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\begin{document}
\begin{center}
  \Large\bfseries Math 73/103 Assignment Three \\
Due Date TBA
\end{center}
\bigskip
\textsc{Clarification:}  Let's review of notation and terminology.   Lebesgue measure, $(\R,\M,m)$, is the complete measure
coming 
from the explicit outer measure $m^{*}$ we defined in lecture.  In
particular, $\M$ is the $\sigma$-algebra of all $m^{*}$-measurable
sets.  A
Lebesgue measurable function $f:\R\to \C$ is just a function such that
$f^{-1}(V)\in\M$ for any open set $V\subset \C$.  We say $f$ is Borel
if $f^{-1}(V)$ is a Borel set in $\R$ for every open set $V$.  We say
$f\in\L^{1}(\R,\M,m)$, or that $f$ is Lebesgue integrable, if $f$ is
measurable and $\int_{\R}|f|\,dm<\infty$.  We have also used the
notation $L^{+}(\R,\M,m)$ for the collection Lebesgue measurable
functions $f$ such that $f\ge 0$ everywhere.

\begin{ques}
  Suppose that $f\in\L^{1}(X,\M,m)$ is a Lebesgue integrable function
  on the real line.  Let $\epsilon>0$.  Show that there is a
  continuous function $g$ that vanishes outside a bounded interval
  such that $\|f-g\|_{1}<\epsilon$.  (Hint: this is easy if $f$ is the
  characteristic function of a bounded interval: draw a
  picture.)\label{ques:1}
\end{ques}

\bigbreak

Another of Littlewood's Principles is that a pointwise convergent
sequence of functions is nearly uniformly convergent.  This is also
known as ``Egoroff's Theorem''.

\begin{ques}
  Prove Egoroff's Theorem: Suppose that $(X,\M,\mu)$ is a finite
  measure space --- that is, $\mu(X)<\infty$.  Suppose that
  $\sset{f_{n}}$ is a sequence of measurable functions converging
  pointwise to a
  measurable function $f:X\to \C$.  Then for all $\epsilon>0$ there is
  a set $E\in\M$ such that $\mu(X\setminus E)<\epsilon$ and $f_{n}\to
  f$ uniformly on $E$.  

  Some suggestions:
  \begin{enumerate}
  \item There is no harm in assuming that $f_{n}\to f$ everywhere.
  \item Let $E_{n}(k)=\bigcup_{m=n}^{\infty} \set{x\in
      X:|f_{m}(x)-f(x)|\ge \frac1k}$.
  \item Show that $\lim_{n\to\infty}\mu(E_{n}(k))=0$.  (You need
    $\mu(X)<\infty$ here.)
  \item Fix $\epsilon>0$ and $k$.  Choose $n_{k}\ge n$ so that
    $\mu(E_{n_{k}}(k))< \frac\epsilon{2^{-k}}$, and let
    \begin{equation*}
      E=\bigcup_{k=1}^{\infty} E_{n_{k}}(k).
    \end{equation*}
  \end{enumerate}
\end{ques}

\bigbreak

Fortunately, Littlewood only had three principles.  The final one is
that every measurable function is nearly continuous.  This is known as
``Lusin's Theorem''.

\begin{ques}
  Prove Lusin's Theorem: Suppose that $f$ is a Lebesgue measurable
  function on $[a,b]\subset \mathbf R$.  Given $\epsilon>0$, show that
  there is a closed subset $K\subset [a,b]$ such that
  $m([a,b]\setminus K)<\epsilon$ and that $f\restr K$ is continuous.
  (Suggestion use problem~\ref{ques:1}, Egoroff's Theorem and that
  fact that the uniform limit of continuous functions is continuous.)
\end{ques}


\begin{ques}
  Recall that a sequence $\sset{f_{n}}$ of measurable functions from
  $(X,\M)$ to $\C$ converges in measure to a measurable function
  $f:X\to\C$ if for all $\epsilon>0$ we have $\lim_{n\to \infty}
  \mu\bigl(E_{n}(\epsilon)\bigr)=0$ where
  \begin{equation*}
    E_{n}(\epsilon)=\set{x\in X:|f_{n}(x)-f(x)|\ge\epsilon}.
  \end{equation*}
Show that as claimed in lecture that if $\sset{f_{n}}$ converges to
$f$ in measure then $\sset{f_{n}}$ has a subsequence
$\sset{f_{n_{k}}}$ which converges to $f$ almost everywhere.

Some suggestions:
\begin{enumerate}
\item Let $n_{k}$ be such that $n\ge n_{k}$ implies
  $\mu(E_{n}(2^{-k}))<2^{-k}$.
\item Let $E_{k}=E_{n_{k}}(2^{-k})$ and $G_{k}=\bigcup_{m\ge k}E_{m}$.
\item Show that $f_{n_{k}}(x)\to f(x)$ if $x\notin
  A:=\bigcup_{k=1}^{\infty} G_{k}$.
\end{enumerate}

\end{ques}

\begin{ques}
  Suppose that $\rho$ is a premeasure on an algebra $\A$ of sets in
  $X$.  Let $\rho^{*}$ be the associated outer measure.
  \begin{enumerate}
  \item Show that $\rho^{*}(E)=\rho(E)$ for all $E\in\A$.
  \item If $\M^{*}$ is the $\sigma$-algebra of $\rho^{*}$-measurable
    sets, show that $\A\subset \M^{*}$.
  \end{enumerate}

\end{ques}

\begin{ques}
  Let $m$ be Lebesgue measure on $[0,1]$ and let $\mu$ be counting
  measure.  Clearly, $m<\!\!<\mu$.  Show that there is no function $f$
  satisfying the conclusion of the Radon-Nikodym Theorem.  Why is this
  not a counter-example to the Radon-Nikodym Theorem.
\end{ques}

\begin{ques}
  Suppose that $f_{n}\to f$ in measure and that there is a $g\in
  \L^{1}(X,\M,\mu)$ is such that $|f_{n}(x)|\le g(x)$ for all $x\in
  X$.  Show that $f_{n}\to f$ in $L^{1}(X,\M,\mu)$.
\end{ques}

\bibliography{references-nov01}
\end{document}


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