analysis--berkeley-202a-hw11.tex

\section{Math 202A - HW11 - Dan Davison - \texttt{ddavison@berkeley.edu}}

\begin{verbatim}
Review 1. Dirichlet's test is not valid here, because assumption (2) fails if \kappa = 0. (-2) You do have a
lower bound on the integral based on the harmonic series but you should be more specific how you got it. If
you're going to argue by picture you should probably draw a picture. (-2) Aidan Backus, Nov 21 at 1:29pm

3. What happens if \mu(X_i) = 0? (-2)
Aidan Backus, Dec 1 at 5:43am
\end{verbatim}

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\includegraphics[width=400pt]{img/analysis--berkeley-202a-hw11-8650.png}
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\begin{verbatim}
1. Dirichlet's test is not valid here, because assumption (2) fails if \kappa = 0. (-2) You do have a lower
bound on the integral based on the harmonic series but you should be more specific how you got it. If you're
going to argue by picture you should probably draw a picture. (-2)
\end{verbatim}

\begin{lemma}[Dirichlet's test for improper integrals]
  Let $I = \int_a^\infty f(x) g(x) \dx$ where $\int$ denotes the Riemann integral. Then $I$ converges if all
  the following are true
  \begin{enumerate}
  \item $f$ is continuous on $[a, \infty]$
  \item $\int_a^x f(t) \dt$ is bounded on $[a, \infty]$
  \item $g$ is differentiable on $[a, \infty]$ with $g' \leq 0$ and $\lim_{x\to\infty} g(x) = 0$.
  \end{enumerate}
\end{lemma}

\begin{proof}
  Note that $f_\kappa$ is an even function, therefore (hint from @ankit in
  Slack) $\int_{-R}^R f_{\kappa}(x) \dx = 2\int_{0}^R f_\kappa(x)$ and
  therefore $I_\infty = \frac{1}{2}\int_0^\infty f_{\kappa}(x) \dx$, if this limit (in the upper bound of the
  integral) exists.

  We now (hint from Piazza) invoke Dirichlet's test with $a = 0$, $g(x) = 1/(1 + x)$,
  and $f(x) = \cos(\kappa x)$, where these variable names refer to the statement in the lemma. We see that $f$
  is continuous on $[0, \infty]$ as required; we see that $\int_0^x f_{\kappa}(t) \dt = \sin(\kappa t)$ is
  bounded on $[0, \infty]$; and we see that $g$ is differentiable on $[0, \infty]$ with $g'(x) = -1/(1 + x)^2$
  and that $g(x) \to 0$ as $x \to \infty$. Therefore we conclude that $\int_0^\infty f_{\kappa}(x) \dx$
  converges and therefore that $f_{\kappa}$ is Riemann integrable for all $\kappa$.

  By definition, $f_{\kappa}$ is Lebesgue integrable if $\int_{-\infty}^\infty |f_{\kappa}| < \infty$.
  Since $|f_{\kappa}|$ is even and non-negative, $f_{\kappa}$ is Lebesgue integrable if and only
  if $\int_{0}^\infty |f_{\kappa}| < \infty$.

  Note that (by considering the graphs of $|\cos(\kappa x)|$ and $1/(1 + x)$)
  \begin{align*}
    \int_{0}^\infty |f_{\kappa}|
    = \int_0^\infty \frac{|\cos(\kappa x)|}{1 + x}
    > \(\int_0^{2\pi}|\cos(\kappa x)|\) \sum_{n=1}^\infty \frac{1}{1 + 2n\pi}.
  \end{align*}
  Clearly $\int_0^{2\pi}|\cos(\kappa x)| > 0$. But $\sum_{n=1}^\infty \frac{1}{1 + 2n\pi}$ is a divergent
  series, therefore $f_{\kappa}$ is not Lebesgue integrable for any value of $\kappa$.
\end{proof}

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\includegraphics[width=400pt]{img/analysis--berkeley-202a-hw11-b6a3.png}
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\begin{proof}
  We will prove this by contradiction. Let $P$ be the proposition
  \begin{quote}
    {\it There exists $y \in \R$ such that $m(\{x \in [0, 1] ~:~ \phi(x) < y\}) > 0$ and $m(\{x \in [0, 1] ~:~ \phi(x) > y\}) > 0$.}
  \end{quote}
  Since $\phi$ is integrable, it is measurable, and therefore these preimages are measurable sets.

  The statement
  \begin{quote}
    ​{\it $\phi(x)$ equals a constant a.e.}
  \end{quote}
  is false if and only if $P$ is true. Therefore we suppose, for a contradiction, that $P$ is true.

  Let $\tau: [0, 1) \to [0, 1)$ be the map defined by $x \mapsto x + \alpha \mod \Z$, and let $y \in \R$ be a
  value satisfying $P$, such that $A = \{x \in [0, 1] ~:~ \phi(x) < y\}$
  and $B = \{x \in [0, 1] ~:~ \phi(x) > y\}$ both have positive measure.

  But recall from HW6 Q6 that $\mu(\bigcup_{n=0}^\infty \tau^n(E)) = 1$ for every set $E \subseteq [0, 1]$,
  where $\tau^n(E)$ is the image of $E$ under the $n$-th iterate of $\tau$.

  Therefore $\mu(\bigcup_{n=0}^\infty \tau^n(A)) = 1$. But we have $\phi(x) = \phi(x + \alpha)$ a.e.
  therefore $m(\{x \in \bigcup_{n=0}^\infty \tau^n(A) ~:~ \phi(x) < y\}) = 1$. %  (should justify this better)

  Similarly $\mu(\bigcup_{n=0}^\infty \tau^n(B)) = 1$ and $m(\{x \in \bigcup_{n=0}^\infty \tau^n(A) ~:~ \phi(x) > y\}) = 1$.

  But this is a contradiction, since we simultaneously have $\phi(x) < y$ a.e. and $\phi(x) > y$ a.e., which
  violates countable additivity of $m$.
\end{proof}

% The problem with the thoughts below, I think, is that the function $\phi$ being described is not measurable.

% This is true because the orbit of the map $x \mapsto x + a$ is dense in $[0, 1]$. We basically just need to show that.

% We must show that $\phi(x)$ equals a constant for almost all $x$.

% Why should that be? Why, for example, can't each orbit have its own constant value?


% \begin{proof}
%   Let $\tau: [0, 1) \to [0, 1)$ be the map defined by $x \mapsto x + \alpha \mod \Z$, and for $x \in [0, 1)$
%   define the sequence $(r_n(x)) = x, \tau(x), \tau^2(x), \ldots$, and
%   let $\orb(x) := \big\{r_n(x) ~:~ n \in \N\big\}$ be the orbit of $x$ under $\tau$. Recall from HW2 and HW6
%   that for each $x \in [0, 1)$ the sequence $(r_n(x))$ is non-repeating, the corresponding orbit $\orb(x)$ is
%   dense in $[0, 1]$, and each orbit is disjoint from every other orbit. Thus each orbit is a countable set and
%   there are uncountably many distinct orbits.

%   We have that the set of points $x$ at which $\phi(\tau(x)) = \phi(x)$ has measure $1$. Therefore we have the
%   following qualitative picture: there are uncountably many disjoint orbits and each orbit is associated with a
%   sequence $\phi(x), \phi(\tau(x)), \phi(\tau^2(x)), \ldots$. The set of points at which this sequence changes
%   value has measure $0$.

%   We must use the fact that $\phi$ is integrable to show that $\phi$ is constant a.e.

%   Let $\eps > 0$ and let $g: [0, 1] \to \R$ be a continuous function such that
%   \begin{align*}
%     \int |g - \phi| < \eps.
%   \end{align*}

%   Let $\eps > 0$. According to Lusin's theorem there exists a closed set $F \subseteq [0, 1]$
%   with $m(F) > 1- \eps$ such that $\phi|_F$ is continuous. So let $F \subseteq [0, 1]$ be such that $m(F) = 1$
%   and $\phi|_F$ is continuous.

%   Note that $F$ is uncountable and therefore it contains points from uncountably many orbits.


%   Probabilistic view:

%   - Pick $x$. It is a member of some orbit and has some value $\phi(x)$. We have $Pr(\phi(x + \alpha) = \phi(x)) = 1$.

%   - Suppose the claim is not true. Then there exist $c$ and $d$ such that $m(\phi = c) = p_c > 0$ and $m(\phi = d) = p_d > 0$.

%   - Pick $x$. With probability $p_c$ we have $\phi(x) = c$ and also $\phi(x + \alpha) = c$.

%   Why might $\phi$ be constant a.e.?

%   - Each orbit is countable, so if one orbit is constant, that's just a measure zero set.

%   - Similarly, if every orbit is constant, but they all have different values that's not constant a.e.

%   - What we need to show is something more like that all orbits ``start​'' with the same value

% \end{proof}

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\includegraphics[width=400pt]{img/analysis--berkeley-202a-hw11-3704.png}
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\begin{verbatim}
3. What happens if \mu(X_i) = 0? (-2)
\end{verbatim}

\begin{proof}
  Since $\mu$ is $\sigma$-finite we can write $X$ as a countable disjoint union $X = \bigcup_{i=1}^\infty X_i$
  with $\mu(X_i) < \infty$ for all $i$.

  Let $\nu$ be a set function such that for every $A \in \mc A$
  \begin{align*}
    \nu(A) = \sum_{i=1}^\infty 2^{-i} \frac{\mu(A \cap X_i)}{\mu(X_i)}.
  \end{align*}
  We claim that $\nu$ is a measure. We have $\nu(\emptyset) = 0$. Let $A_1, A_2, \ldots \in \mc A$ be pairwise
  disjoint. We see that $\nu$ is countable additive since
  \begin{align*}
    \nu\(\bigcup_{j=1}^\infty A_j\)
    &= \sum_{i=1}^\infty 2^{-i} \frac{\mu\(\(\bigcup_{j=1}^\infty A_j\) \cap X_i\)}{\mu(X_i)} \\
    &= \sum_{i=1}^\infty 2^{-i} \frac{\mu\(\bigcup_{j=1}^\infty \(A_j \cap X_i\)\)}{\mu(X_i)} \\
    &= \sum_{i=1}^\infty 2^{-i} \sum_{j=1}^\infty\frac{\mu\(A_j \cap X_i\)}{\mu(X_i)} \\
    &= \sum_{j=1}^\infty \sum_{i=1}^\infty 2^{-i}\frac{\mu\(A_j \cap X_i\)}{\mu(X_i)} \\
    &= \sum_{j=1}^\infty \nu(A_j).
  \end{align*}
  Therefore $\nu$ is a measure. Furthermore $\nu$ is finite since $\nu(X) = \sum_{i=1}^\infty 2^{-i} = 1$.

  If $A \in \mc A$ and $\mu(A) = 0$ then
  \begin{align*}
    0 \leq \nu(A)
    = \sum_{i=1}^\infty 2^{-i} \frac{\mu(A \cap X_i)}{\mu(X_i)}
    < \sum_{i=1}^\infty 2^{-i} \frac{\mu(A)}{\mu(X_i)}
    = 0,
  \end{align*}
  hence $\nu \ll \mu$. Finally, if $A \in \mc A$ and $\nu(A) = 0$ then $\mu(A \cap X_i) = 0$ for all $i$,
  therefore $\mu(A) = \sum_{i=1}^\infty \mu(A \cap X_i) = 0$, since the $X_i$ partition $X$,
  hence $\mu \ll \nu$.
\end{proof}


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\begin{claim*}
  $\rho \ll \mu$
\end{claim*}
\begin{proof}
  We must show that $\rho(A) = 0$ whenever $\mu(A) = 0$ and $A$ is in the $\sigma$-algebra.

  Let $A$ be in the $\sigma$-algebra such that $\mu(A) = 0$. Then $\nu(A) = 0$, since $\nu \ll \mu$.
  Therefore $\rho(A) = 0$, since $\rho \ll \nu$.
\end{proof}

\begin{claim*}
  \begin{align*}
    \frac{\d\rho}{\d\mu} = \frac{\d\rho}{\d\nu}\frac{\d\nu}{\d\mu} \ae
  \end{align*}
\end{claim*}

\begin{proof}
  By definition, $\frac{\d\rho}{\d\mu}$ is a function such that for any measurable set $E$
  \begin{align*}
    \rho(E) = \int_E \frac{\d\rho}{\d\mu} \d\mu.
  \end{align*}
  We will first show that the function $\frac{\d\rho}{\d\nu}\frac{\d\nu}{\d\mu}$ also serves as a derivative
  of $\rho$ with respect to $\mu$, i.e. that for any measurable set $E$
  \begin{align*}
    \rho(E) = \int_E \frac{\d\rho}{\d\nu}\frac{\d\nu}{\d\mu} \d\mu.
  \end{align*}
  Let $E$ be a measurable set.

  Write $f = \frac{\d\rho}{\d\nu}$ and let $f_n$ be a sequence of increasing simple functions converging
  pointwise to $f$. Let $F$ be a measurable set and note that
  \begin{align*}
    \int_E \ind_F \d\nu
    = \nu(E \cap F)
    = \int_{E \cap F} \frac{\d\nu}{\d\mu} \d\mu
    = \int_E \ind_F \frac{\d\nu}{\d\mu} \d\mu.
  \end{align*}
  By linearity of the integral this result applies to the simple functions $f_n$ and we have
  \begin{align*}
    \int_E f_n \d\nu = \int_{E} f_n \frac{\d\nu}{\d\mu}\d\mu,
  \end{align*}
  and by monotone convergence
  \begin{align*}
    \limninf \int_E f_n \d\nu = \int_E f \d\nu = \int_{E} f \frac{\d\nu}{\d\mu}\d\mu.
  \end{align*}
  Substituting its definition $\frac{\d\rho}{\d\nu}$ in place of the symbol $f$ we have
  \begin{align*}
    \int_E \frac{\d\rho}{\d\nu} \d\nu
    = \rho(E)
    = \int_E \frac{\d\rho}{\d\mu} \d\mu
    = \int_{E} \frac{\d\rho}{\d\nu} \frac{\d\nu}{\d\mu}\d\mu.
  \end{align*}

  Therefore
  \begin{align*}
    \int_E \frac{\d\rho}{\d\mu}  - \frac{\d\rho}{\d\nu}\frac{\d\nu}{\d\mu} \d\mu = 0,
  \end{align*}
  and therefore by Bass proposition 8.2
  \begin{align*}
    \frac{\d\rho}{\d\mu} = \frac{\d\rho}{\d\nu}\frac{\d\nu}{\d\mu} \ae
  \end{align*}
\end{proof}

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\begin{proof}

  [Not attempted -- out of time]

\end{proof}