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\begin{document}
\begin{enumerate}
\centerline{\sc Math 597.  Homework 7\\ Bryden Cais}
\medskip

\item\begin{enumerate}
    \item Define $E_j(\eta)=\{x:|f_j|>\eta\}$ and $F_j(\eta)=\{x:|f_j|\leq\eta\}$.  Obviously $E_j(\eta)\cup F_j(\eta)=X$
    for any $j$.  In one direction, suppose that $\mu(E_j(\eta))\rightarrow 0$ as $j\rightarrow\infty$ for any fixed $\eta >0$.
    Then we have
    $$\int_X \frac{|f_j|}{1+|f_j|}\,{\rm d}\mu\leq \int_{E_j(\eta)}\,{\rm d}\mu+\int_{F_j(\eta)}|f_j|\,{\rm d}\mu,$$
    since we have the obvious inequality $|f_j|/(1+|f_j|)\leq \min\{1,|f_j|\}$.  Since $|f_j|\leq \eta$ on $F_j(\eta)$,
    it follows that we have the bound
    $$\int_X \frac{|f_j|}{1+|f_j|}\,{\rm d}\mu\leq \mu(E_j(\eta))+\eta\mu(F_j(\eta))$$
    for each $\eta>0$ and all $j$.  By hypothesis $\mu(E_j(\eta))\rightarrow 0$ (equivalently $\mu(F_j(\eta))\rightarrow \mu(X)<\infty$)
    so that
    $$\lim_{j\rightarrow \infty} \int_X \frac{|f_j|}{1+|f_j|}\,{\rm d}\mu\leq \eta\mu(X).$$
    Since this holds for any $\eta >0$ and $\mu(X)$ is finite, it follows that
    $$\int_X \frac{|f_j|}{1+|f_j|}\,{\rm d}\mu\rightarrow 0.$$

    In the other direction, let $0<\eta <1$ be arbitrary.  Then it is not difficult to see that
    $$\{x: |f_j(x)|>\frac{\eta}{1-\eta}\}=\{x:\frac{|f_j(x)|}{1+|f_j(x)|}>\eta\}.$$
    Moreover, $\eta\rightarrow 0$ if and only if $\eta/(1-\eta)\rightarrow 0$.  Since we may without loss of
    generality restrict to only considering $\eta$ with $0<\eta<1$, it follows that $|f_j|$ tend to 0 in measure
    if and only if $|f_j|/(1+|f_j|)$ tend to 0 in measure.  Now using Chebyschev's inequality from Homework 6,
    we have
    $$\mu(\{x:|f_j|/(1+|f_j|)>\eta\})\leq \frac{1}{\eta}\int_X \frac{|f_j|}{1+|f_j|}\,{\rm d}\mu.$$  Since the integral
    tends to 0 by hypothesis, we see that $|f_j|/(1+|f_j|)$ tends to 0 in measure, and hence $|f_j|$ tends to 0 in measure.
    This completes the proof.

    \item  Let $X=\R$ and $f_j(x)=\frac{1}{j}$.  Then we have
    $$\mu(\{x:f_j(x)>\eta\})=\begin{cases}\infty &\text{if}\ \eta< 1/j\\ 0 & \text{otherwise}\end{cases}.$$
    Since for any $\eta>0$ there exists some $j>0$ with $1/j < \eta,$ it follows that $f_j$ tends to 0 in measure.
    On the other hand,
    $$\int_{\R}\frac{|f_j|}{1+|f_j|}\,{\rm d}\mu=\int_{\R}\frac{1}{j+1}\,{\rm d}\mu=\infty$$
    for all $j$.  It follows that the integral does not tend to $0$ and thus part (a) can fail if $\mu(X)$
    is not finite.

\end{enumerate}

\item Recall that the Lebesgue-Stieljes measure $\nu$ examined in Problem 3 of Homework 4 satisfies $\nu(\C)=\nu([0,1])=1$
and $\nu([0,1]\setminus \C)=0$, where $\C$ is the usual cantor set (of measure 0) on the interval $[0,1]$.  If $\nu$
were induced as
$$\nu(E)=\int_E g\,{\rm d}\mu$$
for some nonnegative measurable function $g$ with $\mu$ being Lebesgue measure, then we would necessarily have
$$1=\nu([0,1])=\int_{[0,1]} g\,{\rm d}\mu=\int_{[0,1]\setminus \C} g\,{\rm d}\mu=\nu([0,1]\setminus\C)=0,$$
since we can ``throw away'' any set of measure 0 from the integral.  This is a contradiction so that there is no such
$g$ inducing $\nu$ as above.

\item Since the $a_j$ are positive, the Monotone Convergence Theorem tells us that
$$\int_{0}^1\sum_{j=1}^{\infty}\frac{a_j}{\sqrt{|x-r_j|}}\,{\rm d}\mu=\sum_{j=1}^{\infty}\int_{0}^1\frac{a_j}{\sqrt{|x-r_j|}}\,{\rm d}\mu.$$
Now observe that
\begin{align*}
    \int_{0}^1\frac{a_j}{\sqrt{|x-r_j|}}\,{\rm d}\mu &= \int_{0}^{r_j}\frac{a_j}{\sqrt{r_j-x|}}\,{\rm d}\mu+
    \int_{r_j}^1\frac{a_j}{\sqrt{x-r_j}}\,{\rm d}\mu\\
    &=-2a_j\sqrt{r_j-x}\,\Bigg]_0^{r_j}+2a_j\sqrt{x-r_j}\,\Bigg]_{r_j}^1\\
    &=2a_j\sqrt{r_j}+2a_j\sqrt{1-r_j}\\
    &\leq 4a_j,
\end{align*}
so that
$$\sum_{j=1}^{\infty}\int_{0}^1\frac{a_j}{\sqrt{|x-r_j|}}\,{\rm d}\mu$$ is finite.
If $\sum_{j=1}^{\infty}\frac{a_j}{\sqrt{|x-r_j|}}$ were {\em not} finite a.e. then the integral
$$\int_{0}^1\sum_{j=1}^{\infty}\frac{a_j}{\sqrt{|x-r_j|}}\,{\rm d}\mu$$ would not be finite;
by our considerations (with MCT) above, this is a contradiction.

\item Suppose that $f(x)$ and $|x|f(x)$ are integrable on $\R$.
\begin{enumerate}
    \item Observe that for fixed $y$ we have $$f(x)|y-x|=\begin{cases}f(x)y-xf(x) & \text{if}\ y-x\geq 0\\ xf(x)-f(x)y & \text{otherwise}\end{cases}.$$
    Since $f(x),|x|f(x)$ are integrable, so are $\mp yf(x)\pm xf(x)$ since $y$ is constant.  It follows that
    $f(x)|y-x|$ is integrable.

    \item Set
    $$\phi_h(x)=f(x)\frac{|y+h-x|-|y-x|}{h}$$ and observe that
    $$|\phi_h(x)|\leq |f(x)|\frac{|h|+|y-x|-|y-x|}{|h|}=|f(x)|,$$
    so that for all $h$, we have a dominating function $|f(x)|$ with $|\phi_h|\leq |f|$.
    It follows from the Dominated Convergence Theorem that
    $$h'(y):=\lim_{h\rightarrow 0}\int_{\R} \phi_h\,{\rm d}\mu=\int_{\R} \lim_{h\rightarrow 0}\phi_h\,{\rm d}\mu.$$
    On the other hand, we have
    $$\lim_{h\rightarrow 0}\phi_h=f(x)\lim_{h\rightarrow 0}\frac{|y-x+h|-|y-x|}{h}=f(x){\rm sgn}(y-x).$$
    The proposed equality follows.

\end{enumerate}

\end{enumerate}
\end{document}