\documentclass[10pt]{amsart}
\usepackage[leqno]{amsmath}
\usepackage{amssymb}
   \topmargin=0in
   \oddsidemargin=0in
   \evensidemargin=0in
   \textwidth=6.5in
   \textheight=8.5in

\newcommand{\Z}{\mathbf{Z}}
\newcommand{\R}{\mathbf{R}}
\newcommand{\C}{\mathcal{C}}
\newcommand{\F}{\mathbf{F}}
\newcommand{\Q}{\mathcal{Q}}
\renewcommand{\H}{\mathcal{H}}
\renewcommand{\a}{\alpha}
\renewcommand{\b}{\beta}
\newcommand{\e}{\epsilon}
\newcommand{\ml}{\mu_{{\rm leb}}^{\ast}}
\newcommand{\lcm}{\mathrm{lcm}}
\newcommand{\Hom}{\mathrm{Hom}}
\DeclareMathOperator{\diam}{diam}

\begin{document}
\begin{enumerate}
\centerline{\sc Math 597.  Homework 9\\ Bryden Cais}
\medskip

\item Let $\mu$ be counting measure and $\nu$ Lebesgue measure on (say) the unit interval.  Obviously we have $\nu \ll \mu$ since the only set $E$
with $\mu(E)=0$ is the empty set $E=\emptyset$, which has Lebesgue measure 0.  However, the unit interval is not $\sigma$-finite, so that we may
not apply the Radon-Nikodym Theorem.  Indeed, there is no measurable function $f$ with $\nu(E)=\int_E f\,\mathrm{d}\mu$.  To see this, let $x$
be a point.  Then we would have $0=\nu(x)=\int_x f\,\mathrm{d}\mu=f(x)$.  It follows that $f$ is identically 0, but then we would have
$1=\nu([0,1])=\int_{[0,1]}f\,\mathrm{d}\mu=0$, a contradiction.

\item Let $\nu,\mu$ be $\sigma$-finite measures on the measure space $(X,\mathcal{A})$ with $\nu\ll\mu$ and set $\lambda=\nu+\mu$.
Then $\nu\ll\lambda$ so we may set $f=\frac{\mathrm{d}\nu}{\mathrm{d}\lambda}$.
\begin{enumerate}
    \item Let $U_i$ be a countable covering of $X$ by measurable sets with $\nu(U_i)$ finite for each $i$.  Let
    $E=\{x\in X: f(x)\geq 1\}$.  Then we have
    $$\nu(E\cap U_i)=\int_{E\cap U_i} f(x)\,\mathrm{d}\lambda\geq \int_{E\cap U_i} \,\mathrm{d}\lambda=\lambda(E\cap U_i)=\mu(E\cap U_i)+\nu(E\cap U_i),$$
    so that we have $\nu(E\cap U_i)\geq \nu(E\cap U_i)+\mu(E\cap U_i)$.  Since $\nu(E\cap U_i)$ is finite, it follows that we have
    $\mu(E\cap U_i)=0$ for each $i$ and hence $\mu(E)=0$.  Thus, $f<1$ $\mu$-almost everywhere.  Similarly, for $\delta >0$ let
    $F_{\delta}=\{x:f(x)<-\delta\}$. Then $\nu(F_{\delta})=\int_{F_{\delta}} f\,\mathrm{d}\lambda < -\delta\lambda(F_{\delta})< 0$
    if $\lambda(F_{\delta})>0$.  Thus, since $\nu$ is positive,we conclude that $\lambda(F_{\delta})=0$ for all $\delta>0$
    and consequently that $\mu(F_{\delta})=0$ for all $\delta >0$.  It follows that $\mu(\cup_{n>0} F_{1/n})=0$ so that
    $f\geq 0$ $\mu$-almost everywhere.

    \item It follows from the definition of an integral as a supremum over all simple functions bounded above by the integrand
    on the set of integration that we have for any $\lambda$-measurable set $E$ and any integrable function $h$ on $E$
    $$\int_E h\,\mathrm{d}\lambda= \int_E h\,\mathrm{d}\mu+\int_E h\,\mathrm{d}\nu.$$
    Moreover, we have that
    $$\int_E hf \,\mathrm{d}\lambda=\int_E h \,\mathrm{d}\nu.$$
    Combining these two relations gives
    $$\int_E hf\,\mathrm{d}\mu+\int_E hf\,\mathrm{d}\nu=\int_E h\,\mathrm{d}\nu$$
    from which we obtain
    $$\int_E hf\,\mathrm{d}\mu=\int_E h(1-f)\,\mathrm{d}\nu.$$
    This gives $f\,\mathrm{d}\mu=(1-f)\,\mathrm{d}\nu$ from which the proposed expression for $\frac{\mathrm{d}\nu}{\mathrm{d}\mu}$ follows.

\end{enumerate}

\item Let $(X,\mathcal{A},\mu)$ be a finite measure space, $f$ a nonnegative $\mu$-measurable function, $\mathcal{B}$ a subalgebra
of $\mathcal{A}$ and $\nu=\mu|_{\mathcal{B}}$.  Define
$$f_n(x)=\min\{f(x),n\}$$ and observe that $f_n\rightarrow f$ pointwise as $n\rightarrow\infty$ and that the sequence $\{f_n\}$
is monotone.  Let
$$\lambda_n(E)=\int_E f_n\,\mathrm{d}\mu.$$
Then since $X$ is finite and $E\subset X$ and $f_n$ is bounded above (and below) for each $n$ we see that $\lambda_n$ is a finite measure
on $(X,\mathcal{B})$.  Moreover, if $\mu(E)=0$ then we have $\lambda_n(E)=0$ so that $\lambda_n\ll\mu$ for each $n$, and since $\nu$
is just a restriction of $\mu,$ we also have $\lambda_n\ll\nu$.  Since $\nu$ is finite (and also $\lambda_n$ is finite by construction)
we may apply the Radon-Nikodym Theorem to obtain a sequence of functions $g_n$ with
$$\int_E g_n\,\mathrm{d}\nu =\lambda_n(E)=\int_E f_n\,\mathrm{d}\mu$$ for each $n$.  Moreover, since the sequence $\{f_n\}$
is monotone, we have that the sequence $\{g_n\}$ is monotone almost everywhere.  We then define
$h_n(x)=\max\{g_1(x),g_2(x),\ldots,g_n(x)\}$ and observe that since $\{g_n\}$ is monotone a.e. we have $h_n=g_n$ a.e. and $\{h_n\}$
monotone.  The limit $\lim_{n\rightarrow\infty} h_n$ therefore exists and we call it $h$.  From our remarks above, we now have
\begin{align*}
    \int_E f\,\mathrm{d}\mu&= \int_E \lim f_n\,\mathrm{d}\mu\\
    &= \lim \int_E f_n\,\mathrm{d}\mu\\
    \intertext{by the Monotone Convergence Theorem,}
    &= \lim \int_E g_n\,\mathrm{d}\nu\\
    &=\lim \int_E h_n\,\mathrm{d}\nu\\
    &= \int_E \lim h_n\,\mathrm{d}\nu\\
    \intertext{again by MCT,}
    &= \int_E h\,\mathrm{d}\nu
\end{align*}
for any $E\in\mathcal{B}$, as required.

\item
\begin{enumerate}
    \item Let $\mu=\lambda-\nu$ be a signed measure on a space $X$ with $\lambda,\nu$ positive.  As usual, we have the decomposition
    $X=X^+\cup X^-$ for $\mu$. By definition, we have
    $$\mu^+(E)=\mu(E\cap X^+)=(\lambda-\nu)(E\cap X^+)=\lambda(E\cap X^+)-\nu(E\cap X^+)\leq \lambda(E\cap X^+)$$
    since $\nu$ is positive.  But since $E\cap X^+\subset E,$ we have $\lambda(E\cap X^+)\leq \lambda(E)$ so that
    $\mu^+\leq \lambda$.  Similarly, we have
    $$-\mu^-(E)=\mu(E\cap X^-)=(\lambda-\nu)(E\cap X^-)=\lambda(E\cap X^-)-\nu(E\cap X^-)\geq -\nu(E\cap X^-)$$
    since $\lambda$ is positive.  This gives $-\mu^-(E)\geq -\nu(E)$ so that $\nu\geq \mu^-$.

    \item Observe that for any signed measures $\mu_1,\mu_2$ we may write
    $\mu_1+\mu_2=(\mu_1^+ +\mu_2^+) -(\mu_1^- + \mu_2^-)$.  Since $\mu_1^+ +\mu_2^+$ and $\mu_1^- + \mu_2^-$ are positive, we may apply
    part (a) to obtain $(\mu_1+\mu_2)^+\leq \mu_1^+ + \mu_2^+$ and $(\mu_1+\mu_2)^-\leq \mu_1^- + \mu_2^-$.
    Adding these inequalities together gives
    $$(\mu_1+\mu_2)^+ + (\mu_1+\mu_2)^- \leq (\mu_1^+ +\mu_1^-)+ (\mu_2^+ + \mu_2^-).$$
    Recalling that $|\mu|=\mu^-+\mu^+$ we obtain $|\mu_1+\mu_2|\leq |\mu_1|+|\mu_2|$ as required.

\end{enumerate}

\end{enumerate}
\end{document}
