From 759b17f8027caded269e1cef1940b877714eecef Mon Sep 17 00:00:00 2001 From: Joshua Potter Date: Tue, 16 May 2023 16:14:45 -0600 Subject: [PATCH] Finish Apostol 1.18-1.19. --- Bookshelf/Apostol.tex | 104 +++++++++++++++++++++++++++++++++++++----- 1 file changed, 93 insertions(+), 11 deletions(-) diff --git a/Bookshelf/Apostol.tex b/Bookshelf/Apostol.tex index 9d78db4..27122aa 100644 --- a/Bookshelf/Apostol.tex +++ b/Bookshelf/Apostol.tex @@ -54,6 +54,15 @@ Such a number $B$ is also known as the \textbf{greatest lower bound}. \end{definition} +\section{\partial{Integrable}}% +\label{sec:def-integrable} + +Let $f$ be a function defined and bounded on $[a, b]$. +$f$ is said to be \textbf{integrable} if there exists one and only one number + $I$ such that \eqref{sec:def-integral-bounded-function-eq2} holds. +If $f$ is integrable on $[a, b]$, we say that the integral + $\int_a^b f(x) \mathop{dx}$ \textbf{exists}. + \section{\partial{Integral of a Bounded Function}}% \label{sec:def-integral-bounded-function} @@ -65,19 +74,18 @@ Let $s$ and $t$ denote arbitrary step functions defined on $[a, b]$ such that \end{equation} for every $x$ in $[a, b]$. If there is one and only one number $I$ such that - $$\int_a^b s(x) \mathop{dx} \leq I \leq \int_a^b t(x) \mathop{dx}$$ + \begin{equation} + \label{sec:def-integral-bounded-function-eq2} + \int_a^b s(x) \mathop{dx} \leq I \leq \int_a^b t(x) \mathop{dx} + \end{equation} for every pair of step functions $s$ and $t$ satisfying \eqref{sec:def-integral-bounded-function-eq1}, then this number $I$ is called the \textbf{integral of $f$ from $a$ to $b$}, and is denoted by the symbol $\int_a^b f(x) \mathop{dx}$ or by $\int_a^b f$. -When such an $I$ exists, the function $f$ is said to be - \textbf{integrable on $[a, b]$}. If $a < b$, we define $\int_b^a f(x) \mathop{dx} = -\int_a^b f(x) \mathop{dx}$, - provided $f$ is integrable on $[a, b]$. + provided $f$ is \nameref{sec:def-integrable} on $[a, b]$. We also define $\int_a^a f(x) \mathop{dx} = 0$. -If $f$ is integrable on $[a, b]$, we say that the integral - $\int_a^b f(x) \mathop{dx}$ \textbf{exists}. The function $f$ is called the \textbf{integrand}, the numbers $a$ and $b$ are called the \textbf{limits of integration}, and the interval $[a, b]$ the @@ -692,8 +700,8 @@ Let $Q$ be a set that can be enclosed between two step regions $S$ and $T$, so 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 (1.1), - then $Q$ is measurable and $a(Q) = c$. + $$a(S) \leq c \leq a(T)$$ for all step regions $S$ and $T$ satisfying + \eqref{sec:exhaustion-property-eq1}, then $Q$ is measurable and $a(Q) = c$. \begin{axiom} @@ -2506,7 +2514,7 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$. \chapter{Upper and Lower Integrals}% \label{chap:upper-lower-integrals} -\section{\unverified{Theorem 1.9}}% +\section{\partial{Theorem 1.9}}% \label{sec:theorem-1.9} Every function $f$ which is bounded on $[a, b]$ has a lower integral @@ -2518,8 +2526,8 @@ Every function $f$ which is bounded on $[a, b]$ has a lower integral \bar{I}(f) \leq \int_a^b t(x) \mathop{dx} \end{equation} for all \nameref{sec:def-step-function}s $s$ and $t$ with $s \leq f \leq t$. -The function $f$ is integrable on $[a, b]$ if and only if its upper and lower - integrals are equal, in which case we have +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 $$\int_a^b f(x) \mathop{dx} = \ubar{I}(f) = \bar{I}(f).$$ \begin{proof} @@ -2567,4 +2575,78 @@ The function $f$ is integrable on $[a, b]$ if and only if its upper and lower \end{proof} +\chapter{The Area of an Ordinate Set Expressed as an Interval}% +\label{chap:area-ordinate-set-expressed-interval} + +\section{Theorem 1.10}% +\label{sec: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 $[a, b]$. +Then $Q$ is measurable and its area is equal to the integral + $\int_a^b f(x) \mathop{dx}$. + +\begin{proof} + + Let $f$ be a nonnegative function, \nameref{sec:def-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{sec:def-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{sec:area-exhaustion-property}, $Q$ is measurable and its area + is equal to $I = \int_a^b f(x) \mathop{dx}$. + +\end{proof} + +\section{Theorem 1.11}% +\label{sec: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{sec:theorem-1.11-eq1} + \{(x, y) | a \leq x \leq b, y = f(x)\}, + \end{equation} + is measurable and has area equal to $0$. + +\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{sec:def-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{sec: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{sec:theorem-1.11}, $Q$ 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{sec:theorem-1.11-eq1}, is equal to set + $Q - Q'$. + By the \nameref{sec: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} + \end{document}