Finish Apostol 1.22.

finite-set-exercises
Joshua Potter 2023-05-17 15:38:16 -06:00
parent 9ac70c15c9
commit dc4932e0e7
1 changed files with 158 additions and 15 deletions

View File

@ -2670,7 +2670,7 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
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:measurability-graph-nonnegative-function-eq1}
\label{sec:theorem-1.11-eq1}
\{(x, y) \mid a \leq x \leq b, y = f(x)\},
\end{equation}
is measurable and has area equal to $0$.
@ -2686,7 +2686,7 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
equal to $0$.
\paragraph{(i)}%
\label{par:measurability-graph-nonnegative-function-i}
\label{par:theorem-1.11-i}
By definition of integrability, there exists one and only one number $I$
such that
@ -2704,10 +2704,9 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
Let $Q$ denote the ordinate set of $f$.
By \nameref{sec:measurability-ordinate-sets}, $Q$ is measurable with area
equal to the integral $I = \int_a^b f(x) \mathop{dx}$.
By \nameref{par:measurability-graph-nonnegative-function-i}, $Q'$ is
By \nameref{par:theorem-1.11-i}, $Q'$ is
measurable with area also equal to $I$.
We note the graph of $f$,
\eqref{sec:measurability-graph-nonnegative-function-eq1}, is equal to set
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
$$a(Q - Q') = a(Q) - a(Q') = I - I = 0.$$
@ -2751,7 +2750,7 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
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}
\label{sec: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}
@ -2770,7 +2769,7 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
& = \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},
By \eqref{sec:theorem-1.12-eq1},
\begin{align*}
\ubar{I}(f)
& \leq \bar{I}(f) \\
@ -2790,7 +2789,7 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
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}
\label{sec: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}
@ -2809,7 +2808,7 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
& = \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},
By \eqref{sec:theorem-1.12-eq2},
\begin{align*}
\ubar{I}(f)
& \leq \bar{I}(f) \\
@ -2822,9 +2821,9 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
\end{proof}
\section{\unverified{%
Calculation of the Integral of a Bounded Monotonic Function}}%
\label{sec:calculation-integral-bounded-monotonic-function}
\section{\partial{%
Calculation of the Integral of a Bounded Increasing Function}}%
\label{sec:calculation-integral-bounded-increasing-function}
\label{sec:theorem-1.13}
\begin{theorem}[1.13]
@ -2832,13 +2831,157 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
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
$$\frac{b - a}{n} \sum_{k=0}^{n-1} f(x_k)
\begin{equation}
\label{sec:theorem-1.13-eq1}
\frac{b - a}{n} \sum_{k=0}^{n-1} f(x_k)
\leq I \leq
\frac{b - a}{n} \sum_{k=1}^n f(x_k)$$
\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{sec: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{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] \\
& = \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{sec:theorem-1.13-eq1} can alternatively be written as
\begin{equation}
\label{sec: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{sec:theorem-1.12}, $f$ is integrable.
Therefore \nameref{sec:theorem-1.9} indicates $f$ satisfies
\begin{equation}
\label{sec: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{sec:theorem-1.13-eq2} and \eqref{sec: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}
\section{\partial{%
Calculation of the Integral of a Bounded Decreasing Function}}%
\label{sec:calculation-integral-bounded-decreasing-function}
\label{sec: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{sec: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{sec: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{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] \\
& = \sum_{k=0}^{n-1} f(x) \left[\frac{b - a}{n}\right].
\end{align*}
Therefore \eqref{sec:theorem-1.14-eq1} can alternatively be written as
\begin{equation}
\label{sec:theorem-1.14-eq2}
\int_a^b s(x) \mathop{dx} \leq I \leq \int_a^b t(x) \mathop{dx}.
\end{equation}
By \nameref{sec:theorem-1.12}, $f$ is integrable.
Therefore \nameref{sec:theorem-1.9} indicates $f$ satisfies
\begin{equation}
\label{sec:theorem-1.14-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{sec:theorem-1.14-eq2} and \eqref{sec:theorem-1.14-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}
\section{\unverified{%
Calculation of the Integral \texorpdfstring{$\int_0^b x^p \mathop{dx}$}{int-x-p}
when \texorpdfstring{$p$}{p} is a Positive Integer}}%
\label{sec:calculation-integral-int-x-p-p-positive-integer}
\label{sec:theorem-1.15}
\begin{theorem}[1.15]
If $p$ is a positive integer and $b > 0$, we have
$$\int_0^b x^p \mathop{dx} = \frac{b^{p+1}}{p+1}.$$
\end{theorem}
\begin{proof}
TODO