diff --git a/chapters/measures_integrals.tex b/chapters/measures_integrals.tex index 0b79507..bff42d8 100644 --- a/chapters/measures_integrals.tex +++ b/chapters/measures_integrals.tex @@ -7,4 +7,5 @@ \pagebreak \subfile{sections/contents_measures.tex} + \subfile{sections/integrals.tex} \end{document} \ No newline at end of file diff --git a/chapters/sections/integrals.tex b/chapters/sections/integrals.tex new file mode 100644 index 0000000..bf29b56 --- /dev/null +++ b/chapters/sections/integrals.tex @@ -0,0 +1,635 @@ +% !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} \ No newline at end of file diff --git a/script.pdf b/script.pdf index ab59dd3..84bdd2a 100644 Binary files a/script.pdf and b/script.pdf differ diff --git a/script.tex b/script.tex index 96c6b8e..f0d3573 100644 --- a/script.tex +++ b/script.tex @@ -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.}