\classheader{2012-10-04} \begin{definition} Given a set $X$ and a $\sigma$-algbera $\cal{A}$ on $X$, a function $\mu:\cal{A}\to[0,\infty]$ is a measure if \begin{enumerate} \item $\mu(\varnothing)=0$ \item ($\sigma$-additivity) For disjoint sets $A_1,A_2,\ldots$, \[\sum_{n=1}^\infty\mu(A_n)=\mu\left(\bigcup_{n=1}^\infty A_n\right)\] \end{enumerate} \end{definition} \begin{examples} $\text{}$ \begin{itemize} \item The counting measure on $X$: for any $\sigma$-algebra $\cal{A}\subseteq P(X)$, we define $\mu(A)=|A|$. \item The Dirac measure: given an $x\in X$, \[\delta_x(A)=\begin{cases}1 & \text{if }x\in A,\\ 0 &\text{if }x\notin A. \end{cases}\] \item An atomic measure is one of the form $\mu=\sum c_j\delta_{x_j}$, so that \[\mu(A)=\sum_{x_j\in A}c_j\] Note that any measure can be broken into an atomic part and a non-atomic part (i.e. a measure for which no points have positive measure). This is somewhat of a hint for the following homework: \end{itemize} \end{examples} \begin{homework} Is there a $\sigma$-algebra $\cal{A}$ which is countably infinite? \end{homework} Some properties of measures include: \begin{itemize} \item $\sigma$-additive $\implies$ additive (just take $A_j=\varnothing$ for large $j$) \item Monotonicity: if $A\subseteq B$, then $\mu(A)\leq\mu(B)$. Note that \[\mu(B)=\mu(A)+\mu(B\setminus A)\] but you should not write this as \[\mu(B\setminus A)=\mu(B)-\mu(A)\] because we could have $\mu(B)=\mu(A)=\infty$. \item Given $A_1\subseteq A_2\subseteq \cdots $, \[\mu\left(\bigcup_{n=1}^\infty A_n\right)=\lim_{n\to\infty}\mu(A_n).\] \begin{proof} The sets $A_i$ are not disjoint, but consider instead the sets $A_1,A_2\setminus A_1, A_3\setminus A_2,\ldots$ which are disjoint. \newpage Then \[\bigcup_{n=1}^\infty A_n=A_1\cup\bigcup_{n=1}^\infty (A_{n+1}\setminus A_n)\] so \begin{align*} \mu\left(\bigcup_{n=1}^\infty A_n\right)&=\mu(A_1)+\mu(A_2\setminus A_1)+\mu(A_3\setminus A_2)+\cdots\\ & = \lim_{n\to\infty}(\mu(A_1)+\mu(A_2\setminus A_1)+\cdots+\mu(A_{n}\setminus A_{n-1}))\\ & = \lim_{n\to\infty} \mu(A_n) \end{align*} \end{proof} \item What about when we have $A_1\supseteq A_2\supseteq \cdots$? It is not true in general that \[\mu\left(\bigcap_{n=1}^\infty A_n\right)=\lim_{n\to\infty}\mu(A_n).\] For example, consider the counting measure on $\N$, and $A_1=\N$, $A_2=\{2,3,\ldots,\}$, $A_3=\{3,4,\ldots\}$, etc. Then their intersection is $\varnothing$ which has $\mu(\varnothing)=0$, even though $\mu(A_n)=\infty$ for all $n$. Howver, if there is at least one $n$ where $\mu(A_n)$ is finite, then the statement is true. \begin{proof} Without loss of generality, we can assume that $\mu(A_1)$ is finite. Then \[\underbrace{\mu(A_1\setminus A_2)+\mu(A_2\setminus A_3)+\cdots}_{\displaystyle\lim_{n\to\infty}\mu(A_1\setminus A_n)}+\mu\left(\bigcap_{n=1}^\infty A_n\right)=\mu(A_1).\] Because $\mu(A_1)$ is finite, we \textit{can} write $\mu(A_1\setminus A_n)=\mu(A_1)-\mu(A_n)$, so \[\lim_{n\to\infty}\mu(A_1\setminus A_n)=\lim_{n\to\infty}(\mu(A_1)-\mu(A_n))\] and therefore \[\mu(A_1)-\lim_{n\to\infty}\mu(A_n)+\mu\left(\bigcap_{n=1}^\infty A_n\right)=\mu(A_1).\] \end{proof} \end{itemize} Let $X$ be a compact metric space and $\cal{A}$ the Borel $\sigma$-algebra on $X$. Is a finite Borel measure $\mu$ determined by the measure of the balls? \begin{homework}[$\ast$] The answer is no; find an example. Federer wasn't able to find an example, after thinking about it for a day, but the construction is simple once you see it. \end{homework} We say that $f:X\to\R$ is simple if it is measurable and takes only finitely many values. Equivalently, there are some disjoint sets $A_1,\ldots,A_n$ and $c_j\in\R$ such that \[f=\sum_{j=1}^n c_j\chi_{A_j}.\] \newpage You can visualize this as \begin{center} \begin{tikzpicture}[scale=0.8] \begin{scope}[every node/.style={inner sep=0pt,outer sep=0pt}] \node (a1) at (0,1) {}; \node (a2) at (1,1) {}; \node (b1) at (1,2) {}; \node (b2) at (2,2) {}; \node (c1) at (2,3) {}; \node (c2) at (3,3) {}; \node (d1) at (3,4) {}; \node (d2) at (4,4) {}; \node (l1) at (0.5,1.15) [label={90:$c_1$}] {}; \node (l2) at (1.5,2.15) [label={90:$c_2$}] {}; \node (l3) at (2.5,3.15) [label={90:$c_3$}] {}; \node (l4) at (3.5,4.15) [label={90:$c_4$}] {}; \end{scope} \fill[gray!0] (0,0) rectangle (a2); \fill[gray!0] (b2) rectangle (1,0); \fill[gray!0] (c2) rectangle (2,0); \fill[gray!0] (d2) rectangle (3,0); \draw[thick] (0,5) -- (0,0) -- (5,0); \draw[thick] (d2.center) -- (d1.center) -- (c2.center) -- (c1.center) -- (b2.center) -- (b1.center) -- (a2.center) -- (a1.center); \draw[thick] (a2.center) -- (1,0); \draw[thick] (b2.center) -- (2,0); \draw[thick] (c2.center) -- (3,0); \draw[thick] (d2.center) -- (4,0); \node (w) at (0.5,-0.0) [label={270:$A_1$}] {}; \node (x) at (1.5,-0.0) [label={270:$A_2$}] {}; \node (y) at (2.5,-0.0) [label={270:$A_3$}] {}; \node (z) at (3.5,-0.0) [label={270:$A_4$}] {}; \end{tikzpicture} \end{center} We define for a simple function $f\geq 0$ \[\int_A f\,d\mu=\sum_{j=1}^n c_j\mu(A\cap A_j).\] If $g\geq 0$ is an arbitrary measurable function, then we define \[\int_Ag\,d\mu=\sup_{\substack{f\text{ simple}\\ f\leq g}}\;\int_A f\,d\mu.\] Thus, we are approximating our function $f$ from below by simple functions. \begin{center} \begin{tikzpicture}[scale=0.8] \begin{scope} \clip (0,0) rectangle (5,5); \draw[red,very thick,line cap=round] plot [smooth] coordinates {(0,1.7) (0.5,1.3) (1,2.1) (1.5,2.3) (2,3.1) (2.5,4.2) (3,4.1) (3.5,4.3) (4,4.1)}; \end{scope} \begin{scope}[every node/.style={inner sep=0pt,outer sep=0pt}] \node (a1) at (0,1) {}; \node (a2) at (1,1) {}; \node (b1) at (1,2) {}; \node (b2) at (2,2) {}; \node (c1) at (2,3) {}; \node (c2) at (3,3) {}; \node (d1) at (3,4) {}; \node (d2) at (4,4) {}; %\node (l1) at (0.5,1.15) [label={90:$c_1$}] {}; %\node (l2) at (1.5,2.15) [label={90:$c_2$}] {}; %\node (l3) at (2.5,3.15) [label={90:$c_3$}] {}; %\node (l4) at (3.5,4.15) [label={90:$c_4$}] {}; \end{scope} \fill[gray!40] (0,0) rectangle (a2); \fill[gray!40] (b2) rectangle (1,0); \fill[gray!40] (c2) rectangle (2,0); \fill[gray!40] (d2) rectangle (3,0); \draw[thick] (0,5) -- (0,0) -- (5,0); \draw[thick] (d2.center) -- (d1.center) -- (c2.center) -- (c1.center) -- (b2.center) -- (b1.center) -- (a2.center) -- (a1.center); \draw[thick] (a2.center) -- (1,0); \draw[thick] (b2.center) -- (2,0); \draw[thick] (c2.center) -- (3,0); \draw[thick,line cap=rect] (d2.center) -- (4,0); %\node (w) at (0.5,-0.0) [label={270:$A_1$}] {}; %\node (x) at (1.5,-0.0) [label={270:$A_2$}] {}; %\node (y) at (2.5,-0.0) [label={270:$A_3$}] {}; %\node (z) at (3.5,-0.0) [label={270:$A_4$}] {}; \end{tikzpicture} \end{center} We need to restrict to nonnegative functions, even though $\pm\infty$ are not being allowed as values of our functions in this definition, because (for example) we cannot integrate $g(x)=\frac{1}{x}$ on $\R$ with this definition; there is no simple function $f\leq g$. This illustrates the difference between being finite everywhere and being bounded. For an arbitrary measurable function $g$, we set \[g^+=\max(0,g),\quad g^-=-\min(0,g)\] so that $g=g^+-g^-$ and then define \[\int_A g\,d\mu = \int_A g^-\,d\mu-\int_A g^-\,d\mu.\] From now on, when I write a function, it will be assumed to be measurable, even if I don't say so. Given a measure $\mu$ and an $f\geq 0$, we can make a new measure $\nu$ defined by \[\nu(A)=\int_Af\,d\mu.\] \begin{theorem}[Monotone Convergence Theorem] Given a sequence of functions \[0\leq f_1\leq f_2\leq f_3\leq \cdots\] then for $f=\lim_{n\to\infty}f_n$, \[\int f\,d\mu=\lim_{n\to\infty} \int f_n\,d\mu.\] \end{theorem} \begin{proof} It is easy to see that \[\int f_1\leq \int f_2\leq\int f_3\leq\cdots\] so there is some $\alpha=\lim_{n\to\infty} \int f_n$. We need to show that $\alpha\leq\int f$ and $\alpha\geq \int f$. The former is trivial because $f_n\leq f$ for all $n$. For the latter, consider the definition of the integral. We want to show that for any simple $g\leq f$, we have $\int g\leq \alpha$. Unfortunately, it is not true that for any such $g$, there is an $n$ such that $g\leq f_m$ for all $m\geq n$. It isn't even true pointwise, since we could have $g(x)=f(x)$ and $f_n(x)0$ that is smaller than all the $c_i$, define \[g_\epsilon=\begin{cases} 0&\text{if }g=0,\\ g-\epsilon & \text{if }g>0.\end{cases}\] As $\epsilon\to 0$, we have that $\int g_\epsilon \to \int g$, so in fact it is enough to show that $\int g_\epsilon \leq \alpha$ for all $\epsilon$. Define \[B_n=\{x\in X\mid f_n(x)\geq g_\epsilon(x)\}.\] We have that $\bigcup B_n = X$, and that $B_1\subseteq B_2\subseteq\cdots$. For any given $\epsilon>0$, define a measure $\nu$ by \[\nu(A)=\int_A g_\epsilon.\] Then \[\nu(X)=\nu\left(\bigcup B_n\right)=\lim_{n\to\infty}\nu(B_n)\] and \[\nu(B_n)=\int_{B_n} g_\epsilon\leq\int_{B_n} f_n\leq \int_X f_n.\] Thus, \[\int g_\epsilon=\nu(X)=\lim_{n\to\infty}\nu(B_n)\leq\lim_{n\to\infty} \int_X f_n = \alpha.\qedhere\] \end{proof} \begin{corollary}[Beppo-Levi] Given a sequence of functions $f_n\geq 0$, then \[\int \sum_{n=1}^\infty f_n=\sum_{n=1}^\infty \int f_n.\] \end{corollary} \begin{proof} \[\sum_{n=1}^\infty \int f_n = \lim_{N\to\infty} \sum_{n=1}^N \int f_n=\lim_{N\to\infty}\int \sum_{n=1}^N f_n \stackrel{\text{MCT}}{=}\int\lim_{N\to\infty}\sum_{n=1}^N f_n=\int \sum_{n=1}^\infty f_n.\qedhere\] \end{proof} \begin{corollary}[Fatou's Lemma] Given a sequence of functions $f_n\geq 0$, \[\int \liminf_{n\to\infty} f_n\leq\liminf_{n\to\infty} \int f_n.\] \end{corollary} \begin{homework} Find an example where this inequality is strict. \end{homework} \begin{proof} By definition, \[\liminf_{n\to\infty} f_n = \lim_{n\to\infty}\; \underbrace{\inf\{f_n,f_{n+1},f_{n+2},\ldots\}}_{g_n}.\] We have that $0\leq g_n\leq g_{n+1}\leq\cdots$ and therefore \[\int\liminf_{n\to\infty} f_n =\int \lim_{n\to\infty} g_n = \lim_{n\to\infty} \int g_n\leq \liminf_{n\to\infty}\int f_n\] because $g_n\leq f_n$. \end{proof} \begin{corollary}[Lebesgue Theorem] Given a sequence of functions $f_n$ such that $|f_n|\leq g$, where $g$ is a function such that $\int g<\infty$, \[\int \lim_{n\to\infty} f_n = \lim_{n\to\infty} \int f_n.\] \end{corollary} \begin{homework} Find a sequence of functions $f_n$ converging pointwise to a function $f$ which do not satisfy the conclusion of this theorem. \end{homework} \begin{proof} Let $f=\lim_{n\to\infty} f_n$. In fact, a stronger statement is true: as $n\to\infty$, \[\left|\int f_n - \int f\right|\leq\int |f_n-f| \to 0.\] We have that $|f_n-f|\leq 2g$. Let $h_n=2g-|f_n-f|$, so that $h_n\geq 0$. Apply Fatou: \[\int\underbrace{\liminf_{n\to\infty} h_n}_{2g}\leq \liminf_{n\to\infty} \int h_n.\] Therefore \[\int 2g\leq\liminf_{n\to\infty}\left(\int 2g-|f_n-f|\right)=\int 2g +\liminf_{n\to\infty}\left(-\int |f_n-f|\right).\] Therefore \[\limsup_{n\to\infty} \left(\int |f_n-f|\right)\leq 0\] which implies that in fact \[\int |f_n-f|\to 0.\qedhere\] \end{proof}