, $t_i = s_i$. + + By definition of the \nameref{sec:def-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*} + +\end{proof} + \end{document} diff --git a/preamble.tex b/preamble.tex index 476ab0a..287de70 100644 --- a/preamble.tex +++ b/preamble.tex @@ -23,30 +23,27 @@ % Environments % ======================================== +\newcommand{\divider}{\vspace{10pt}\hrule\vspace{10pt}} +\newcommand{\header}[2]{\title{#1}\author{#2}\date{}\maketitle} + \newenvironment{axiom}{% \paragraph{\normalfont\normalsize\textit{Axiom.}}} {\hfill$\square$} \newenvironment{definition}{% \paragraph{\normalfont\normalsize\textit{Definition.}}} {\hfill$\square$} -\newcommand{\divider}{% - \vspace{10pt} - \hrule - \vspace{10pt}} -\newcommand{\header}[2]{% - \title{#1} - \author{#2} - \date{} - \maketitle} -\newcommand{\note}[1]{% + +\newcommand{\admonition}[2]{% \begin{center} \doublebox{ \begin{minipage}{0.95\textwidth} \vspace{2pt} - \hl{Note:} #1 + \hl{#1} #2 \vspace{2pt} \end{minipage}} \end{center}} +\newcommand{\note}[1]{\admonition{Note:}{#1}} +\newcommand{\todo}[1]{\admonition{TODO:}{#1}} % ======================================== % Status