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\coverheader{\large R.~de Santiago}{\large Math 561}{\large }
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\begin{document}
%%%%% COVER PAGE %%%%%
\begin{center} California State University, Long Beach \\
Department of Mathematics and Statistics \vspace{1cm} \\

{\Large \bf Real Analysis Comprehensive Examination } \\
Sep.~2025 \vspace{1cm} \\

{\bf Choose six of the nine problems, and mark the problems that you want graded on the table below. }\vspace{1cm}\\
%\vsword{}
%\gradetable
\begin{tabular}{ |c|c| } 
 \hline
 Question & To Be Graded (\checkmark) \\ \hline
 1 &   \\ \hline
 2 &   \\ \hline
  3 &   \\ \hline
   4 &   \\ \hline
 5 &   \\ \hline
  6 &   \\ \hline
 7 &   \\ \hline
  8 &   \\ \hline
 9 &   \\ \hline
 

\end{tabular}
\end{center}
Notation:
\begin{itemize}
    \item If $A\subseteq\R$,  $|A|$ denotes the Lebesgue Outer Measure of $A$.
    \item $\mathcal{B}$ denotes the Borel measurable subsets of $\R$
    \item $\lambda$ denotes the Lebesgue Measure on the Real line
    \item $(X,\mathcal{S},\mu)$ will denote a measure space
    \item $\mathcal{L}^1(X,\mathcal{S},\mu):=\set{ f \text{ measurable}  :  \int_X|f|\, d\mu<\infty}$
    \item $\|f\|_1:=\int_X |f|\, d\mu$
    
\end{itemize}
\newpage
\begin{questions}

\vspace{\stretch{1}}
\question If $E\subseteq \R$, we define 
$|E|_*:=\sup \set{|A|: A\subseteq E, \text{ and }A\text{ is closed and bounded}}$.  
    \begin{parts}
         \part Show that $|E|_*\leq |E|$ for all $E\subseteq \R$.
      \part Show that if $E$ is bounded and $E$ is Lebesgue measurable, then $|E|_*=|E|$.
      \part Show that if $|E|<\infty$  and $E$ is Lebesgue measurable, then $|E|_*= |E|$. Hint: Notice 
      $$|E|= \sum_{n=0}^\infty|E\cap ([-n-1,-n) \cup[n,n+1))|.$$
      \part Show that if $|E|=\infty$ and $E$ is Lebesgue measurable, then $|E|_*= |E|$. Hint: a finite union of closed sets is closed.
    \end{parts}
\vspace{\stretch{1}}
\question Let $(X,\mathcal{S},\mu)$ be a measure space.
    \begin{parts}
        \part Show that if $A\cap B=\emptyset$ and $A,B\in \mathcal{S}$, then $\mu(A\cup B)= \mu(A)+\mu(B)$.
        \part Show that if $A,B\in \mathcal{S}$, $A\subseteq B$, and $\mu(B)<\infty$, then $\mu(B\setminus A)=\mu(B)-\mu(A)$. Briefly explain why the hypothesis that $\mu(B)<\infty$ required here.
        \part Let $\set{A_i}_{i=1}^\infty$ be sets in $\mathcal{S}$ with $\mu(\cup_{i=1}^\infty A_i)<\infty$.  Show that
        \[\mu\left( \bigcap_{j=1}^\infty \bigcup_{i=1}^j A_i\right)\geq \limsup \mu(A_i).\]
    \end{parts}
\vspace{\stretch{1}}
\newpage
\question For $f\in \mathcal{L}^1([0,1], \lambda)$, we define $F:[0,1]\to \R$ by
\[F(x):=\int_{[0,x] }f \, d\lambda\]
    \begin{parts}
        \part Show that $F$ is continuous on $[0,1]$.
        \part Suppose that $f_n\in\mathcal{L}^1([0,1], \lambda)$ is a sequence such that $\lim_{n\to\infty}\|f_n-f\|_1=0$ and define
        \[F_n(x):=\int_{[0,x] }f_n \, d\lambda\]
        Show that $F_n\to F$ uniformly on $[0,1]$.
    \end{parts}
\vspace{\stretch{1}}
\question
    Let $(X,\mathcal{S},\mu)$ be a measure space with $\mu(X)=1$. Given $f\in \mathcal{L}^1(X,\mathcal{S},\mu)$, set
    \begin{itemize}
        \item $a=\int_X f\, d\mu$.
        \item $s=\sqrt{\int_X |f-a|^2\, d\mu}$.
    \end{itemize}
    \begin{parts}
        \part Prove that if $g:X\to \R$ is measurable, then
        for all $k>0$ \textcolor{red}{$t>0$} we have
        \[\mu(\set{x\in X: |g(x)|\geq t})\leq \frac{1}{t}\int_{X} |g|\, d\mu \]

        \part Show that for all $k>0$
        \[\mu(\set{x\in  X : |f(x)-a|\geq ks  } )\leq \frac{1}{k^2}\]
        Hint: $\set{x\in  X : |f(x)-a|\geq ks  }=\set{x\in  X : |f(x)-a|^2\geq k^2s^2  }$.
    \end{parts}
\vspace{\stretch{1}}
\question Let $(X,\mathcal{S},\mu)$  be a measure space and  $f_n,f:X\to \R$ measurable.
    \begin{parts}
        \part State Fatou's Lemma.
        \part State the Dominated Convergence Theorem.
        \part Derive the Dominated Convergence Theorem as a consequence of Fatou's lemma.
    \end{parts}
\vspace{\stretch{1}}\newpage
\question Compute
\[\lim_{n\to\infty } \int_0^n\left(1-\frac{x}{n}\right)^nx\, d\lambda(x)\]
You may freely use the fact that $|1-x/n|^n\leq e^{-x}$ for all $x\in [0,n]$.

\vspace{\stretch{1}}
%\textcolor{red}{My Solution\\
%Let $f(x)=e^{-x}x$.  Notice that $f(x)\geq0$ for all $x\in [0,\infty)$ and that (using integration by parts) $\int_{[0,\infty) }xe^{-x}\, d\lambda(x) =1$.  Thus, $f\in\mathcal{L}^1([0,\infty)]) $. Let $f_n(x)= (1-(x/n))^nx\chi_{[0,x]}(x)$. Then by the hint, $|f_n(x)|\leq f(x)$ for all $n$ and $x\in [0,\infty)$.  Since the pointwise limit of $f_n(x)= e^{-x}x$,  by the Lebesgue dominated convergence theorem, we see  that
%\[\lim_{n\to\infty}\int_0^n\left(1-\frac{x}{n}\right)^nx\, d\lambda(x)=\lim_{n\to\infty }\int_{[0,\infty)} f_n\, d\lambda=\int_{[0,\infty)} f\, d\lambda=1\]}


\question Let $f_n:[0,1]\to \R$ be the following function:
\[
f_n(x)= \begin{cases}
2n^2 x&  x\in [0,\frac{1}{2n}]\\
-2n^2x+2n & x\in (\frac{1}{2n},\frac{1}{n}]\\
0 & \text{otherwise}
\end{cases}
\]
\begin{center}


\includegraphics[scale=0.4,alt={A sequence of graphs, each in the shape of a triangle, The base of each triangle is from 0 to $1/n$, the vertex of the triangle is at $(\frac{1}{2n}, n)$ }]{graphsForSequencesofTriangles.png}

    
\end{center}
    \begin{parts}
        \part Does there exist a function $f$ so that $f_n\to f$ pointwise?
        \part Does there exist a function $f$ so that $f_n\to f$ pointwise almost everywhere?
        \part Does there exist a function $f$ so that $f_n\to f$ uniformly?
        \part Does there exist a function $f$ so that $f_n\to f$ in $\|\cdot\|_1$?
    \end{parts}
\vspace{\stretch{1}}

\question Let $(X,S,\mu)$ be a measure space.
\begin{parts}
    \part State Egoroff's theorem.

    \part Let $f_n,f:X\to \R$ be measurable \textcolor{red}{$f_n\to f$  pointwise} and $\mu(X)<\infty$.  Show that there exists a sequence of measurable sets $\set{E_k}_{k=1}^\infty$ so that
        \begin{itemize}
            \item $\mu(X\setminus \bigcup_{k=1}^\infty E_k)=0$, and
            \item $f_n\to f$ uniformly on $E_k$.
        \end{itemize}
  %  \textcolor{red}{Alternate question: Let $f_n,f:X\to \R$ be measurable and $X=\bigcup_{k=1}^\infty X_k $ where $\mu(X_k)<\infty$ for all $k\in\N$. Show that for all $\varepsilon>0$, there exists a a set $E\in S$ so that 
%    \begin{itemize}
%        \item $\mu(X\setminus E)<\varepsilon $, and 
%        \item $f_n\to f$ uniformly on $E$.
%    \end{itemize}}
\end{parts}
\vspace{\stretch{1}}
\end{questions}


\end{document}