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\begin{document}
\begin{enumerate}
\centerline{\sc Math 597.  Homework 6\\ Bryden Cais}
\medskip

\item Let $f:E\rightarrow R^{\ast}$ be measurable and assume that $|f|>0$
a.e. on $E$.  Then for any $\e> 0$ there exists a measurable set
$E_{\e}\subset E$ and a positive number $\delta_{\e}$ such that
$\mu(E\setminus E_{\e})\leq \e$ and $|f|>\delta_{\e}$ on $E_{\e}$.

Without loss of generality, we may suppose that $f$ is finite a.e. since
if it is not, then we already have $|f|>\delta_{\e}$ on some $X\subset E$
where $f$ is not finite on $X$, so we need only be concerned with the subset
of $E$ on which $f$ is a.e. finite.

Now define
$$g_n(x)=\begin{cases}\frac{1}{f(x)} & \text{if}\ |f(x)|\geq 1/n\\ 1 &
\text{otherwise}\end{cases}.$$
Clearly, $g_n\rightarrow 1/f$ pointwise a.e. and since $|f|>0$ a.e,
we have $1/f$ finite a.e.  Since $E$ is of finite measure, for any $\e>0$,
the Egorov Theorem guarantees the existence of a set $E_{\e}\subset E$
on which the convergence $g_n\rightarrow 1/f$ is uniform.  That is,
for any $\delta >0$ there exists an $N>0$ such that for all $n\geq N$,
$|g_n(x)-1/f(x)|<\delta$ for all $x\in E_{\e}$.
Now by construction, we have $|g_n|< n$ so that for any $\delta>0$
there exists an $N$ such that for all $n\geq N$ we have
$|g_n-f| < \delta$, so that $|f|< |g_n|+\delta \leq N+\delta$.
Thus, for any $\e>0$ we have a set $E{\e}$ with $\mu(E\setminus E_{\e})<\e$
and $1/|f| > \delta_{\e}$ on $E_{\e}$ with $\delta_{\e}=1/(N+\delta)$.


\item Let $f_n:E\rightarrow R^{\ast}$ be a sequence of measurable functions
with $f_n\rightarrow f$ a.e on $E$.  Suppose that $|f_n|>0$ and $|f|>0$
a.e.  As above, we may suppose without loss of generality that $f_n,f$
are finite a.e.  Let $\e>0$ be arbitrary.  Then by (1) above, there
exists a set $E_{\e}\subset E$ and $\delta_{\e}>0$ with
$\mu(E\setminus E_{\e})<\e/3$ and $|f|>\delta_{\e}$ on $E_{\e}$.

Now by the Egorov Theorem,
there exists a set $E^{(0)}\subset E_{\e}$ with $\mu(E_{\e}\setminus E^{(0)})<\e/3$
and $f_n\rightarrow f$ uniformly on $E_{\e}$.  Thus, there exists a positive $N$
such that for all $n\geq N$ we have $|f_n-f|<\delta_{\e}/3$ on $E^{(0)}$, so that
for all $n\geq N$ we have $|f_n|>\delta_{\e}/3$.

Next, we may repeatedly apply problem (1) above to the functions $f_i$ for $1\leq i <N$
to obtain sets $E^{(i)}\subset E^{(i-1)}$ for $i=1\ldots N-1$ and positive numbers
$\delta_{i}$ with $\mu(E^{(i)}\setminus E^{(i-1)}) < \e/3N$ and $|f_i|>\delta_{i}$
on $E^{(i)}$.  Let $\delta=\min\{\delta_{\e}/3,\delta_1,\ldots,\delta_N\}$.  Then $\delta>0$
and by construction we have $|f_i|>\delta$ on $E^{(N)}$.  Moreover, we have
$\mu(E\setminus E^{(N)})\leq \e/3+\e/3+N\cdots \e/3n=\e$, as required.

\item Let $f_n:E\rightarrow \R^{\ast}$ be a sequence of measurable functions and suppose that
$f_n(x)$ is finite a.e. on $E$.  Let $E_{n,m}=\{x:f_n(x)\geq m\}$.  By hypothesis,
$\lim_{m\rightarrow \infty}\mu(E_{m,n})=0$ for all $n$.  Thus, for each $n$ there exists
a positive integer $m_n$ such that for all $m\geq m_n$ we have $\mu(E_{m,n})<1/2^n$.
Let $k_n=2^n m_n$, and let $F\subset E$ be the set $F=\{x:\lim_{n\rightarrow \infty}|f_n(x)/k_n|> 0\}$.
Define $F_n=\{x:|f_n(x)/k_n| > 2^{-n}\}$.  Observe that $x\in F$ implies that $x\in F_n$ for all sufficiently
large $n$, and hence that $x\in \bigcap_{n=1}^{\infty} \bigcup_{j=n}^{\infty} F_j$, i.e. $F\subseteq \lim\sup F_n$.
On the other hand $F_n=\{x:|f_n(x)/k_n|>2^{-n}\}=\{x:|f_n(x)|>m_n\}$ so that $\mu(F_n)<2^{-n}$, and hence
$\sum_n \mu(F_n)=1 <\infty$.  It follows that $\mu(\lim\sup F_n)=0$ and hence $\mu(F)=0$
whence $f_n/k_n\rightarrow 0$ a.e.

\item Recall the definition
$$\int_X f\,{\rm d}\mu=\sup \left\{\int_X \varphi\,{\rm d}\mu:\varphi\ \text{is simple}\right\}.$$
Set
$$\varphi_{\alpha}(x)=\begin{cases}\alpha & \text{if}\ f(x)> \alpha\\ 0 & \text{otherwise}\end{cases}.$$
Certainly, $\varphi$ is simple and we have $f \geq \varphi$ on $X$ by construction and since $f\geq 0$ on $X$.  It follows
that
$$\int_X f\,{\rm d}\mu \geq \int_X \varphi\,{\rm d}\mu=\alpha\mu\{x:f(x)>\alpha\}.$$
Chebyschev's inequality now follows.

\item Set $A_j=\bigcup_{a=0}^{2^j-1} [2a/2^{j+1},(2a+1)/2^{j+1})$ and $B_j=\bigcup_{a=0}^{2^j-1} [(2a+1)/2^j,(2a+2)/2^j)$.
Observe that for each $j$ we have
$$R_j(x)=\begin{cases}-1 & \text{if}\ x\in A_j\\ 1 & \text{if}\ x\in B_j\end{cases}.$$
\begin{enumerate}
    \item
    \begin{enumerate}
        \item From the above description, since $A_j\cup B_j=[0,1]$, we see that $R_j^1=1$ on $[0,1]$ and hence
        $$\int_0^1 R_j^2\,{\rm d}\mu=1.$$

        \item Without loss of generality, suppose that $j<k$.  Then for each $a$, we have
        $$\int_{a/2^j}^{(a+1)/2^j}R_jR_k\,{\rm d}\mu=\sum_{b=2^{k-j}a}^{2^{k-j}(a+1)-1} \int_{b/2^k}^{(b+1)/2^k}R_jR_k\,{\rm d}\mu=
        \sum_{b=2^{k-j}a}^{2^{k-j}(a+1)-1} (-1)^{b+a}=0.$$
        It follows that
        $$\int_{0}^1 R_jR_k\,{\rm d}\mu=0$$
        for $j\neq k$.

        \item We have
        \begin{align*}
            \int_0^1 (R_1+R_2+\cdots+R_n)^2\,{\rm d}\mu &= \int_0^1 \sum_{j=1}^n R_j^2+2\sum_{1\leq j<k\leq n}R_jR_k\,{\rm d}\mu\\
            &= \sum_{j=1}^n \int_0^1 R_j^2\,{\rm d}\mu+2\sum_{1\leq j<k\leq n}\int_0^1 R_jR_k\,{\rm d}\mu\\
            &=n,
        \end{align*}
        where we have used parts ({\em i}) and ({\em ii}).

    \end{enumerate}
    \item Since $(R_1+\cdots+R_n)^2\geq 0$ on $[0,1]$, we have by problem 4 that
    $$\mu(\{x\in X: (R_1+\cdots+R_n)^2 >n^2\e^2\})\leq \frac{1}{ n^2\e^2}\int_{0}^1(R_1+\cdots+R_n)^2\,{\rm d}\mu=\frac{1}{n\e^2}$$
    by part (a), ({\em iii}).

    \item It follows from part (b) that $\frac{R_1+\cdots +R_n}{n}$ tends to 0 in measure as $n\rightarrow\infty$;
    indeed, if it did not, we would have that
    $$\mu(\{x\in X: R_1+\cdots+R_n >n\e\})$$ does not tend to 0, and hence that
    $$\mu(\{x\in X: (R_1+\cdots+R_n)^2 >(n\e)^2\})$$ does not tend to 0 as $n\rightarrow\infty$, and this contradicts part (b).

    \item On the other hand, the set
    $$S=\left\{x\in (0,1):\frac{R_1+\cdots+R_n}{n}\rightarrow 0\right\}$$ has empty interior.  Indeed, let $x\in S$ and let $\e>0$
    be arbitrary.  Pick $m$ so that $2^{-m}<\e$ and let $x=.a_1a_2\ldots $ be the binary expansion of $x$.
    Now set $y=.a_1a_2\ldots a_m 10110111011110111110\ldots$, where the pattern of 1's and 0's at the end consists of $j$ 1's
    followed by a 0 for $j=1,2,3,\ldots$.  By construction, $y\in (x-\e,x+\e)$.  On the other hand, it is evident
    by construction that $y\not\in S$.  Since $\e$ was arbitrary, it follows that $S$ has empty interior.
\end{enumerate}
\end{enumerate}
\end{document}
