\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

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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.

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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 $f_n(x)<f(x)$ for all $n$. We will need a different argument.

Given a simple function $g\leq f$, let its values be $c_1,\ldots,c_N$. For any $\epsilon>0$ 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}




