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\newenvironment{sol}[0]{\begin{solution}\setlength{\parskip}{1em}}{\end{solution}}

\begin{document}

\begin{center}
\begin{large}
Math 70100: Functions of a Real Variable I \\
Homework 10, due Wednesday, November 19th.
\end{large}
\end{center}

{\bf Name:} Insert your name here.
\vspace{1em}

\noindent
{\bf Remark on conventions:} By convention, a {\em measurable set} in $\R$ is a Lebesgue measurable set. If $E \subset \R$ is (Lebesgue) measurable and $f:E \to \R$ is a function, then
by convention $f$ is {\em measurable} if it is $({\mathcal L}, {\mathcal B})$-measurable, i.e.,
if the preimage of every Borel measurable set in $\R$ is a Lebesgue measurable subset of $E$.

\begin{questions}
\question {\em (Modified from Royden-Fitzpatrick, \S 2.7 \# 46)} Let $X$ and $Y$ be topological spaces. Prove that every continuous function $f:X \to Y$ is Borel measurable. That is, prove that the preimage
of a Borel set in $Y$ is a Borel set in $X$. ({\em Hint:} The collection of sets $E$ for which $f^{-1}(E)$ is Borel is a $\sigma$-algebra containing the open sets.)
\begin{sol}
Insert answer here.
\end{sol}

\question Let $E \subset \R$ be Lebesgue measurable and $f:E \to \R$ be a function. Let $g:E \to \R$ be another function. We say $f=g$ {\em almost everywhere} if there is a subset $Z \subset E$ of Lebesgue measure zero so that $f(x)=g(x)$ for all $x \in E \smallsetminus Z$. 
Show that if $f=g$ almost everywhere and $g$ is measurable, then $f$ is measurable.
\begin{sol}
Insert answer here.
\end{sol}

\question Let $\{E_i \subset \R\}$ be a countable collection of measurable sets,
and let $E=\bigcup_i E_i$. Let $f:E \to \R$ be a function. 
Show that $f$ is measurable if and only if $f|_{E_i}$ is measurable for each $i$.
\begin{sol}
Insert answer here.
\end{sol}

\question {\em (Royden-Fitzpatrick \S 3.1 \#9)} Let $f_n$ be a sequence of measurable functions defined on a measurable set $E \subset \R$. Define $E_0$ to
be the set of points $x$ in $E$ at which $\{f_n(x)\}$ converges. Is the set $E_0$ measurable?
If so, prove it. Otherwise, explain how to produce a counterexample.
\begin{sol}
Insert answer here.
\end{sol}

\question {\em (Modified from Pugh: Chapter 6 \# 54)}\\
{\bf Egoroff's theorem.} Let $E \subset \R$ be a measurable
set of finite Lebesgue measure, and let $\{f_n\}$ be a sequence of measurable functions which converges
almost everywhere (i.e., there is a set $Z \subset E$ of zero Lebesgue measure so that $\{f_n(x)\}$ converges when $x \in E \smallsetminus Z$). Then for each $\epsilon>0$ there is a measurable set $S \subset E$ with $\lambda(E \smallsetminus S)<\epsilon$ such that $\{f_n(x)\}$ converges uniformly for $x \in S$.

Prove Egoroff's theorem by using the following steps. {\em Setup:} Let $E \subset \R$ be a measurable
set of finite Lebesgue measure,
and suppose $\{f_n:E \to \R\}$ is a sequence of measurable functions which converges almost everywhere. Thus, there is a zero set $Z$
and a $f:E \smallsetminus Z \to \R$ so that $\{f_n(x)\}$ converges to $f(x)$ for $x \in E \smallsetminus Z$. 
\begin{parts}
\part For $k, \ell \in \N$, set 
$$X(k,\ell)=\{x \in E \smallsetminus Z~:~ \forall n \geq k, ~|f_n(x)-f(x)|<1/\ell\}.$$
Observe these sets are measurable. Show that for each $\ell$, $E\smallsetminus Z=\bigcup_k X(k,\ell)$.
\begin{sol}
Insert answer here.
\end{sol}
\part Given $\epsilon>0$, show that there is a sequence $\{k_\ell \in \R~:~ \ell \in \N\}$ so that
by defining $X_\ell=X(k_\ell,\ell)$, we have $\lambda(E \smallsetminus Z \smallsetminus X_\ell)<\frac{\epsilon}{2^\ell}.$
\begin{sol}
Insert answer here.
\end{sol}
\part Let $X=\bigcap_\ell X_\ell$. Show that $\lambda\big(E \smallsetminus X\big)<\epsilon$
and that $\{f_n|_X\}$ converges uniformly to $f|_X$. (Don't give two meanings to $\epsilon$!)
\begin{sol}
Insert answer here.
\end{sol}
\end{parts}

\question Show that Egoroff's theorem is not necessarily true when $E$ has infinite measure.
\begin{sol}
Insert answer here.
\end{sol}

\question {\em (Pugh, Chapter 6 \#50)} Construct a monotone function $f:[0,1] \to \R$ whose discontinuity set is exactly $\Q \cap [0,1]$
or show that such a function can not exist. Prove your answer is correct.
\begin{sol}
Insert answer here.
\end{sol}
\end{questions}

\end{document}
