\classheader{2012-10-02}

The course will cover a mixture of real analysis and probability. Homeworks will be due on the Thursday of the following week. Homeworks will be 10\% of the grade and the midterm and final together will be the other 90\%. The exams will be in class.

The abstract setting for measure theory is as follows. We have a set $X$, with power set $P(X)$. A subset $\cal{A}\subseteq P(X)$ is called a $\sigma$-algebra when
\begin{enumerate}
\item $\varnothing\in \cal{A}$,
\item for any $A\in \cal{A}$, its complement $(X\setminus A)\in \cal{A}$, and 
\item for any $A_1,A_2,\ldots,\in \cal{A}$, their union $\bigcup_{n=1}^\infty A_n\in \cal{A}$.
\end{enumerate}
Some immediate consequences are:
\begin{itemize}
\item We also have $X\in\cal{A}$.
\item $\cal{A}$ is also closed under countable intersections; for any $A_1,A_2,\ldots\in\cal{A}$, $\bigcap_{n=1}^\infty A_n\in\cal{A}$. This is because
\[X\setminus\bigcap_{n=1}^\infty A_n=\bigcup_{n=1}^\infty(X\setminus A_n).\]
\item If $A,B\in\cal{A}$, then $A\setminus B\in\cal{A}$ because
\[A\setminus B= A\cap (X\setminus B).\]
\end{itemize}
The most extreme cases of $\sigma$-algebras are $\{\varnothing,X\}$ and $P(X)$.

A more interesting example is when $X$ is a topological space and $\cal{A}$ is the $\sigma$-algebra generated by all open sets of $X$. This is called the Borel $\sigma$-algebra on $X$.

The $\sigma$-algebra generated by a collection of sets is the smallest $\sigma$-algebra that contains all of those sets. This can be constructed by considering all of the $\sigma$-algebras containing those sets, and taking their intersection. The intersection of $\sigma$-algebras can easily be seen to be a $\sigma$-algebra.

Let's consider a fundamental topological space, $\R$. What can we say about Borel sets in $\R$?
\begin{itemize}
\item Open sets are Borel.
\item Closed sets, being complements of open sets, are Borel.
\item Countable intersections of open sets (called $G_\delta$ sets) are Borel.
\item Countable unions of closed sets (called $F_\sigma$ sets) are Borel.
\end{itemize}
Some examples of $F_\sigma$ sets that are neither open nor closed are $[0,1)$ and $\Q$. Their complements are necessarily $G_\delta$ sets, and will also be neither open nor closed.


\begin{homework}
Is $\Q$ a $G_\delta$ set?
\end{homework}
Continuing our list of Borel sets in $\R$,
\begin{itemize}
\item Countable unions of $G_\delta$ sets (called $G_{\delta\sigma}$ sets) are Borel.
\item Countable intersections of $F_\sigma$ sets (called $F_{\sigma\delta}$ sets) are Borel.
\item Countable intersections of $G_{\delta\sigma}$ sets (called $G_{\delta\sigma\delta}$ sets) are Borel.
\item Countable unions of $F_{\sigma\delta}$ sets (called $F_{\sigma\delta\sigma}$ sets) are Borel.
\item $\cdots$
\end{itemize}
It is a theorem that each of these new classes is strictly bigger than the previous one; there is no finite step when we get no new sets. We do not even get all Borel sets when we look at all sequences of $\delta$'s and $\sigma$'s. We must go all the way to $\omega_1$ (the first uncountable ordinal) to get all Borel sets. Most of the time, we do not think beyond the first few steps here, but in descriptive set theory this is studied in more detail.

A pair $(X,\cal{A}$) of a set $X$ together with a $\sigma$-algebra $\cal{A}$ on $X$ is called a measure space (more precisely, a measurable space). The elements of $\cal{A}$ are called $\cal{A}$-measurable sets, or just measurable sets if the $\sigma$-algebra is understood.

\begin{homework}
What is the $\sigma$-algebra generated by the half-open intervals $[a,b)$? How is it related to the Borel $\sigma$-algebra - smaller, bigger, not comparable?
\end{homework}

\begin{definition}
Given two measurable spaces $(X,\cal{A})$ and $(Y,\cal{B})$, a map $f:X\to Y$ is called measurable if $f^{-1}(B)\in\cal{A}$ for all $B\in\cal{B}$.
\end{definition}

Since it is difficult to understand what the Borel sets of $\R$ are, this would seem to be a difficult condition to check. But in fact, when $(Y,\cal{B})=(\R,\text{Borel})$, a map $f:X\to \R$ is $\cal{A}$-measurable if
\[\{x\in X\mid f(x)<c\}\in\cal{A}\quad\text{ and }\quad \{x\in X\mid f(x)>c\}\in\cal{A}\]
for all $c\in\R$. This is because the open half-lines already generate the entire Borel $\sigma$-algebra. More generally, if $\cal{C}$ is a collection of subsets of $Y$ that generate the $\sigma$-algebra $\cal{B}$, then a map $f:X\to Y$ is measurable if $f^{-1}(C)\in\cal{A}$ for all $C\in\cal{C}$.

What about the image of Borel sets, instead of preimages? Is the image of a Borel set Borel?

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The projection of an open set is open, and the projection of a countable union is the countable union of the projections. However, as you can see in the above image, the projection of the complement need not be the complement of the projection. Lebesgue famously made the mistake of assuming the projection of any Borel set is Borel, but in fact there is a $G_\delta$ set in $[0,1]^2$ whose projection is not Borel.

Given a measurable space $(X,\cal{A})$, and Borel measurable functions $f_1,f_2:X\to\R$, then
\[f_1+f_2,\quad f_1-f_2,\quad f_1\cdot f_2,\quad f_1/f_2\]
are all Borel measurable (the last, of course, under the assumption that $f_2(x)\neq 0$ for all $x$).
\begin{proof}
The function $f_1+f_2$ can be obtained as the composition
\[X\xrightarrow{\;\;F=(f_1,f_2)\;\;}\R\times\R\xrightarrow{\;+\;}\R\]
The composition of measurable functions is measurable, so it suffices to show that $F$ and $+$ are measurable. The function $+$ is measurable (in fact, it is continuous). Because sets of the form $G_1\times G_2$, where $G_1,G_2\subseteq\R$ are open, generate the Borel $\sigma$-algebra on $\R\times\R$, it is enough to show that $F^{-1}(G_1\times G_2)\in\cal{A}$ for any open $G_1,G_2\subseteq\R$. But this is clear, because
\[F^{-1}(G_1\times G_2)=f_1^{-1}(G_1)\cap f^{-1}(G_2).\qedhere\]
\end{proof}
\begin{definition}
The extended real line is $\overline{\R}=\R\cup\{\pm\infty\}$. The open balls around $+\infty$ are sets of the form $(a,\infty]$ and the open balls around $-\infty$ are sets of the form $[-\infty,a)$.
\end{definition}
\begin{homework}
If $f_1,f_2,\ldots:X\to\R$ are Borel measurable, prove that $g=\sup(f_n):X\to\overline{\R}$ is Borel measurable.
\end{homework}
