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\begin{document}
\begin{enumerate}
\centerline{\sc Math 597.  Homework 3\\ Bryden Cais}
\medskip

\item Let $\mathcal{Q}$ consist of $X,\emptyset,$ and all singletons of $X$.  Define the set functions $\lambda_i:\mathcal{Q}\rightarrow\R$
by
\begin{align*}
    \lambda_1(X) &= \infty & \lambda_1(\emptyset)&= 0 & \lambda_1(E)&=1\\
    \intertext{for all $E\neq X,\emptyset$ in $\mathcal{Q}$, and}
    \lambda_2(X) &= 1 & \lambda_2(E)&= 0 & &\\
\end{align*}
for all $E\neq X$ in $\mathcal{Q}$.  Then we claim that for any set $U\subset X$,
\begin{align*}
    \mu_{\lambda_1}^{\ast}(U)&=\begin{cases}\# U & \text{if}\ \# U <\infty\\ \infty & \text{otherwise}\end{cases},\\
    \mu_{\lambda_2}^{\ast}(U)&=\begin{cases}0 & \text{if $U$ is countable}\\ 1 & \text{otherwise}\end{cases}.
\end{align*}
Indeed, if $U$ is countable we have
$$\mu_{\lambda_1}^{\ast}(U)=\inf\left\{\sum \lambda_1(Q_j):U\subset \bigcup Q_j\right\}=\sum_{x\in U} \lambda_1(x)=\sum_{x\in U} 1,$$
which is clearly equal to $\# U$ if $\# U$ is finite and $\infty$ otherwise.  On the other hand, if $U$ is not countable, then the only
way to cover $U$ by a countable union of sets in $\mathcal{Q}$ is by $X$ itself (possibly with additional terms, but at any rate,
$X$ must be part of the cover) so that $\mu_{\lambda_1}^{\ast}(U)=\infty$, as claimed.

Again, for countable $U$ we have
$$\mu_{\lambda_2}^{\ast}(U)=\inf\left\{\sum \lambda_2(Q_j):U\subset \bigcup Q_j\right\}=\sum_{x\in U} \lambda_2(x)=\sum_{x\in U} 0=0,$$\
while for uncountable $U$, we must have $X$ as an element of our cover $Q_j$ so that
$$\mu_{\lambda_2}^{\ast}(U)=1$$ (since we can take the cover consisting only of $X$).

Let us now determine the $\sigma$-algebra of $\mu_{\lambda_i}^{\ast}$ for $i=1,2$.  We have already seen in class that for $i=1$,
every subset of $X$ is measurable.  Indeed, for any $U\subset X$ we have two cases:  if $\mu_{\lambda_1}^{\ast}(A)<\infty$
then certainly $\mu_{\lambda_1}^{\ast}(A\cap E),\mu_{\lambda_1}^{\ast}(A\setminus E)<\infty$ by monotonicity, and
$$\mu_{\lambda_1}^{\ast}(A\setminus E)+\mu_{\lambda_1}^{\ast}(A\cap E)=\# (A\setminus E)+\#(A\cap E)=\# A \leq \mu_{\lambda_1}^{\ast}(A).$$
On the other hand, if $\mu_{\lambda_1}^{\ast}(A)=\infty$ then there is nothing to prove.
Now we claim that the $\sigma$-algebra of $\mu_{\lambda_2}^{\ast}$ measurable sets is generated (under countable unions and complemantations)
by all countable subsets of $X$.  For it is clear that every set that is obtainable in this way is either countable, or
its complement in $X$ is countable.  (One must check a few trivial things: for example, obviously countable unions of countable sets are
countable.  The key is that $(X\setminus E_1)\cup (E_2)=X\setminus((X\setminus E_1)\cap E_2)$ which is the complement of a countable
set if $E_1,E_2$ are countable.)  Observe that for any $A\subset X$,if $E$ or $X\setminus E$ is countable then
$$\mu_{\lambda_2}^{\ast}(A\setminus E)+\mu_{\lambda_2}^{\ast}(A\cap E)=\mu_{\lambda_2}^{\ast}(A).$$
On the other hand, if $E$ and $X\setminus E$ are uncountable then for any $A\subset X$, since $\mu_{\lambda_2}^{\ast}(A)\leq 1$
we have
$$\mu_{\lambda_2}^{\ast}(X\setminus E)+\mu_{\lambda_2}^{\ast}(X\cap E)=1+1> \mu_{\lambda_2}^{\ast}(X)$$
so that $E$ is not measurable.  Another way to characterize this $\sigma$-algebra (which ic plain from the description above) is as the
collection of all subsets of $X$ that are either countable or have countable complement in $X$.

\item It is clear that $\mu_{\lambda}^{\ast}(U)=\lambda(U)$ when $U\in\mathcal{Q}$, so that we need only determine
$\mu_{\lambda}^{\ast}(U)$ when $U$ is a set consisting of one element.  But the only way to cover a one element set of $X$
is by a set in $\mathcal{Q}$ of measure at least $1$.  On the other hand, every element of $X$ is contained in one of the
two-element sets in $\mathcal{Q}$ so that any (nonempty) singleton has measure $1$.  We claim that no subsets of $X$ other than $X,\emptyset$
are measurable.
For let $E\subset X$ consist of one element and let $A\supset E$ be any two element set.  Then
$\mu_{\lambda}^{\ast}(A\cap E)+\mu_{\lambda}^{\ast}(A\setminus E)=1+1>\mu_{\lambda}^{\ast}(A)=1.$
If $E$ is a two element set, let $A\neq E$ be any two element set having nontrivial intersection with $E$.
Then, again,
$\mu_{\lambda}^{\ast}(A\cap E)+\mu_{\lambda}^{\ast}(A\setminus E)=1+1>\mu_{\lambda}^{\ast}(A)=1.$
Clearly $X,\emptyset$ are measurable as for any $A$ we have $A\cap X=A$ and $A\setminus X=\emptyset$ while
$A\cap \emptyset =\emptyset$ and $A\setminus\emptyset=A$.

\item Let $\mathcal{Q}$ be the collection of squares in $\R^2$ together with the empty set and set $\lambda(Q)=\diam Q$.
Assume that $\mu_{\lambda}^{\ast}(Q)=\lambda(Q)$.  Define $Q_j$ as in the problem set for $1\leq j\leq 4$.
\begin{enumerate}
    \item Since $\bigcup_{j=1}^4 Q_j$ is covered by the square $[0,3]\times [0,3]$, we have
    $$\mu_{\lambda}^{\ast}\left(\bigcup_{j=1}^4 Q_j\right)\leq \diam([0,3]\times [0,3])=3\sqrt{2}.$$
    On the other hand, $\mu_{\lambda}^{\ast}(Q_j)=\sqrt{2}$ for each $j$ by assumption so that
    $$\mu_{\lambda}^{\ast}\left(\bigcup_{j=1}^4 Q_j\right)\leq 3\sqrt{2}<4\sqrt{2}=\sum_{j=1}^4 \mu_{\lambda}^{\ast}(Q_j).$$

    \item Observe that $\mu_{\lambda}^{\ast}$ is translation invariant.  Indeed, if $Q_j$ is any covering of
    $E$ with $\sum_{j}\lambda(Q_j)=\alpha$ then $x+Q_j$ is a covering of $x+A$ with $\lambda(x+Q_j)=\lambda(Q_j)$
    and hence $\sum_{j}\lambda(x+Q_j)=\alpha$.  Conversely, we may replace $x$ by $-x$ to obtain a covering
    of $E$ from a covering of $x+E$.  This gives a bijection of sets
    $$\left\{\sum_{j}\lambda(Q_j):E\subset \bigcup Q_j\right\}\leftrightarrow \left\{\sum_{j}\lambda(Q'_j):x+E\subset \bigcup Q'_j\right\}$$
    from which it follows that $\mu_{\lambda}^{\ast}(E)=\mu_{\lambda}^{\ast}(x+E)$.
    Now if $Q_1$ were measurable, then $Q_2$ would be also, since it is just a translation of $Q_1$.  For suppose not, and
    let $A$ be such that
    $$\mu_{\lambda}^{\ast}(A\setminus Q_2)+\mu_{\lambda}^{\ast}(A\cap Q_2)\neq \mu_{\lambda}^{\ast}(A).$$
    Then $A'=A-(2,0)$ satisfies
    $$\mu_{\lambda}^{\ast}(A\setminus Q_2)+\mu_{\lambda}^{\ast}(A\cap Q_2)=\mu_{\lambda}^{\ast}(A'\setminus Q_1)+\mu_{\lambda}^{\ast}(A'\cap Q_1)$$
    and $\mu_{\lambda}^{\ast}(A)=\mu_{\lambda}^{\ast}(A'),$ so that $Q_1$ is not measurable.  Thus, if $Q_1$ is measurable, then
    $Q_i$ is measurable for $1\leq i\leq 4$ and also $\bigcup_{i=1}^4 Q_i$ is measurable.  But part (a) contradicts this, as
    countable additivity of measurable sets holds for $\mu_{\lambda}^{\ast}$.  It follows that $Q_1$ is not measurable.
\end{enumerate}

\end{enumerate}
\end{document}
