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\title{Problem Set\\ MTH 70200 \\ Real Analysis}
\date{February 20, 2024}

\theoremstyle{definition}
\newtheorem{prob}{Problem}

\begin{document}
  \maketitle
  \begin{prob}
   (Quals Fall 2016) Prove that an arbitrary collection of pairwise disjoint Lebesgue measurable subsets of 
   $\RR$, each of which has positive Lebesgue measure, is at most countable.  
  \end{prob}
  
  \begin{prob} (Folland 2.3) Let $(f_n)_{n\geq 1}$ be a sequence of measurable functions on $X$.
    Show that $\{x\colon \text{$\lim_{n\to\infty}f_n(x)$ exists}\}$ is a measurable set.
  \end{prob}
  \begin{prob} (Folland 2.8) Show that if $f\colon\RR\to\RR$ is monotone, then $f$ is
    Borel measurable.
  \end{prob}
  \begin{prob} (Folland 2.9, parts (a), (b), and (c) only)
    Let $f\colon[0,1]\to[0,1]$ be the Cantor function. You can use the properties proven about it
    at the end of Section~1.5 or in class, e.g., that it is increasing and locally constant
    on $C^c$, where $C$ is the Cantor set. Let $g(x)=f(x)+x$.
    \begin{enumerate}[a.]
      \item Prove that $g$ is a bijection from $[0,1]$ to $[0,2]$, and that $h=g^{-1}$
        is continuous from $[0,2]$ to $[0,1]$.
      \item Prove that $m(g(C))=1$, where $m$ is Lebesgue measure.
    \end{enumerate}
      Let $A\subseteq g(C)$ be a Lebesgue nonmeasurable set. You don't have to prove that
        $A$ exists. To see that it does, just take the nonmeasurable set we constructed on the first
        day of class and intersect it with $g(C)$. If the resulting set were measurable, then
        $g(C)$ would be the union
        of countably many disjoint translates of it, a contradiction since $g(C)$ has strictly positive
        but noninfinite Lebesgue measure.
    \begin{enumerate}[a.,resume]
      \item Let $B=g^{-1}(A)$. Prove that $B$ is Lebesgue measurable but not Borel.
    \end{enumerate}
    \textit{Commentary:} The result of this problem is that $h$ is a continuous function and  $h^{-1}(B)$
    is nonmeasurable. That is, $h$ pulls back a Lebesgue measurable set to set that is not
    Lebesgue measurable. This illustrates why we take the $\sigma$-algebra for the codomain
    of real-valued functions to be the Borel $\sigma$-algebra rather than the larger
    set of Lebesgue measurable sets. Otherwise even continuous functions could be nonmeasurable.
  \end{prob}
  
  \begin{prob}
    Let $f\colon\RR\to\RR$ be bi-Lipschitz, meaning that there exists $C>0$ such that for all $x,y\in\RR$,
    \begin{align*}
      C^{-1}\abs{x-y}\leq\abs{f(x)-f(y)}\leq C\abs{x-y}.
    \end{align*}
    Show that if $E$ is Lebesgue measurable, then $f^{-1}(E)$ is Lebesgue measurable.    
    (You can take for granted that the Lebesgue measurable sets are the completion of the Borel
    sets with respect to Lebesgue measure---we never proved this in class but it's a consequence
    of some straightforward exercises from Chapter~1.4. Thus every Lebesgue measurable set 
    can be represented as the union of a Borel measurable set and a set of Lebesgue measure zero.)
    \textit{Commentary:} This problem shows that the concern of continuous functions pulling back
    measurable sets to nonmeasurable sets doesn't happen with a more stringent continuity condition.
  \end{prob}
\begin{prob}
  Let $(X,\mathcal{M},\mu)$ be a complete measure space. Show that if $f$ is measurable
  and $f=g$ $\mu$-a.e., then $g$ is measurable.
\end{prob}

  
\end{document}

2.3, 2.8, 2.9abc
