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\begin{document}
\begin{enumerate}
\centerline{\sc Math 597.  Homework 4\\ Bryden Cais}
\medskip

\item \begin{enumerate}
    \item For any $\e >0$ there exists a dyadic cover $Q_j$ of $E$ such that $\mu(E)+\epsilon\geq \sum \mu(Q_j)$.  Moreover, we have
    $$\mu(E)\leq \sum \mu(Q_j\cap E)$$ by countable subadditivity of $\mu$ so that by hypothesis (since the $Q_j$ are dyadic) we have
    $$\mu(E)\leq \sum \mu(Q_j\cap E)\leq c\sum \mu(Q_j)\leq c(\mu(E)+\e).$$  It follows that we have, for any $\e >0$
    $\mu(E)\leq c\mu(E)+\e$ so that
    $$\mu(E)\left(\frac{1-c}{c}\right)\leq \e.$$
    If $c=0$ we are done since then $\mu(E)\leq \e$ for any $\e >0$.  Otherwise,
    $0<c<1$ so that, $(1-c)/c$ is some fixed nonzero constant.  Since $\e >0$ is arbitrary, we must have $\mu(E)=0$.

    \item Let $F=[0,1]\setminus E$.  Then since $E$ is measurable, do is $F$.  Moreover, for any dyadic interval $I$ we have
    $$\mu(F\cap I)=\mu(([0,1]\cap I)\setminus (E\cap I))=\mu(I)-\mu(E\cap I)\leq |I|-\tilde{c}|I|=(1-\tilde{c})|I|.$$
    Now obviously $\tilde{c}\leq 1$ and $\tilde{c}>0$ by assumption so that $0\leq 1-\tilde{c} <1$.  We now apply part (a)
    and deduce that $\mu(F)=0$.  It follows that $\mu(E)=\mu([0,1])=\mu(F)=1$.
\end{enumerate}

    \item Define $I_n^k=[\tfrac{k}{2^n},\tfrac{k+1}{2^n}]$ and for any $0<\a <1$ and any closed interval $I$, let $\varphi_{\a}(I)$
    be the cantor set constructed on $I$ with constant $1-\a,$ so that $\varphi_{\a}(I)$ has measure $\a |I|$.
    We define $E_j$ and $\a_j$ recursively as follows:  let $\a_0=1/2$ and $E_0=\varphi_{1/2}([0,1])$, so that $E_0$ has measure
    $1/4$.  Supposing that $\a_j$ is defined for $0\leq j <n$ define $E_j$ for $0\leq j <n$ as
    $$E_j = \bigcup_{k=0}^{2^j-1} \varphi_{\a_j}(I_j^k)$$  and define
    $$\beta_n^l=\mu\left(I_n^l\cap \bigcup_{0\leq j <n} E_j\right).$$
    We claim that $\beta_n^l < |I_n^l|$.  Indeed, observe that $E_j$ is closed in $[0,1]$ and contains no intervals; this follows
    directly from the cantor process.  It follows that $\bigcup_{0\leq j <n} E_j$ is closed and contains no intervals since a {\em finite}
    union of closed sets that contain no intervals is again closed and contains no intervals.  Since $I_n^l$ is closed in $[0,1]$,
    the intersection $I_n^l\cap \bigcup_{0\leq j <n} E_j$ is closed in $I_n^l$ in the subspace topology.  Therefore,
    $I_n^l\setminus \bigcup_{0\leq j <n} E_j$ is open in $I_n^l$ in the subspace topology and nonempty (since $\bigcup_{0\leq j <n} E_j$
    contains no intervals).  Since any nonempty open set contains an interval, it follows that $I_n^l\setminus \bigcup_{0\leq j <n} E_j$
    contains an interval and hence that $\beta_n^l < |I_n^l|$.  We therefore set
    $$\e_n^l=|I_n^l|-\beta_n^l.$$  We have just shown that $\e_n^l >0$.  We now define
    $$\a_n=\frac{1}{2^n}\min_{\substack{0\leq q \leq 2^n -1\\ 0\leq r \leq n}} \{\e_r^q,\frac{1}{2^{n+2}}\}.$$
    Observe that $\a_n >0$ since the minimum is taken over a finite set.

    We have thus defined the sets $E_j$ and the constants $\a_j$ recursively.  Let $E=\bigcup_{j=0}^{\infty} E_j$.
    We claim that $\mu(E\cap I_n^l) < |I_n^l|$.  Indeed, we have
    \begin{align}
    \mu(E\cap I_n^l)\leq \mu\left(I_n^l\cap \bigcup_{0\leq j <n} E_j\right)+\sum_{j\geq n}\mu(E_j\cap I_n^l)=\beta_n^l+\sum_{j\geq n}\mu(E_j\cap I_n^l).
    \label{junk}
    \end{align}
    by countable subadditivity.  We now estimate $\mu(E_j\cap I_n^l)$ for $j\geq n$.  Since $E_j$ is a union of sets of the form
    $\varphi_{\a_j}(I_j^m)$, each set having measure $\a_j|I_j^m|$, and $j\geq n$, we see that $I_n^l\cap E_j$ is a union of
    $2^{j-n}$ sets of the form $\varphi_{\a_j}(I_j^m)$, each set having measure $\a_j |I_j^m|=\frac{a_j}{2^j}$.  Therefore,
    $$\mu(E_j\cap I_n^l)=2^{-n}\a_j.$$  It follows that
    \begin{align}
    \sum_{j\geq n} \mu(E_j\cap I_n^l)&=2^{-n}\sum_{j\geq n} \a_j\leq 2^{-n}\sum_{j\geq n} 2^{-j} \e_n^l=\frac{1}{2^{2n-1}}\e_n^l\label{crap}
    \end{align}
    since
    $$\a_j =\frac{1}{2^j}\min_{\substack{0\leq q \leq 2^j -1\\ 0\leq r \leq j}} \{\e_r^q,\frac{1}{2^{j+2}}\} \leq \frac{1}{2^j}\e_n^l$$
    for $j\geq n$ since, afterall, $\e_n^l$ is one of the terms in the minimum.
    Combining (\ref{crap}) with (\ref{junk}), we have
    $$\mu(E\cap I_n^l)\leq \beta_n^l + \frac{\e_n^l}{2^{2n-1}} < |I_n^l|$$ by the definition of $\e_n^l$ and $\b_n^l$.
    Since any dyadic interval $I$ is a countable disjoint union of intervals of the form $I_n^l$, it follows that for
    any dyadic interval $I$ we have $\mu(E\cap I) < |I|$.

    Observe that $E\cap I_n^l$ contains the set $\varphi_{\a_n}(I_n^l)$ of measure $\a_n |I_n^l| >0 $ so by monotonicity of $\mu$,
    we have $0<\mu(E\cap I_n^l)$.  Again, $0<\mu (E\cap I)$ for any dyadic interval $I$ follows as above.
    It is also obvious from this that $0< \mu(E)$.  On the other hand, we have
    $$\mu(E)\leq \sum_{0\leq j}\mu(E_j)\leq \sum_{\substack{0\leq j\\ 0\leq l\leq 2^j-1}} \mu(\varphi_{\a_j}(I_j^l))\leq
     \sum_{0\leq j} \a_j\leq \sum_{0\leq j}\frac{1}{2^{2j+2}}<1,$$
     so we are done.

    \item Let $f$ be the cantor ternary function extended by $f(x)=0$ for $x<0$ and $f(x)=1$ for $x>1$.  Then $f$ is right-continuous as
    we saw in class and $\C$ is measurable since it is Borel (again from class).
    \begin{enumerate}
        \item  Let $\e>0$ be arbitrary.  Since $f$ is right-continous there exists a $\delta >0$ such that $|f(x)-f(y)|<\e$
        whenever $|x-y|<\delta$.  Now we may cover $x$ by the set $I=(x-\delta/2,x+\delta/2]$, and we have $\mu_f(I)<\epsilon$.
        Since $\e$ was arbitrary, it follows that $\mu_f(\{x\})=0.$

        \item From lecture, since $f$ is right continuous we have $\mu_f((0,1])=f(1)-f(0)=1$.  Since $\mu_f(0)=0$
        by part (a), it follows that $\mu_f([0,1])=1$.

        \item We may cover $[0,1]\setminus \C$ by sets of the form $(\a_j,\b_j]$.  Here $(a_j,b_j)$ are all the intervals thrown away
        in the construction of $\C$.  From the definition of $f$ and right continuity, we have $f(b_j)=f(a_j)$ for all $j$
        and therefore $\mu_f((a_j,b_j])=0$.  It follows that $\mu_f([0,1]\setminus \C)=0$.

        \item Since $\C$ is $\mu_f$ measurable (being Borel), we have
        $$\mu_f([0,1])=\mu_f([0,1]\cap \C)+\mu_f([0,1]\setminus \C)=\mu_f(\C)$$ by part (c).
        Then by part (a), it follows that $\mu_f(\C)=1$.
    \end{enumerate}
\end{enumerate}
\end{document}
