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\begin{document}
\begin{enumerate}
\centerline{\sc Math 597.  Homework 10\\ Bryden Cais}
\medskip

\item  Let $B_n$ denote the unit ball in $\R^n$ and let $C_n$ be the volume of $B_n$.  Let $\mu$ be the usual Lebesgue measure on $\R^n$.
Then by Prop. III.15.1 we have
\begin{align*}
    \int_{B_n} \| x\|^p &= p\int_0^{\infty} t^{p-1} \mu(\{x\in B_n : \|x\|>t\})\,\mathrm{d}t\\
    &=p\int_0^{1} t^{p-1} \mu(\{x\in B_n : \|x\|>t\})\,\mathrm{d}t\\
    &=p\int_0^{\infty} t^{p-1} (C_n-C_nt^n)\,\mathrm{d}t\\
    &=pC_n\int_0^{\infty} (t^{p-1}-t^{p+n-1})\,\mathrm{d}t\\
    &=pC_n\left\{\frac{1}{p}-\frac{1}{p+n}\right\}\\
    &=\frac{nC_n}{p+n}.
\end{align*}
Similarly, we have
\begin{align*}
    \int_{\R^n-B_n} \| x\|^p &= p\int_0^{\infty} t^{p-1} \mu(\{x\in \R^n-B_n : \|x\|>t\})\,\mathrm{d}t,
\end{align*}
but for all $t$, we have $\mu(\{x\in \R^n-B_n : \|x\|>t\})=+\infty$ so that the integral is infinite.

\item Let $f,g$ be integrable on $[a,b]$ and set $F=\int_a^x f\,\mathrm{d}\mu,\ G=-\int_x^b g\,\mathrm{d}\mu$.
\begin{enumerate}
    \item Since $F,G$ are integrals of $f,g$, they are of bounded variation and
    absolutely continuous.  Moreover, $F^{'},G^{'}$ exist almost everywhere and are equal a.e. to $f,g$
    respectively.  It follows that $FG$ is of bounded variation and absolutely continuous
    (a result from class).  Therefore, $(FG)^{'}$ exists a.e. and we have
    $(FG)^{'}=F^{'}g+G^{'}f$ a.e.  Moreover, we have
    \begin{align*}
        (FG)^{'}&=(FG)(a)+\int_a^x (FG)^{'}\,\mathrm{d}\mu\\
        &=(FG)(b)+\int_b^x (FG)^{'}\,\mathrm{d}\mu.
    \end{align*}
    Using $F(a)=G(b)=0$ we see that
    $$\int_a^x (FG)^{'}\,\mathrm{d}\mu +\int_x^b (FG)^{'}\,\mathrm{d}\mu=\int_a^b (FG)^{'}\,\mathrm{d}\mu=0.$$
    Using the fact that $(FG)^{'}=F^{'}g+G^{'}f$ a.e. it follows that
    $$\int_a^b Fg\,\mathrm{d}\mu=-\int_a^b Gf\,\mathrm{d}\mu.$$
    Of course, this is formally just integration by parts for differentiable functions $F,G$.

    \item  No, the proposed formula is false without further restrictions.  For example, let
    $$F(x)=\begin{cases}0 & \text{if}\ x\in [0,1/2]\\ 1 & \text{if}\ x\in (1/2,1]\end{cases},$$
    and
    $$G(x)=\begin{cases}1 & \text{if}\ x\in [0,1/2]\\ 0 & \text{if}\ x\in (1/2,1]\end{cases}.$$
    Then we have $F(0)=G(1)=0$ and
    $$\int_0^1 F\,\mathrm{d}\mu_G= F(1/2)\mu_G(\{1/2\})=0\cdot (-1)=0,$$
    while
    $$-\int_0^1 G\,\mathrm{d}\mu_F= G(1/2)\mu_F(\{1/2\})=-1\cdot 1=-1.$$

\end{enumerate}

\item Let $f(x)=x\sin\left(\frac{1}{x^{\alpha}}\right)$ for $\alpha> 0$.  We consider $f$ on the interval $[0,1]$
and show that it is of bounded variation for $\alpha <1$.  It is clear that $f$ has a single extremum in any interval
$[\frac{1}{((j+1)\pi)^{1/\alpha}},\frac{1}{(j\pi)^{1/\alpha}}]$ for $j\geq 1$.  Let $m_{2n}$ for $n\geq 0$
denote the local maxima values of $f$ and $m_{2n+1}$ for $n\geq 0$ denote the values of $f$ at the local minima.
We easily have the estimates
$$\frac{1}{((j+1)\pi)^{\alpha}}\leq |m_{j-1}|\leq \frac{1}{(j\pi)^{\alpha}}$$
for $1\le j$.  Moreover, since $f$ is piecewise monotone on the intervals between any consecutive extreme points,
we have
$$V_f[0,1]=\sum_{n\geq 0}(m_{2n}-m_{2n+1})$$
so that
$$\sum_{j\geq 2} \frac{1}{(j\pi)^{1/\alpha}}\leq V_f[0,1] \leq \sum_{j\geq 1} \frac{1}{(j\pi)^{1/\alpha}}.$$
It follows that $f$ is of bounded variation if and only if the sum $\sum_{j >0} \frac{1}{j^{1/\alpha}}$ converges.
By the integral test, this is the case if and only if $\alpha <1,$ as claimed.

\item Let $\mu$ be a finite signed Borel measure on $[0,1]$.
\begin{enumerate}
    \item Let $E'=\{x\in [0,1]:\mu(\{x\})\neq 0\}$.  Since $\mu$ is finite, so are $\mu^+$ and $\mu^-$ and hence so is $|\mu|$.
    Clearly, $\mu(x)\neq 0$ implies $|\mu|(x) >0$.  By monotonicity and finite additivity of $|\mu|$ we then have
    $$|\mu|([0,1])\geq \sup\{\sum_{x\in S\subset E'} |\mu|(x): S\ \text{finite}\}=\sum_{x\in E'}|\mu|(x).$$
    It follows from HW 1 that $E'$ is countable and that $\sum_{x\in E'}|\mu|(x)$ is finite.
    since
    $$|\mu|(x)=\mu_1(x)+\mu_2(x)=|\mu_1(x)|+|\mu_2(x)|\geq |\mu_1(x)-\mu_2(x)|=|\mu(x)|,$$
    we see that $\sum_{x\in E'}|\mu(x)|<\infty.$

    \item It follows that $\mu_{pp}(E):=\sum_{x\in E}\mu(\{x\})$ is a finite Borel measure on $[0,1]$
    and that $\mu_{pp} \bot \mu_{\rm Leb}$.  Indeed, we have $|\mu_{pp}|(E^{'c})=0$ and $\mu_{\rm Leb}(E^{'}=0)$.
    By Thm. III.18.2 we can write $\mu=\mu_0+\mu_1$ with $\mu_0 \bot \mu_{\rm Leb}$ and $\mu_1\ll \mu_{\rm Leb}$.
    Thus let $F$ be such that $|\mu_0|(F^c)=\mu_{\rm Leb}(F)=0$.

    We claim that $\mu_{pp}(F^c)=0$.  Indeed, by definition,
    $$\mu_{pp}(F^c)=\sum_{x\in F^c}\mu(\{x\}).$$
    Since $\mu=\mu_0+\mu_1$ and $\mu_1 \ll \mu_{\rm Leb},$ we see that $\mu_1(\{x\})=0$ for any point $x$ so that
    $\mu(\{x\})=\mu_0(\{x\}).$  Therefore, for any $x\in F^c$ we have $|\mu_0|(\{x\})\leq |\mu_0|(F^c)=0$
    where we have used the monotonicity of $|\mu_0|$.  Thus, for all $x\in F^c$ we have $\mu(\{x\})=0$ and
    therefore $\mu_{pp}(F^c)=0$ as claimed.  Thus $\mu_3=\mu_0-\mu_{pp}$ satisfies $\mu_3(F^c)=0$
    and of course we still have $\mu_{\rm Leb}(F)=0$ so that $\mu_3 \bot \mu_{\rm Leb}$.
    Now by our above discussion, we have seen that for any point $x$ we have $\mu(\{x\})=\mu_0(\{x\})$.
    Similarly, by definition we have $\mu_{pp}(\{x\})=\sum_{x\in \{x\}} \mu(x)=\mu(\{x\})=\mu_0(\{x\}) $
    so that $\mu_3(\{x\})=0$ for all points $x$.

\end{enumerate}

\end{enumerate}
\end{document}