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\begin{document}
\begin{enumerate}
\centerline{\sc Math 597.  Homework 5\\ Bryden Cais}
\medskip

\item \begin{enumerate}
    \item Let $f$ be non-decreasing, right continuous function.  Set
    \begin{align*}
        S_1 &= \left\{\sum (f(b_i)-f(a_i)) : E\subset \bigcup_i (a_i,b_i]\right\},\\
        S_2 &= \left\{\sum (\b_i-\a_i) : f(E)\subset \bigcup_i [\a_i,\b_i]\right\},
    \end{align*}
    and recall that
    $\mu_f^{\ast(E)}=\inf S_1 $ and $\ml(f(E)) = \inf S_2$ where in each case the infimum is taken over
    all coverings by intervals of the sets $E,f(E)$ respectively (note that it is inconsequential that
    we are taking closed coverings for the Lebesgue measure).  Now since $f$ is non-decreasing, if $E \subset \bigcup_i (a_i,b_i]$
    then $f(E)\subset \bigcup_i [f(a_i),f(b_i)]$ (since for any interval $I_i=(a_i,b_i]$ we have $f(a_i)\leq f(x) \leq f(b_i)$ for all
    $x\in I_i)$.  It follows that every covering by intervals $(a_i,b_i]$ gives a covering $[f(a_i),f(b_i)]$ of $f(E)$ so that
    every element of $S_1$ is an element of $S_2$.  It follows that $\inf S_1 \geq \inf S_2$ or
    $$\mu_f^{\ast}(E)\geq \ml(f(E)).$$

    \item Let $f:[0,1]\rightarrow \R$ be given by $$f(x)=\begin{cases}1 & \text{if}\ x=1\\ 0 & \text{otherwise}\end{cases}.$$
    Observe that $f$ is nondecreasing and right continuous.  However we see that
    $$\mu_f^{\ast}((0,1])=f(1)-f(0)=1$$ while
    $$\ml(f(E))=\ml(\{0,1\})=0.$$


    \end{enumerate}

\item Let $f(x,y)=\min\{x,y\}$ and define $$\lambda((x_1,x_2]\times (y_1,y_2]) = f(x_2,y_2)+f(x_1,y_1)-f(x_1,y_2)-f(x_2,y_1).$$
Let $\Q=\{\emptyset\}\cup \{(x_1,x_2]\times (y_1,y_2]: x_1 < x_2,\ y_1 < y_2\}.$  We determine a simplified geometric description
of $\mu_f^{\ast}$.  Observe that for any rectangle $R$ lying entirely above the line $x=y$ we have $f(x_i,y_i)=\min\{x_i,y_i\}=x_i$ for all vertices
$x_i,y_i$ of $R$, and therefore
$$\lambda(R) = f(x_2,y_2)+f(x_1,y_1)-f(x_1,y_2)-f(x_2,y_1) = x_2+x_1-x_1-x_2 = 0.$$
Similarly, for any rectangle $R$ lying completely below the line $y=x$, we have via a similar computation that $\lambda(R)=0$.
Therefore, the only rectangles $R$ having nonzero pre-measure are those that intersect the line $y=x$.  Now if a rectangle
$R$ intersects the line $y=x$, we may assume that $R$ is in fact a square and the line $y=x$ lies on the diagonal: indeed, we may ``lop''
off the other portions of $R$ that lie entirely above or below the diagonal.  Now for a square $R$ whose diagonal lies on the line $y=x$
it is not hard to see that
$$\lambda(R) = f(x_2,y_2)+f(x_1,y_1)-f(x_1,y_2)-f(x_2,y_1)= x_2 + y_1 - x_1 - y_1 = x_2-x_1, $$
so that $\lambda(R)=\frac{1}{\sqrt{2}}|\diam R|$.  It follows that $\mu_f^{\ast}(E)=\frac{1}{\sqrt{2}}\mu(E\cap\{y=x\})$
where $\mu$ is Lebesgue measure, i.e., $\mu_f^{\ast}(E)$ gives the ``$1/\sqrt{2}$ times the amount of $E$ that lies on the
line $y=x$.''


\item \begin{enumerate} Fix $E\subset [0,1]$.  Let $$f(x)=\begin{cases}x & \text{if}\ x\in E\\ x+5 &\text{if} x\not\in E\end{cases}.$$
    \item Observe that
    $$f^{-1}(\a)= \begin{cases}\emptyset & \text{if}\ x\not\in E\cup ([0,1]\setminus E +5)\\ \a & \text{if}\ \a\in E \\ \a-5 &
    \text{if}\ \a\in [0,1]\setminus E + 5\end{cases},$$
    so that in any case $f^{-1}(\alpha)$ is empty or a single point for all $\a\in \R$, whence $f^{-1}(\alpha)$ is measurable
    for all $\a\in \R$.

    \item On the other hand, $f$ need not be measurable.  Indeed, if $E$ is not measurable then $[0,1]\setminus E$ is not measurable
    and we have $[0,1]\setminus E= \{x: f(x)\geq 5\}$ is not measurable so $f$ is non-measurable.


    \end{enumerate}

\item Recall from class that we defined a continuous function $f:[0,1]\rightarrow [0,1]$ with continuous inverse that mapped a set
$S$ of measure $1$ to a set of measure 0.  Let $E\subset S$
be any non-measurable set.  Observe that $f(E)\subset f(S)$ must have outer measure 0 and is hence measurable. Define
$$g(x)=\begin{cases}\phantom{-}1 &\text{if}\ x\in f(E)\\ -1 &\text{otherwise}\end{cases}.$$
Observe that since $f(E)$ is measurable,
$$\{x: g(x) \geq c\}$$ is measurable (since it is either $f(E)$ or $[0,1]$), and hence $g$ is measurable.
However, we have
$$\{x: g\circ f (x)\geq 0\} = f^{-1}(\{y: g(y)\geq 0\}) = f^{-1}f(E)=E,$$
which is non-measurable (where we have used that $f^{-1}$ is continuous and $f$ is bijective).  It follows that
$g\circ f$ is not a measurable function.

\end{enumerate}
\end{document}