\documentclass[hw]{dpwexam-new}
\usepackage{mathtools,paralist}
\mathtoolsset{showonlyrefs} 
%\usepackage{amsstdmathmacs}
\def\set#1{\lbrace\,#1\,\rbrace}
\def\R{\mathbf{R}}
\def\pp{\mathcal{P}}
\def\qq{\mathcal{Q}}
\def\rr{\mathcal{R}}
\def\mm{\mathcal{M}}
\def\lint{{{\mathcal R\underline\int}\,\,}}
\def\uint{{{\mathcal R\overline\int}\,\,}}
\def\rint{\mathcal R\int}

\showanswerstrue

\begin{document}
\begin{center}\large\bfseries
  Homework for Math 103 \\
  Assignment One -- TBA \\
\end{center}
\bigbreak
\begin{ques}
  Show that the \emph{countable} union of sets of measure zero in
  $\R$ has measure zero.
\end{ques}



\begin{ques}
  Suppose $f:[a,b]\to\R$ is bounded, and let $\pp$ and $\qq$ be
  partitions of $[a,b]$.  Prove that $L(f,\pp)\le U(f,\qq)$, where
  $L(f,\pp)$ and 
  $U(f,\qq)$ are the lower and upper Riemann sums, respectively, for
  $f$ on $[a,b]$.  (Hint: the result is trivial if $\pp=\qq$; now
  let $\rr=\pp\cup\qq$.)
\end{ques}


\begin{ques}
  Prove that a bounded function $f:[a,b]\to\R$ is Riemann integrable
  on $[a,b]$
  if and only if for all $\epsilon>0$ there is a partition $\pp$ of
  $[a,b]$ such that 
  \begin{equation*}
    U(f,\pp)-L(f,\pp)<\epsilon.
  \end{equation*}
\end{ques}



\begin{ques}
(\emph{Rudin}: page~31~\#1)
  Suppose that $(X,\mm)$ is a measurable space.  Show that if $\mm$ is
  countable, then $\mm$ is finite.  (Hint: since $\mm$ is countable,
  you can show that $\omega_{x}=\bigcap\set{E:\text{$E\in\mm$ and
  $x\in E$}}$ belongs to $\mm$.  The sets $\set{\omega_{x}}_{x\in X}$
  partition $X$.)
\end{ques}


\begin{ques}
  Let $X$ be an uncountable set and let $\mm$ be the collection of
  subsets $E$ of $X$ such that either $E$ or $E^{c}$ is countable.
  Prove that $\mm$ is a $\sigma$-algebra.
\end{ques}


\begin{ques}
  Recall from calculus that if $\set{a_{n}}$ is a sequence of
  nonnegative real numbers, then $\sum_{n=1}^{\infty} a_{n} = \sup_{n}
  s_{n}$, where $s_{n} = a_{1}+\dots+ a_{n}$.  (Note the value
  $\infty$ is allowed.)
  \begin{enumerate}
  \item Show that $\sum_{n=1}^{\infty} a_{n} =\sup\set{\sum_{k\in
        F}a_{k}: \text{$F$ is a finite subset of
        $\mathbf{Z}^{+}=1,2,3,\dots$}}$.  (The point of this problem
    is that if  $I$ is a (not necessarily countable) 
   set, and if $a_{i}\ge0$ for all $i\in I$,
    then we can define $\sum_{i\in I} a_{i} = \sup\set{\sum_{k\in
        F}a_{k}: \text{$F$ is a finite subset of
        $I$}}$, and our new definition coincides with the usual one
    when both make sense.)
    

  \item Now let $X$ be a set and $f:X\to[0,\infty)$ a function.  For
    each $E\subset X$, define
    \begin{equation*}
      \nu(E) := \sum_{x\in E} f(x).
    \end{equation*}
  Show that $\nu$ is a measure on $\bigl(X,\pp(X)\bigr)$.  (In
  lecture, we
  considered the special cases of \emph{counting measure}, where
  $f(x)=1$ for all $x\in X$, and the \emph{delta measure at $x_{0}$}, where
  $f(x_{0})=1$ for some $x_{0}\in X$ and $f(x)=0$ otherwise.  Another
  important example is the case where $\sum_{x\in X}f(x)=1$.  Then $f$
  is a (discrete) probability distribution on $X$ and $\nu(E)$ is the
  probability of the event $E$ for this distribution.)
  
  \item Let $X$, $f$, and $\nu$ be as in part~(b).  Show that if
    $\nu(E)<\infty$, then $\set{x\in E:f(x)>0}$ is countable.  Hint:
    if $\set{x\in E:f(x)>0}$ is uncountable, then for some $m\in
    \mathbf{Z}^{+}$, the set $\set{x\in E:f(x)>\frac1m}$ is infinite.
    (Note that this last result says that discrete probability
    distributions ``live on'' countable sample spaces.)
  \end{enumerate}
\end{ques}

\begin{ques}
  (\emph{Rudin}: page 31 \#3) Prove that if $f$ is a real-valued
  function on a measurable space $(X,\mm)$ such that $\set{x:f(x)\ge
  r}$ is measurable for all rational $r$, then $f$ is measurable.
\end{ques}
\begin{ques}
  (\emph{Rudin}: page 31 \#5) Suppose that
  $f,g:(X,\mm)\to[-\infty,\infty]$ are measurable functions.  Prove
  that the sets
  \begin{equation*}
    \set{x:f(x)<g(x)}\quad\text{and}\quad\set{x:f(x)=g(x)}
  \end{equation*}
are measurable.  (Remark: if $h=f-g$ were defined, then this problem
would be much easier (why?).  The problem is that $\infty-\infty$ and
$-\infty+\infty$ make no sense, so $h$ may not be everywhere defined.)
\end{ques}


\end{document}

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