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chapters/measures_integrals.tex
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\pagebreak
\subfile{sections/contents_measures.tex}
\subfile{sections/integrals.tex}
\end{document}

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% !TeX root = ../../script.tex
\documentclass[../../script.tex]{subfiles}
\begin{document}
\section{Integrals}
Let $\measure$ be a measure space.
The most important example is on $\realn$ with the Lebesgue-$\sigma$-algebra and the Lebesgue measure.
We have one technical requirement, and that is that $\measure$ is a $\sigma$-finite measure space, i.e.
\[
\exists \anyseqdef[E]{\setfam}: ~~\bigcup_{n \in \natn} E_n = \Omega \text{ and } \mu(E_n) < \infty ~~\forall n \in \natn
\]
On $\realn$ this requirement is fulfilled by defining $E_n = [-n, n]$.
\begin{rem}[Notation]
Let $\Phi(x)$ be a statement depending on $x \in \Omega$. We write $[\Phi]$ for
\[
\set[\Phi(x)]{x \in \Omega}
\]
Example: $y \in \cmpln$
\[
[f = y] = \set[f(x) = y]{x \in \Omega} = \inv{f}(y)
\]
We write "$\Phi$ holds" for "$\Phi(x)$ holds $\forall x \in \Omega$". For example "$f > g$" instead of "$f(x) > g(x) ~~\forall x \in \Omega$".
$\Phi$ is said to hold "almost everywhere" (a.e.) if the set
\[
\set[\neg\Omega(x)]{x}
\]
is a null set. For example, "$f > g$ almost everywhere" means $\mu([f \le g]) = 0$.
The sequence $(f_n)$ converges pointwise a.e. towards $f$ if
\[
\left[\limn f_n \ne f\right] = \set[\limn f_n(x) \ne f(x)]{x \in \Omega}
\]
is a null set.
\end{rem}
\begin{defi}
Let $A \in \setfam$, then
\begin{align*}
\charfun_A: \Omega &\longrightarrow \realn \\
\omega &\longmapsto \begin{cases}
1, & x \in A \\
0, & \text{else}
\end{cases}
\end{align*}
is said to be the characteristic function of $A$. $A$ is the support of $\charfun_A$.
With this we can define the space of simple functions
\[
X = \set[n \in \natn, A_i \in \setfam, ~\mu(A_i) < \infty, ~a_i \in \cmpln]{\sum_{i=1}^{n} a_i \charfun_A}
\]
$X^+$ notates the non-negative, simple functions.
\end{defi}
\begin{rem}
\begin{enumerate}[(i)]
\item Let $A, B \in \setfam$
\begin{align*}
\charfun_{A \cap B} &= \charfun_A \cdot \charfun_B \\
\charfun_{A \cup B} &= \charfun_A + \charfun_B - \charfun_{A \cap B} = \charfun_A + \charfun_B - \charfun_A \charfun_B
\end{align*}
\item The set $X$ is a vector space, and the product of characteristic functions is another characteristic function, i.e.
\[
f, g \in X \implies f \cdot g \in X
\]
Thus $X$ is an algebra.
\item If $A_1, \cdots, A_n$ is a decomposition of $\Omega$, which means they are disjoint and
\[
\bigcup_{i=1}^n A_i = \Omega
\]
then
\[
(\charfun_i) = \charfun_{\Omega} = \sum_{i=1}^n \charfun_{A_i}
\]
\item The representation of simple functions as a linear combination is not unique
\[
\charfun_{[0, 2]} + \charfun{[2, 3]} = \charfun{[0, 1]} + \charfun{[1, 3]}
\]
\item One can easily see that simple functions can only assume finitely many values, and their support $[f \ne 0]$ has a finite measure.
The canonical representation is
\[
f = \sum_{y = f(\Omega)} g \cdot \charfun_{[f = y]}
\]
\end{enumerate}
\end{rem}
\begin{defi}[Integrals of simple functions]
Let $f \in X$ in canonical representation
\[
f = \sum_{i=1}^{\infty} a_i \charfun_{A_i}
\]
Then we define
\[
\int f \dd{\mu} := \sum_{i=1}^n a_i \mu(A_i)
\]
\end{defi}
\begin{rem}
This sum is always finite, the only $A_i$ with infinite measure is that where $a_i = 0$
\[
a_i \cdot A_i = 0 \cdot \infty = 0
\]
Let $f = \sum_{j=1}^m b_j \charfun_{B_j}$ be another representation of $f$, so $B_1, \cdots, B_m$ is a decomposition.
If $A_i \cap B_j \ne \varnothing$ i.e.
\[
\exists x \in A_i \cap B_j: ~~f(x) = a_i = b_j
\]
Then
\begin{align*}
\int f \dd{\mu} &= \sum_{i=1}^n a_i \mu(A_i) = \sum_{i=1}^n a_i \mu\underbrace{\left(A_i \cap \bigcup_{j=1}^m B_j \right)}_{\bigcup_{j=1}^m (A_i \cap B_j)} = \sum_{i=1}^n a_i \sum_{j=1}^m \mu(A_i \cap B_j) \\
&= \sum_{i=1}^n \sum_{j=1}^m b_j \mu(A_i \cap B_j) = \sum_{j=1}^m b_j \mu \left(\left(\bigcup_{i=1}^n A_i \right) \cap B_j \right) \\
&= \sum_{j=1}^m b_j \mu(B_j)
\end{align*}
\end{rem}
\begin{thm}
Let $f, g$ be simple functions, $\alpha \in \cmpln$. Then
\[
\int (f + \alpha g) \dd{\mu} = \int f \dd{\mu} + \alpha \int g \dd{\mu}
\]
If $f, g$ are real-valued and $f \le g$ a.e., then
\[
\int f \dd{\mu} \le \int g \dd{\mu}
\]
And especially if $f = g$ a.e.
\[
\int f \dd{\mu} = \int g \dd{\mu}
\]
Finally, the triangle inequality holds
\[
\abs{\int f \dd{\mu}} \le \int \abs{f} \dd{\mu}
\]
\end{thm}
\begin{proof}
Let $f, g$ be in canonical representation
\begin{subequations}
\begin{multicols}{2}
\noindent
\begin{equation} f = \sum_{i=1}^n a_i \charfun_{A_i} \end{equation}
\begin{equation} g = \sum_{j=1}^m b_j \charfun_{B_j} \end{equation}
\end{multicols}
\end{subequations}
\noindent Then
\begin{equation}
\begin{split}
f + \alpha g &= \sum_{i=1}^n a_i \charfun_{A_i} + \alpha \sum_{j=1}^m b_j \charfun_{B_j} \\
&= \sum_{i=1}^n a_i \charfun_{A_i} \underbrace{\left(\sum_{j=1}^m \charfun_{B_j}\right)}_{\charfun} + \alpha \sum_{j=1}^m b_j \charfun_{B_j} \underbrace{\left(\sum_{i=1}^n \charfun_{A_i}\right)}_{\charfun} \\
&= \sum_{i=1}^n \sum_{j=1}^m (a_i + \alpha b_j) \charfun_{A_i \cap B_j}
\end{split}
\end{equation}
$A_i \cap B_j$ with $i \in \set{1, \cdots, n}, j \in \set{1, \cdots, m}$ is a decomposition of $\Omega$
\begin{equation}
\bigcup_{\substack{i = 1 \\ j = 1}}^{\substack{m \\ n}} A_i \cap B_j = \bigcup_{i=1}^n A_i \cap \underbrace{\left( \bigcup_{j=1}^m B_j\right)}_{\Omega} = \Omega
\end{equation}
This means that
\begin{align*}
\int (f + \alpha g) \dd{\mu} &= \sum_{i=1}^n \sum_{j=1}^m (a_i + \alpha b_j) \mu(A_i \cap B_j) \\
&= \sum_{i=1}^n a_i \mu\left(A \cap \left(\bigcup_{j=1}^m B_j\right)\right) + \alpha \sum_{j=1}^m b_j \mu\left(\left(\bigcup_{i=1}^n A_i\right) \cap B_j \right) \\
&= \int f \dd{\mu} + \alpha \int g \dd{\mu}
\end{align*}
Now let $f \ge 0$ almsot everywhere, i.e. $[f < 0]$ is a null set. If $a_i < 0$, then $A_i \subset [f < 0]$, and then $\mu(A_i) = 0$
and thus the integral is a sum over non-negative values, so it is non-negative itself.
If $f \le g$ a.e., then $g - f \ge 0$ a.e.:
\begin{equation}
0 \le \int (g - f) \dd{\mu} = \int g \dd{\mu} - \int f \dd{\mu}
\end{equation}
Finally to show the triangle inequality
\begin{equation}
\abs{\int f \dd{\mu}} = \abs{\sum_{i=1}^n a_i \mu(A_i)} \le \sum_{i=1}^n \abs{a_i} \mu(A_i) = \int \abs{f} \dd{\mu}
\end{equation}
\end{proof}
\begin{rem}
From linearity follows, that $f$ can be in any representation
\[
f = \sum_{i=1}^n a_i \charfun_{A_i}
\]
and the integral will still be
\[
\int f \dd{\mu} = \sum_{i=1}^n a_i \mu(A_i)
\]
\end{rem}
\begin{rem}
Notice how the integrals so far did not have any integration variables. The integrals map functions (not their values!) to numbers.
If the integration variable is of concern, we can write
\[
\int f(x) \dd{\mu(x)}
\]
For Lebesgue integrals we define
\[
\int f(x) \dd{x} = \int_{-\infty}^{\infty} f(x) \dd{x}
\]
\end{rem}
\begin{defi}
$f: \Omega \rightarrow \Omega$ is said to be measurable, if there is a sequence of simple functions $\anyseqdef[f]{X}$ that converge pointwise towards $f$.
\end{defi}
\begin{rem}
\begin{enumerate}[(i)]
\item For real-valued functions $f$
\[
f \text{ measurable} \iff [f \le y] \in \setfam ~~\forall y \in \setfam
\]
\item Simple functions and characteristic functions are measurable.
\item Continuous functions are Lebesgue measurable.
\item Sums, products, quotients (if existant) of measurable sets are measurable.
\item If $\seq{f}$ is a sequence of measurable functions, then
\begin{align*}
\sup_{n \in \natn} f_n && \limsupn f_n && \limn f_n
\end{align*}
are measurable if they exist.
\end{enumerate}
\end{rem}
All functions from now on will be considered measurable.
\begin{defi}
Let $f: \Omega \rightarrow [0, \infty)$, then
\[
\int f \dd{\mu} := \sup\set[g \in X^+, ~g < f]{\int g \dd{\mu}}
\]
\end{defi}
\begin{rem}
\begin{enumerate}[(i)]
\item This integral can be $\infty$.
\item If $f$ is a non-negative, simple function, then $\forall h$ that are non-negative, simple functions with $h \le f$
\[
\int h \dd{\mu} \le \int f \dd{\mu}
\]
The old integral (integral over simple functions) is identical to this one.
\item Let $f, g$ be non-negative and $f \le g$ a.e. Define $A = [f \le g]$.
Then for all simple $h < f$ we have
\[
h \cdot \charfun_A \le g
\]
and
\[
\int h \dd{\mu} = \int h \cdot \charfun_A \dd{\mu} \le \int g \dd{\mu}
\]
Which implies
\[
\int f \dd{\mu} = \sup_h \int h \dd{\mu} \le \int g \dd{\mu}
\]
Especially
\[
\int f \dd{\mu} = \int g \dd{\mu} \text{ if } f = g \text{ a.e.}
\]
\item If $[f > 0]$ is a null set, then $f$ is the zero function a.e. and
\[
\int f \dd{\mu} = 0
\]
The inverse is also true
\[
\int f \dd{\mu} = 0 \text{ and } f \ge 0 \implies f = 0 \text{ a.e.}
\]
Let $A_k := [f \ge \rec{k}] \in \setfam$, then
\[
\rec{k} \charfun_{A_k} \le f ~~\forall k \in \natn
\]
Since $\int f \dd{\mu} = 0$, this implies
\begin{align*}
&\int \rec{k} \charfun_{A_k} \dd{\mu} = \rec{k} \mu(A_k) = 0 \\
\implies &\mu(A_k) = 0 ~~\forall k \in \natn
\end{align*}
The $A_k$ are monotonically increasing, and thus due to the continuity of the measure
\[
0 = \limn \mu(A_k) = \mu\left(\bigcup_{k \in \natn} [f \ge \rec{k}]\right) = \mu([f = 0])
\]
\item The definition means $\exists \anyseqdef[f]{X^+}$ such that $f_n \le f$
\[
\int f_n \dd{\mu} \conv{n \rightarrow \infty} \int f \dd{\mu}
\]
Define $g_n = \max \set{f_1, \cdots, f_n}$. These are also simple functions and $f_n \le g_n \le f ~~\forall n \in \natn$.
\[
\implies \int f_n \dd{\mu} \le \int g_n \dd{\mu} \le \int f \dd{\mu}
\]
And thus
\begin{gather*}
\int f_n \dd{\mu} \conv{} \int f \dd{\mu} \\
\downarrow \\
\int g_n \dd{\mu} \conv{} \int f \dd{\mu}
\end{gather*}
The sequence $g_n$ is monotonic.
\item Let $\seq{g}$ be convergent to $g: \Omega \rightarrow [0, \infty)$. Then
\[
g \le f \implies \int g \dd{\mu} \le \int f \dd{\mu}
\]
$\forall n \in \natn$ we have $g_n \le g$, and thus
\[
\limn \int g_n \dd{\mu} \le \int g \dd{\mu}
\]
$\forall f \ge 0$ there exists a monotonically increasing sequence of simple function such that
\[
\int g_n \dd{\mu} \conv{} \int f \dd{\mu}
\]
and thus $g = f$ a.e.
\item
\begin{align*}
\int (cf) \dd{\mu} = c \int f \dd{\mu} ~~c \in [0, \infty) \\
\int f \dd{\mu} + \int g \dd{\mu} \le \int (f + g) \dd{\mu}
\end{align*}
\end{enumerate}
\end{rem}
\begin{thm}[Monotone Convergence Theorem]
Let $f \ge 0$ and $\seq{f}$ a monotonically increasing sequence of functions converging pointwise to $f$ a.e.
Then
\[
\limn \int f_n \dd{\mu} = \int f \dd{\mu}
\]
\end{thm}
\begin{proof}
First, let $\limn f_n = f$ everywhere. Since $(f_n)$ is monotonic, this must also hold for $\int f_n \dd{\mu}$, so
\begin{equation}
\limn \int f_n \dd\mu \le \int f \dd\mu
\end{equation}
First, consider the special case $\anyseqdef[A]{\setfam}$ monotonically increasing, with
\begin{equation}
\bigcup_{n \in \natn} A_n = \Omega
\end{equation}
Then
\begin{equation}\label{eq:750}
\limn \int f \charfun_{A_n} \dd\mu = \int f \dd\mu
\end{equation}
For $f = \charfun_B$
\begin{equation}
\begin{split}
\limn \int \underbrace{\charfun_B \charfun_{A_n}}_{\charfun_{B \cap A_n}} \dd\mu &= \limn \mu(B \cap A_n) \\
&= \mu(\bigcup_{n \in \natn} B \cap A_n) \\
&= \mu(B) = \int \charfun_B \dd\mu
\end{split}
\end{equation}
Since both sides are lienear in $f$ (at least for simple functions), the equality holds for arbitrary simple functions.
Now let $f \ge 0$ be arbitrary and $h \in X^+$, such that for $\epsilon > 0$
\begin{equation}
\int h \dd\mu \ge \int f \dd\mu - \frac{\epsilon}{2}
\end{equation}
and thus $h \le f$. From this it follows that
\begin{equation}
\exists N \in \natn ~\forall n \ge N: ~~\int h \charfun_{A_n} \dd\mu \ge \int h \dd\mu - \frac{\epsilon}{2}
\end{equation}
And thus
\begin{equation}
\forall n \ge N: ~~\int h \charfun_{A_n} \dd\mu \ge \int f \dd\mu - \epsilon
\end{equation}
which proves \Cref{eq:750} for arbitrary $f \ge 0$.
Now let $c \in (0, 1)$, and set
\begin{equation}
A_n = [f_n \ge cf]
\end{equation}
Since $f_n$ are monotonic, the $A_n$ are as well, and
\begin{equation}
\bigcup_{n \in \natn} A_n = \Omega
\end{equation}
Then
\begin{equation}
\int f_n \dd\mu \ge \int f_n \charfun_{A_n} \dd\mu \ge \int c f \charfun_{A_n} \dd\mu = c \int f \charfun_{A_n} \dd\mu
\end{equation}
Thus
\begin{equation}
c \int f \charfun_{A_n} \dd\mu \conv{n \rightarrow \infty} c \int f \dd\mu
\end{equation}
Which in turn implies
\begin{equation}
\limn \int f_n \dd\mu \ge c \int f \dd\mu
\end{equation}
For $c \rightarrow 1$ we have
\begin{equation}
\limn \int f_n \dd\mu = \int f \dd\mu
\end{equation}
And if $f_n \rightarrow f$ only a.e.
\begin{equation}
A = [\limn f_n = f]
\end{equation}
then $\Omega \setminus A$ is a null set.
\begin{equation}
\begin{split}
\limn \int f_n \dd\mu &= \limn \int f_n \charfun_A \dd\mu \\
&= \int f \charfun_A \dd\mu \\
&= \int f \dd\mu
\end{split}
\end{equation}
\end{proof}
\begin{eg}
Calculate the integral of $f(y) = y \charfun_{[0, x]}(x)$
\[
f_n = \sum_{k=0}^{2^n-1} k \frac{1}{2^n} \charfun_{[k \frac{x}{2^n}, (k+1) \frac{x}{2^n}]}
\]
is a monotonically increasing sequence which converges to $f$ on $\realn \setminus \set{x}$.
\begin{align*}
\int f_n \dd\mu = \sum_{k=0}^{2^n-1} k \frac{x}{2^n} \cdot \left(\frac{x}{2^n}\right) = &\frac{x^2}{2^{2n}} \sum_{k=0}^{2^n - 1} k \\
= &\frac{x^2}{2^{2n}} \frac{2^n(2^n - 1)}{2 \cdot 2^n} \\
= &\frac{x^2}{2} \frac{2n - 1}{2^n} \\
\conv{} &\frac{x^2}{2}
\end{align*}
\begin{center}
\begin{tikzpicture}
\draw[thick, ->] (0, 0) -- (5, 0) node[right] {$y$};
\draw[thick, ->] (0, 0) -- (0, 5) node[above] {$f(y)$};
\draw [dashed] (4, 0) node[below] {$x$} -- (4, 4);
\draw [dashed] (3, 0) -- (3, 3);
\draw [dashed] (2, 0) -- (2, 2);
\draw [dashed] (1, 0) -- (1, 1);
\draw (1, 1) -- (2, 1);
\draw (2, 2) -- (3, 2);
\draw (3, 3) -- (4, 3) node[right] {$f_n$};
\draw [red, thick] (0, 0) -- (4, 4) node [above right] {$f$};
\end{tikzpicture}
\end{center}
\end{eg}
\begin{eg}
Consider $f_n = n \charfun_{(0, \frac{1}{n})}$. This sequence converges pointwise to the zero function. But
\[
\int f_n \dd\mu = n \cdot \rec{n} = 1 \ne 0
\]
This is due to $f_n$ not being monotonic increasing.
\end{eg}
\begin{eg}
Let $\anyseqdef[a]{\cmpln}$, and define
\[
f_n = a_n \charfun_{[n, n+1]}
\]
This sequence converges pointwise to $0$, but
\[
\int f_n \dd\lambda = a_n
\]
depends on $\seq{a}$ and can converge to any value (or even diverge).
\end{eg}
\begin{defi}
A function $f: \Omega \rightarrow \cmpln$ is said to be integrable if
\[
\int \abs{f} \dd\mu < \infty
\]
A sequence of simple functions $\seq{f}$ is said to be an approximating sequence of $f$ if
\[
\int \abs{f - f_n} \dd\mu \conv{n \rightarrow \infty} 0
\]
\end{defi}
\begin{cor}
Let $f, g \ge 0$
\[
\int (f + g) \dd\mu = \int f \dd\mu + \int g \dd\mu
\]
\end{cor}
\begin{proof}
Let $\seq{f}, \seq{g} \subset X^+$ be monotone sequences with $f_n \rightarrow f$, $g_n \rightarrow g$ almost everywhere.
Then $(f_n + g_n)$ is monotonically increasing as well and converge pointwise to $(f + g)$ almost everywhere.
\begin{equation}
\begin{split}
&\left[\limn f_n \ne f\right] \text{ null set, } \left[\limn g_n \ne g \right] \text{ null set} \\
\implies &\left[\limn f_n \ne f\right] \cup \left[\limn g_n \ne g\right] \text{ null set}
\end{split}
\end{equation}
This implies
\begin{equation}
\begin{split}
\int (f + g) \dd\mu = \limn \int (f_n + g_n) \dd\mu &= \limn \int f_n \dd\mu + \limn \int g_n \dd\mu \\
&= \int f \dd\mu + \int g \dd\mu
\end{split}
\end{equation}
\end{proof}
\begin{rem}
\begin{enumerate}[(i)]
\item The set of integrable functions is a vector space, because for $f, g$ integrable and $\alpha \in \cmpln$
\begin{align*}
\int \abs{f + \alpha g} \dd\mu &\le \int \abs{f} + \abs{\alpha}\abs{g} \dd\mu \\
&= \int \abs{f} \dd\mu + \abs{\alpha} \int \abs{g} \dd\mu < \infty
\end{align*}
However, $f \cdot g$ is not integrable in general!
\item Let $f \ge 0$ be integrable, and $\anyseqdef[f]{X^+}$ such that $f_n \rightarrow f$ pointwise a.e.
\[
\limn \int f_n \dd\mu = \int f \dd\mu < \infty
\]
$\forall n \in \natn$:
\[
\int \abs{f - f_n} \dd\mu = \int (f - f_n) \dd\mu = \int f \dd\mu - \int f_n \dd\mu \conv{n \rightarrow \infty} 0
\]
\item Let $f: \Omega \rightarrow \realn$ be a function. Decompose the function into a positive and a negative part:
\begin{align*}
f_+ := f \cdot \charfun_{[f \ge 0]} && f_- := -f \cdot \charfun_{[f \le 0]}
\end{align*}
$f_+, f_- \ge 0$, and
\begin{align*}
f = f_+ - f_- && \abs{f} = f_+ + f_-
\end{align*}
\item $\abs{\Re f} \le \abs{f}$, $\abs{\Im f} \le \abs{f}$. If $f$ is integrable, then $\Re(f)$ and $\Im(f)$ are also integrable.
\item Let $f, g$ be arbitrary, and $\seq{f}, \seq{g}$ approximating sequences for $f$ and $g$. Then for $\alpha \in \cmpln$:
\begin{align*}
\int \abs{f + \alpha g - (f_n + \alpha g_n)} \dd\mu \le &\int \abs{f - f_n} \dd\mu + \alpha \int \abs{g - g_n} \\
= &\int \abs{f - f_n} \dd\mu + \abs{\alpha} \int \abs{g - g_n} \dd\mu \\
\conv{n \rightarrow \infty} &0
\end{align*}
Thus $f_n + \alpha g_n$ is an approximating sequence for $f + \alpha g$
\item Consider
\[
f = \left(\left(\Re f\right)_+ - \left(\Re f\right)_-\right) + i\left(\left(\Im f\right)_+ - \left(\Im f\right)_-\right)
\]
If $f$ is integrable, then all the terms are integrable as well and have approximating sequences. Thus, $f$ has an approximating sequence.
\item Now let $\seq{f}$ be an approximating sequence for $f$. Let $\epsilon > 0$, then
\[
\exists N \in \natn ~\forall n \ge N: ~~\int \abs{f - f_n} \dd\mu < \frac{\epsilon}{2}
\]
$\forall n, m \ge N$
\begin{align*}
\abs{\int f_n \dd\mu - \int f_m \dd\mu} &= \abs{\int (f_n + f_m) \dd\mu} \\
&\le \int \abs{f_n - f_m} \dd\mu \\
&\le \int \left( \abs{f_n - f} + \abs{f - f_m}\right) \dd\mu \\
&< \epsilon
\end{align*}
Which means $(\int f_n \dd\mu)$ is a Cauchy sequence, so it converges to $I \in \cmpln$
\item Let $\seq{g}$ be another approximating sequence for $f$
\begin{align*}
\abs{\int f_n \dd\mu - \int g_n \dd\mu} &\le \int \abs{f_n - g_n} \dd\mu \\
&\le \int \abs{f_n - f} \dd\mu + \int \abs{f - g_n} \dd\mu \conv{n \rightarrow \infty} 0
\end{align*}
So the integral is invariant to the choice of approximating sequence.
\end{enumerate}
\end{rem}
\begin{defi}
Let $f$ be integrable, and define
\[
\int f \dd\mu = \limn \int f_n \dd\mu
\]
for some approximating sequence $(f_n)$ of $f$.
\end{defi}
\begin{rem}
If $f$ is a simple function, then $\seq{f}_{n \in \natn}$ is an approximating sequence.
The new integral definition is compatible with the integral for simple functions. and with the integral for non-negative functions.
\end{rem}
\begin{thm}
Let $f, g$ be integrable.
\begin{enumerate}[(i)]
\item \[ \forall \alpha \in \cmpln: ~~\int (f + \alpha g) \dd\mu = \int f \dd\mu + \alpha \int g \dd\mu \]
\item If $f \le g$ a.e., then
\[
\int f \dd\mu \le \int f \dd\mu
\]
and
\[
f = g \text{ a.e.} \implies \int f \dd\mu = \int g \dd\mu
\]
\item \[ \abs{\int f \dd\mu} \le \int \abs{f} \dd\mu \]
\end{enumerate}
\end{thm}
\begin{proof}
Let $\seq{f}$, $\seq{g}$ be approximating sequences for $f$ and $g$.
Then $(f_n + \alpha g_n)$ is an approximating sequence for $(f + \alpha g)$.
\begin{equation}
\int (f + \alpha g) \dd\mu = \limn \int (f_n + \alpha g_n) \dd\mu = \underbrace{\limn \int f_n \dd\mu}_{\int f \dd\mu} + \underbrace{\limn \int g_n \dd\mu}_{\int g \dd\mu}
\end{equation}
To prove the second statement, let $f \le g$ a.e. Then $(g - f)_- = 0$ a.e.
\begin{equation}
\implies \int (g - f)_- = 0
\end{equation}
And thus
\begin{equation}
\begin{split}
\int g \dd\mu - \int f \dd\mu &= \int (g - f) \dd\mu \\
&= \int (g - f)_+ \dd\mu - \int (g - f)_- \dd\mu \ge 0
\end{split}
\end{equation}
The final statement is proven by applying the reverse triangle inequality
\begin{equation}
\int \abs{\abs{f} - \abs{f_n}} \dd\mu \le \int \abs{f - f_n} \dd\mu \conv{n \rightarrow \infty} = 0
\end{equation}
This means if $(\abs{f_n})$ is an approximating sequence for $\abs{f}$, then
\begin{equation}
\int \abs{f} \dd\mu = \limn \int \abs{f_n} \dd\mu \ge \limn \abs{\int f_n \dd\mu} = \abs{\int f \dd\mu}
\end{equation}
\end{proof}
\begin{rem}
For $A \subset \setfam$ we define
\[
\int_A g \dd\mu := \int g \charfun_A \dd\mu
\]
$g \charfun_A$ can be integrable, even if $g$ isn't. The above integral doesn't depend on the behavior of $g$ outside of $A$.
We use $\int_A g \dd\mu$ even if $g$ isn't defined outside of $A$.
Integrals are independent from the behavior on null sets, so
\[
\int_0^1 \rec{x} \dd x = 0
\]
is perfectly fine, even though the integrand is not defined for $x = 0$.
\end{rem}
\begin{eg}
Let $\Omega = \natn$, $\setfam = \powerset(\Omega)$ and $\mu$ the counting measure. Let $f : \natn \rightarrow [0, \infty)$, then
\[
f_N = f \charfun_{\set{1, \cdots, N}} = \sum_{n=1}^N f(n) \charfun_{\set{n}}
\]
is a sequence of monotonically increasing, simple functions that converge to $f$ pointwise.
\[
\int f \dd\mu = \limes{N}{\infty} \int f_N \dd\mu = \limes{N}{\infty} \sum_{n=1}^N f(n) \mu(\set{n}) = \sum_{n=1}^{\infty} f(n)
\]
Thus we can conclude
\[
f: \natn \rightarrow \cmpln \text{ integrable} \iff \int \abs{f} \dd\mu = \sum_{n=1}^{\infty} \abs{f(n)} < \infty
\]
and
\[
\int f \dd\mu = \sum_{n=1}^{\infty} f(n)
\]
\end{eg}
\end{document}

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@ -2,7 +2,7 @@
\usepackage{amsmath, amssymb, amstext, physics}
\usepackage{amsthm, stackrel, xifthen, mathtools, graphicx}
\usepackage[makeroom]{cancel}
\usepackage{hyperref, cleveref}
\usepackage{hyperref, cleveref, bbm}
\usepackage{enumerate, subcaption}
\usepackage[shortlabels]{enumitem}
\usepackage{multicol}
@ -94,6 +94,7 @@
\newcommand{\measureable}{(\Omega, \setfam)}
\newcommand{\measure}{(\Omega, \setfam, \mu)}
\newcommand{\intervals}{\mathcal{I}}
\newcommand{\charfun}{\mathbbm{1}}
\newcommand{\reader}{Left as an exercise for the reader.}
\newcommand{\noproof}{Without proof.}