\documentclass[hw]{dpwexam-new}
\usepackage{amsmath}
%\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 -- Due September 24 \\
\end{center}
\bigbreak
\begin{ques}
  Show that the \emph{countable} union of sets of measure zero in
  $\R$ has measure zero.
\end{ques}

\begin{ans}
Suppose that $E_{n}$ has measure zero for $n=1,2,\dots$, and let
$E=\bigcup E_{n}$.  Let $\epsilon>0$.  By assumption, there are
intervals $I_{n,m}$ such that $E_{n}\subset \bigcup_{m=1}^{\infty}
I_{n,m}$ and $\sum_{m=1}^{\infty}
\ell(I_{n,m})<\frac{\epsilon}{2^{-n}}$.  Then
$E\subset\bigcup_{n,m=1}^{\infty} I_{n,m}$ and
$\sum_{n,m=1}^{\infty}\ell(I_{n.m}
)<\sum_{n=1}^{\infty}\frac{\epsilon}{2^{-n}} = \epsilon$.  This
suffices.
\end{ans}

\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{ans}
  The following are relatively easy to prove for any partition $\pp$
  and any prtition $\rr$ such that $\pp\subset \rr$: $L(f,\pp)\le
  U(f,\pp)$, $L(f,\pp)\le L(f,\rr)$,
  and $U(f,\rr)\le U(f,\pp)$.  
Thus if $\pp$, $\qq$, and $\rr$ are as in the problem, then
  \begin{align*}
    L(f,\pp) & \le L(f,\rr) \\
    & \le U(f,\rr) \\
    & \le U(f,\qq).
  \end{align*}

However, the challenge is the write up
  your solutions elegantly.  Here are two suggestions for proving that
  $L(f,\rr)\ge L(f,\pp)$.
\smallbreak
\noindent
{\bf Method I---Brute Force:}  Let
  $\pp=\set{a=t_{0}<\cdots < t_{n}=b}$ and $\rr=\set{a=s_{0}<\cdots <
    s_{m}}$.  Since $\pp\subset\rr$, for any $k$, there is a unique
  $i$ such that $[s_{k-1},s_{k}]\subset [t_{i-1},t_{i}]$.  Furthermore,
  \begin{equation}\label{eq:1}\tag{\dag}
    t_{i}-t_{i-1}=\sum_{[s_{k-1},s_{k}]\subset [t_{i-1},t_{i}]} s_{k}-s_{k-1}.
  \end{equation}
Thus if
\begin{equation*}
  m_{i}:=\inf_{t\in[t_{i-1},t_{i}]}f(t) \quad\text{and}\quad
  n_{k}=\inf_{t\in [s_{k-1},s_{k}]}f(t),
\end{equation*}
then we have $m_{i}\le n_{k}$ whenever $[s_{k-1},s_{k}]\subset
[t_{i-1},t_{i}]$.  Thus
\begin{align*}
  L(f,\rr)&= \sum_{k=1}^{m} n_{k}(s_{k}-s_{k-1}) \\
&= \sum_{i=1}^{n} \;\Bigl(\;\sum_{[s_{k-1},s_{k}]\subset [t_{i-1},t_{i}]} n_{k}(s_{k}
-s_{k-1}) \Bigr) \\
&\ge \sum_{i=1}^{n} m_{i}\;\Bigl(\;\sum_{[s_{k-1},s_{k}]\subset
  [t_{i-1},t_{i}]} s_{k}-s_{k-1} \Bigr)\\
\intertext{which, by \eqref{eq:1}, is}
&= \sum_{i=1}^{n} m_{i}(t_{i}-t_{i-1}) \\
&=L(f,\pp).
\end{align*}
\medbreak
\noindent
{\bf Method II---Simple:} Suppose that $\rr$ refines
$\pp=\set{a=t_{0}<\dots < t_{n}=b}$ by adding a
single point $s$ where $t_{j-1}<s<t_{j}$.  Define $m_{i}$ as in
``Method I'' and let
\begin{equation*}
  n_{j1}=\inf_{t\in[t_{j-1},s]}f(t)\quad\text{and}\quad
  n_{j2}=\inf_{t\in[s,t_{j}]}f(t), 
\end{equation*}
and note that $m_{j}\le n_{1j}+n_{2j}$.  Then
\begin{align*}
  L(f,\rr)&=\sum_{i=1}^{j-1}m_{i}(t_{i}-t_{i-1}) + n_{j1}(s-t_{i-1}) +
  n_{2j} (t_{i}-s) + \sum_{i=j+1}^{n} m_{i}(t_{i}-t_{i-1}) \\
&\le \sum_{i=1}^{n}m_{i}(t_{i}-t_{i-1}) \\
&=L(f,\pp).
\end{align*}
Now the general result follows from a simple induction.


\end{ans}

\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{ans}
  Note that for any partition $\pp$, $L(f,\pp) \le \lint_{a}^{b}f$ and
  $U(f,\pp) \ge \uint_{a}^{b}$
  Suppose that $f$ is Riemann integrable.  Then given $\epsilon>0$
  there are \emph{partitions} $\pp$ and $\qq$ such that 
  \begin{gather*}
    {\lint}_{a}^{b}f - L(f,\pp) <
    \frac{\epsilon}2,\quad\text{and} \\
    U(f,\qq) - {\uint}_{a}^{b}f < \frac\epsilon2.
  \end{gather*}
Now let $\rr=\pp\cup\qq$.  Then using the previous problem, we see
that the two inequalities above hold with $\pp$ and $\qq$ replaced by
$\rr$.  In particular, since $f$ is integrable, $\lint_{a}^{b}f =
\uint_{a}^{b} f = \rint_{a}^{b}f$ and 
\begin{equation*}
  U(f,\rr)-L(f,\rr) < \frac\epsilon2 + \rint_{a}^{b}f -\rint_{a}^{b}f
  + \frac\epsilon2 = \epsilon.
\end{equation*}

Now assume that for all $\epsilon>0$ a partition $\pp$ exists as
stated in the problem.  The previous problem implies that
\begin{equation*}
  \lint_{a}^{b}f \le \uint_{a}^{b}f.
\end{equation*}
Let $\epsilon>0$ be given, and choose $\pp$ such that
$U(f,\pp)-L(f,\pp) < \epsilon$.  Thus
\begin{equation*}
  \lint_{a}^{b}f \ge L(f,\pp)> U(f,\pp)-\epsilon \ge \uint_{a}^{b}f -
  \epsilon. 
\end{equation*}
Thus,
\begin{equation*}
  0\le \uint_{a}^{b}f - \lint_{a}^{b}f < \epsilon.
\end{equation*}
Since $\epsilon$ is arbitrary, it follows that $\uint_{a}^{b}f=
\lint_{a}^{b}f$, and $f$ is Riemann integrable on $[a,b]$ as required.
\end{ans}

\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{ans}
  Notice that if $E\in\mm$ and if $x\in E$, then $\omega_{x}\subset
  E$.  On the other hand, if $F\in\mm$ and if $x\notin F$, then $x\in
  \omega_{s}\setminus F$, and $\omega_{x}\subset \omega_{x}\setminus
  F$ so $\omega_{x}\cap F=\emptyset$.  Thus if
  $\omega_{x}\cap\omega_{y}\not= \emptyset$, then $x\in\omega_{y}$ and
  $\omega_{x}\subset \omega_{y}$.  By symmetry, $\omega_{y}\subset
  \omega_{x}$ and $\omega_{x}=\omega_{y}$.  This shows that
  $\set{\omega_{x}}_{x\in X}$ partitions $X$.  If $x\in F\in\mm$, then
  $\omega_{x} \subset F$ and $F=\bigcup_{x\in F}\omega_{x}$.  Thus the
  elements of $\mm$ are in one-to-one correspondence with the
  (distinct) subsets of $\set{\omega_{x}}_{x\in X}$.  If this set is
  finite, then so is $\mm$.  If it is infinite, then it has at least
  as many subsets as does $\mathbf Z$ --- and there are uncountably
  many of these.
\end{ans}

\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{ans}
  Since $\mm$ certainly contains $X$ and is closed under taking
  complements, the only issue is to show that $\mm$ is closed under
  countable unions.  Suppose that $\set{E_{n}}_{n=1}^{\infty}\subset
  \mm$.  If all the $E_{n}$ are countable, then the countable union
  $E=\bigcup_{n} E_{n}$ of countable sets is countable and $E\in\mm$.
  If $E_{k}^{c}$ is countable, then note that $E^{c}\subset E_{k}^{c}$
  must also be countable.  Thus in all cases, $E\in\mm$ and $\mm$ is a
  $\sigma$-algebra. 
\end{ans}

\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.)
    \begin{ans}
      Let $I=\sup\set{\sum_{k\in
        F}a_{k}: \text{$F$ is a finite subset of
        $\mathbf{Z}^{+}=1,2,3,\dots$}}$.
      Since $F=\set{1,\dots,n}$ is finite,
      \begin{align*}
        I &:= \sup\set{\sum_{k\in F} a_{k}:F=\set{1,\dots,n}}\\
&\ge\sup s_{n} = \sum_{n=1}^{\infty} a_{n}.
      \end{align*}
Let $\epsilon>0$.  Choose a finite set $F\subset\mathbf{Z}^{+}$ such
that $\sum_{k\in F}a_{k}> I-\epsilon$.  Let $m=\max_{k\in F}k$.  Then
$s_{m} \ge \sum_{k\in F} a_{k}>I-\epsilon$.  Thus 
\begin{equation*}
  \sum_{n=1}^{\infty}a_{n}=\sup s_{n}\ge s_{m}>I-\epsilon.
\end{equation*}
Since $\epsilon$ was arbitary, $\sum_{n=1}^{\infty}a_{n}\ge I$.  Thus
$\sum_{n=1}^{\infty}a_{n}= I$ as claimed.
    \end{ans}

  \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.)
  \begin{ans}
    Let $\set{E_{n}}$ be disjoint sets and $E=\bigcup_{n}E_{n}$.
    Suppose that $\nu(E)=\infty$.  Fix $M>0$.   Choose a finite set $F\subset E$ such that $M
    < \sum_{x\in F}f(x)$.  Let $F_{n}=\set{x\in F:x\in E_{n}}$.
 Since all but finitely may $F_{n}$ are empty,
    \begin{equation*}
      M < \sum_{x\in E}f(x) =\sum_{n} \sum_{x\in
      F_{n}}f(x) \le \sum_{n=1}^{\infty} \nu(E_{n}).
    \end{equation*}
Since $M$ is arbitray, $\sum_{n=1}^{\infty}
\nu(E_{n})=\infty=\nu(E)$.  So we can assume from here on that
$\nu(E)<\infty$. 
Let
    $\epsilon>0$.  Choose a finite set $F\subset E$ such that $\nu(E) -\epsilon
    < \sum_{x\in E}f(x)$.  Let $F_{n}=\set{x\in F:x\in E_{n}}$.
    Since all but finitely may $F_{n}$ are empty,
    \begin{equation*}
      \nu(E) -\epsilon < \sum_{x\in E}f(x) =\sum_{n} \sum_{x\in
      F_{n}}f(x) \le \sum_{n=1}^{\infty} \nu(E_{n}).
    \end{equation*}
Since $\epsilon$ is arbitrary, $\nu(E)\le \sum_{n=1}^{\infty}
\nu(E_{n})$.  

Notice that if $\nu(E)<\infty$, the $\nu(E_{k})<\infty$ for all $k$.
Let $\epsilon>0$.  Since $\sum_{n}\nu(E_{n}) =
\sup_{n}\sum_{k=1}^{n}\nu(E_{k})$ it will suffice to show that for any $n$
\begin{equation*}
  \nu(E)+\epsilon>\sum_{k=1}^{n}\nu(E_{k}).
\end{equation*}
Choose finite sets $F_{k}\subset E_{k}$ such that 
\begin{equation*}
  \nu(E_{k})-\frac{\epsilon}n <\sum_{x\in F_{k}}f(x).
\end{equation*}
Put $F=\bigcup_{k=1}^{n}F_{k}$.  Note that $F$ is finite, and since
the $F_{k}$ are disjoint, 
\begin{equation*}
  \nu(E) \ge \sum_{x\in F}f(x) = \sum_{k=1}^{n}\sum_{x\in F_{k}} f(x)
  > -\epsilon + \sum_{k=1}^{n} \nu(E_{k}).
\end{equation*}
That's it.
  \end{ans}

  \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}
\begin{ans}
  Since $\set{x:f(x)=g(x)}$ is the complement of
  $\set{x:f(x)<g(x)}\cup \set{x:g(x)<f(x)}$ it suffice to see that
  $\set{x:f(x)<g(x)}$ is measurable.  But
  \begin{equation*}
    \set{x:f(x)<g(x)}=\bigcup_{r\in\mathbf{Q}}\set{x:f(x)<r<g(x)},
  \end{equation*}
and each $\set{x:f(x)<r<g(x)}=g^{-1}\bigl((r,\infty]\bigr) \cap
f^{-1}\bigl([-\infty,r)\bigr)$ is measurable.
\end{ans}

\end{document}

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