Finish Apostol 1.18-1.19.

finite-set-exercises
Joshua Potter 2023-05-16 16:14:45 -06:00
parent 7d68ab1624
commit 759b17f802
1 changed files with 93 additions and 11 deletions

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@ -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}