diff --git a/chapters/measures_integrals.tex b/chapters/measures_integrals.tex index d25da0b..1f7f583 100644 --- a/chapters/measures_integrals.tex +++ b/chapters/measures_integrals.tex @@ -9,4 +9,5 @@ \subfile{sections/contents_measures.tex} \subfile{sections/integrals.tex} \subfile{sections/int_real_nums.tex} + \subfile{sections/product_measures.tex} \end{document} \ No newline at end of file diff --git a/chapters/sections/product_measures.tex b/chapters/sections/product_measures.tex new file mode 100644 index 0000000..a7872a3 --- /dev/null +++ b/chapters/sections/product_measures.tex @@ -0,0 +1,311 @@ +% !TeX root = ../../script.tex +\documentclass[../../script.tex]{subfiles} + +\begin{document} +\section{Product measures and Fubini's Theorem} + +\begin{eg} + Let + \[ + f: [0, 1] \times [0, 1] \longrightarrow [0, \infty) + \] + Question: What is the volume (or the $\lambda^2$ measure) under the graph of $f$, i.e. + \[ + \set[0 \le z \le f(x, y)]{(x, y, z) \in \realn^3} + \] + The possibilities are + \begin{gather*} + \int f \dd\lambda^2 \\ + \int_0^1 \int_0^1 f(x, y) \dd x \dd y ~~\text{ or }~~ \int_0^1 \int_0^1 f(x, y) \dd y \dd x + \end{gather*} +\end{eg} +From now on we define $\measure$ and $(\Phi, \setfamb, \nu)$ to be measure spaces. +\begin{defi} + The product $\sigma$-algebra $\setfam \otimes \setfamb$ is the smallest $\sigma$-algebra on $\Omega \times \Phi$ that contains + all sets of type $A \times B$ for $A \in \setfam, B \in \setfamb$. + + Examples for $A \times B$ are + + \begin{center} + \begin{minipage}{.49\linewidth} + \begin{center} + \begin{tikzpicture} + \draw[very thick, ->] (0, 0) -- (5, 0); + \draw[very thick, ->] (0, 0) -- (0, 5); + + \draw (1, 1) -- (4, 1) -- (4, 4) -- (1, 4) -- (1, 1); + + \draw[dashed] (0, 4) -- (1, 4); + \draw[dashed] (0, 1) -- (1, 1); + \draw[dashed] (1, 0) -- (1, 1); + \draw[dashed] (4, 0) -- (4, 1); + + \node[below] at (2.5, 0) {$A$}; + \node[left] at (0, 2.5) {$B$}; + \end{tikzpicture} + \end{center} + \end{minipage} + \begin{minipage}{.49\linewidth} + \begin{center} + \begin{tikzpicture} + \draw[very thick, ->] (0, 0) -- (5, 0); + \draw[very thick, ->] (0, 0) -- (0, 5); + + \draw (1, 1) -- (2, 1) -- (2, 2) -- (1, 2) -- (1, 1); + \draw (3, 1) -- (4, 1) -- (4, 2) -- (3, 2) -- (3, 1); + \draw (1, 3) -- (2, 3) -- (2, 4) -- (1, 4) -- (1, 3); + \draw (3, 3) -- (4, 3) -- (4, 4) -- (3, 4) -- (3, 3); + + \draw[dashed] (0, 4) -- (1, 4); + \draw[dashed] (0, 3) -- (1, 3); + \draw[dashed] (0, 2) -- (1, 2); + \draw[dashed] (0, 1) -- (1, 1); + \draw[dashed] (4, 0) -- (4, 1); + \draw[dashed] (3, 0) -- (3, 1); + \draw[dashed] (2, 0) -- (2, 1); + \draw[dashed] (1, 0) -- (1, 1); + + \node[left] at (0, 3.5) {$B$}; + \node[left] at (0, 1.5) {$B$}; + \node[below] at (1.5, 0) {$A$}; + \node[below] at (3.5, 0) {$A$}; + \end{tikzpicture} + \end{center} + \end{minipage} + \end{center} + \pagebreak NO examples for $A \times B$ are + + \begin{center} + \begin{minipage}{.49\linewidth} + \begin{center} + \begin{tikzpicture} + \draw[very thick, ->] (0, 0) -- (5, 0); + \draw[very thick, ->] (0, 0) -- (0, 5); + + \draw (1, 1) -- (2, 1) -- (2, 2) -- (1, 2) -- (1, 1); + \draw (3, 1) -- (4, 1) -- (4, 2) -- (3, 2) -- (3, 1); + \draw (1, 3) -- (2, 3) -- (2, 4) -- (1, 4) -- (1, 3); + + \draw[dashed] (0, 4) -- (1, 4); + \draw[dashed] (0, 3) -- (1, 3); + \draw[dashed] (0, 2) -- (1, 2); + \draw[dashed] (0, 1) -- (1, 1); + \draw[dashed] (4, 0) -- (4, 1); + \draw[dashed] (3, 0) -- (3, 1); + \draw[dashed] (2, 0) -- (2, 1); + \draw[dashed] (1, 0) -- (1, 1); + + \node[left] at (0, 3.5) {$B$}; + \node[left] at (0, 1.5) {$B$}; + \node[below] at (1.5, 0) {$A$}; + \node[below] at (3.5, 0) {$A$}; + \end{tikzpicture} + \end{center} + \end{minipage} + \begin{minipage}{.49\linewidth} + \begin{center} + \begin{tikzpicture} + \draw[very thick, ->] (0, 0) -- (5, 0); + \draw[very thick, ->] (0, 0) -- (0, 5); + + \draw (2.5, 2.5) circle[radius=1.5]; + + \node[below] at (0, 0) {\quad}; + \end{tikzpicture} + \end{center} + \end{minipage} + \end{center} + A measure $\vartheta$ defined on $\setfam \otimes \setfamb$ is said to be a product measure of $\mu, \nu$ if + \[ + \vartheta(A \times B) = \mu(A) \nu(B) ~~A \in \setfam, B \in \setfamb + \] +\end{defi} + +\begin{rem} + Product measures always exist. For $\sigma$-finite measure spaces they are unique. Notation: + \[ + \mu \otimes \nu ~~\text{ or }~~ \mu^2 = \mu \otimes \mu + \] +\end{rem} + +\begin{eg} + $\realn$ with Lebesgue measure $\lambda$. $\lambda^2$ is the product measure on $\realn^2$. + \begin{align*} + \lambda^2([a, b] \times [c, d]) &= \lambda([a, b])\lambda([c, d]) \\ + &= (b - a)(d - c) + \end{align*} + This means the product measure characterizes the area. Analogously this can be extended for $\lambda^3$, $\lambda^4$ etc. +\end{eg} + +\begin{eg} + Consider + \begin{align*} + f: \realn^2 &\longrightarrow \realn \\ + f &= \sum_{n=0}^{\infty} \left(\charfun_{[n, n+1)^2} - \charfun_{[n+1, n+2) \times [n, n+1)}\right) + \end{align*} + + \begin{center} + \begin{tikzpicture} + \draw[->, very thick] (0, 0) -- (0, 3.5) node[above] {$y$}; + \draw[->, very thick] (0, 0) -- (5.5, 0) node[right] {$x$}; + + \foreach \x in {1,...,5} \draw (\x, -0.1) -- (\x, 0.1); + \foreach \y in {1,...,3} \draw (-0.1, \y) -- (0.1, \y); + + \draw (0, 1) -- (3, 1); + \draw (1, 2) -- (4, 2); + \draw (2, 3) -- (4, 3); + \draw (1, 2) -- (1, 0); + \draw (2, 3) -- (2, 0); + \draw (3, 3) -- (3, 1); + \draw (4, 3) -- (4, 2); + + \node at (4.5, 2.5) {$\cdots$}; + \node at (0.5, 0.5) {$1$}; + \node at (1.5, 0.5) {$-1$}; + \node at (1.5, 1.5) {$1$}; + \node at (2.5, 1.5) {$-1$}; + \node at (2.5, 2.5) {$1$}; + \node at (3.5, 2.5) {$-1$}; + + \draw[dashed] (0, 0.8) -- (5, 0.8) node[right] {$=0$}; + \draw[dashed] (0.4, 0) -- (0.4, 3) node[above] {$=1$}; + \draw[dashed] (1.8, 0) -- (1.8, 3) node[above] {$=0$}; + \end{tikzpicture} + \end{center} + \begin{align*} + \iint f(x, y) \dd x \dd y = 0 && \iint f(x, y) \dd y \dd x = 1 + \end{align*} + But the integral $\int f \dd\lambda^2$ doesn't exist + \[ + \int \abs{f} \dd\lambda^2 = \sum_{n=0}^{\infty} 2 = \infty + \] +\end{eg} + +\begin{thm}[Tonelli's Theorem] + Let $f: \Omega \times \Phi \rightarrow [0, \infty)$ be measurable (in terms of $\setfam \otimes \setfamb$). + Then the functions + \[ + \omega \longmapsto f(\omega, \phi) + \] + are measurable for almost all $\phi \in \Phi$. Analogously + \[ + \phi \longmapsto f(\omega, \phi) + \] + is measurable for almost all $\omega \in \Omega$. + \begin{align*} + \phi &\longmapsto \int f(\omega, \phi) \dd\mu(\omega) \text{ measurable} \\ + \omega &\longmapsto \int f(\omega, \phi) \dd\mu(\phi) \text{ measurable} + \end{align*} + and + \begin{align*} + \int f(\omega, \phi) \dd (\mu \otimes \nu) (\omega, \phi) &= \iint f(\omega, \phi) \dd\mu(\omega) \dd\nu(\phi) \\ + &= \iint f(\omega, \phi) \dd\nu(\phi) \dd\mu(\omega) + \end{align*} + Furthermore, $f$ is integrable in terms of $\mu \otimes \nu$ is one of the above integrals is finite. +\end{thm} +\begin{proof} + Without proof. +\end{proof} + +\begin{cor}[Cavalieri's Principle] + Let $A \subset \setfam \otimes \setfamb$. Define + \[ + A_{\omega} = \set[(\omega, \phi) \in A]{\phi \in \Phi} + \] + Then + \[ + \omega \longmapsto \nu(A_{\omega}) + \] + is measurable and + \[ + (\mu \otimes \nu)(A) = \int \nu(A_{\omega}) \dd\mu(A_{\omega}) + \] +\end{cor} +\begin{proof} + It is easy to see + \begin{equation} + (\omega, \phi) \in A \iff \phi \in A_{\omega} + \end{equation} + Thus we can see + \begin{equation} + \charfun_A(\omega, \phi) = \charfun_{A_{\omega}} (\phi) + \end{equation} + And then + \begin{equation} + \begin{split} + (\mu \otimes \nu)(A) &= \int \charfun_A(\omega, \phi) \dd{(\mu \otimes \nu)}(\omega, \phi) \\ + &= \iint \underbrace{\charfun_A (\omega, \phi)}_{\charfun_{A_{\omega}}(\phi)} \dd{\nu}(\phi) \dd{\mu}(\omega) \\ + &= \int \nu(A_{\omega}) \dd{\mu}(\omega) + \end{split} + \end{equation} +\end{proof} + +\begin{thm}[Fubini's Theorem] + Let $f: \Omega \times \Phi \rightarrow \field$ be measurable with measures $\mu, \nu$, which is integrable in terms of $\mu \otimes \nu$. + Then the functions $\omega \mapsto f(\omega, \phi)$ are measurable and integrable for $\nu$-almost every $\phi \in \Phi$, + and the functions $\phi \mapsto f(\omega, \phi)$ are measurable and integrable for $\mu$-almost every $\omega \in \Omega$. + The functions + \begin{align*} + \omega &\longmapsto \int f(\omega, \phi) \dd{\nu}(\phi) && \phi \longmapsto \int f(\omega, \phi) \dd{\mu}(\omega) + \end{align*} + are measurable and integrable, and + \begin{align*} + \int f(\omega, \phi) \dd{(\mu \otimes \nu)} &= \iint f(\omega, \phi) \dd{\nu}(\phi) \dd{\mu}(\omega) \\ + &= \iint f(\omega, \phi) \dd{\mu}(\omega) \dd{\nu}(\phi) + \end{align*} +\end{thm} +\begin{proof} + Without proof. +\end{proof} + +\begin{cor} + Let $a_i, b_i \in \realn$, $a_i < b_i ~~\forall i \in \set{1, \cdots, n}$. + \[ + D = [a_1, b_1] \times [a_2, b_2] \times \cdots \times [a_n, b_n] + \] + Let $f: D \rightarrow \realn$ be continuous or bounded. Then + \[ + \int_D f \dd{\lambda^n} = \int\limits_{a_1}^{b_1} \cdots \int\limits_{a_n}^{b_n} f(x_1, \cdots, x_n) \dd{x_n} \cdots \dd{x_1} + \] + and the order of integration is irrelevant. +\end{cor} +\begin{proof} + $f$ is bounded by $k \in \realn$ (continuous with compact domain) + \begin{equation} + \int_D \abs{f} \dd{\lambda^n} \le \int_D k \dd{\lambda^n} = k \cdot (b_1 - a_1)(b_2 - a_2) \cdots (b_n - a_n) < \infty + \end{equation} + $f$ is $\lambda^n$-integrable. By applying Fubini's theorem the desired statement follows. +\end{proof} + +\begin{eg} + Calculate the center of mass of + \[ + A = \set[x \ge y^2 \wedge x \le 1]{(x, y) \in \realn^2} + \] + + \begin{center} + \begin{tikzpicture}[scale=2] + \draw[->, thick] (-0.5, 0) -- (3.5, 0); + \draw[->, thick] (0, -1.2) -- (0, 1.2); + + \draw[domain=-1:1, smooth, variable=\y] plot({3*\y*\y}, {\y}); + \draw[dashed] (3, -1) -- (3, 1); + \end{tikzpicture} + \end{center} + The center of mass is defined by + \[ + \int \begin{pmatrix} + x \\ y + \end{pmatrix} \underbrace{\dd{\lambda^2}(x, y)}_{\dd A} + \] + In component form this is + \begin{align*} + \int_A x \dd{\lambda^2}(x, y) &= \int x \charfun_A(x, y) \dd{\lambda^2}(x, y) \\ + &= \int_{[0, 1] \times [-1, 1]} x \charfun_A(x, y) \dd{\lambda^2}(x, y) \\ + &= \int_0^1 \int_{-1}^1 x \charfun_{[-\sqrt{x}, \sqrt{x}]}(y) \dd{y}\dd{x} \\ + &= \int_0^1 x \cdot 2 \cdot \sqrt{x} \dd{x} = \frac{4}{5} + \end{align*} + Meaning the center of mass is at $(\frac{4}{5}, 0)$. +\end{eg} +\end{document} \ No newline at end of file diff --git a/script.pdf b/script.pdf index feb5eaa..ce95a2c 100644 Binary files a/script.pdf and b/script.pdf differ diff --git a/script.tex b/script.tex index f0d3573..9f90fd8 100644 --- a/script.tex +++ b/script.tex @@ -91,6 +91,7 @@ \renewcommand{\kbrdelim}{)} \newcommand{\setfam}{\mathcal{A}} +\newcommand{\setfamb}{\mathcal{B}} \newcommand{\measureable}{(\Omega, \setfam)} \newcommand{\measure}{(\Omega, \setfam, \mu)} \newcommand{\intervals}{\mathcal{I}}