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\begin{document}
\begin{enumerate}
\centerline{\sc Math 597.  Homework 2\\ Bryden Cais}
\medskip
\item Suppose that $\Q$ is of type $G_{\delta}$, i.e. that we can write $\Q=\bigcap_{i=1}^{\infty} U_i$ where each $U_i$ is open.
Observe that since $\Q$ is dense in $\R$ and $\Q\subset U_i$ for all $i$, each $U_i$ is dense in $\R$, so that $\Q$ is a countable
intersection of dense, open sets.
For any $\alpha\in \Q$, let $S_{\alpha}=(-\infty,\alpha)\cup (\alpha,\infty)$.  Clearly, $S_{\alpha}$ is open and dense in $\R$.
Then $Q\setminus\{\alpha\}=\Q\cap S_{\alpha}$ is a countable intersection of dense open sets, and hence
$$\bigcap_{\alpha\in\Q}Q\setminus\{\alpha\}=\emptyset$$ is a countable intersection of dense open sets, and by Baire's Theorem,
is dense.  This is absurd, so that $\Q$ is not of type $G_{\delta}$.

\item Recall that Axiom ({\em iv}) in the definition of a measure is that $\mu(E)<\infty$ for some set $E$.  We show that the axiom
$\mu(\emptyset)=0$ is equivalent to this.  One direction is trivial, as we just set $E=\emptyset.$ In the other direction,
suppose that $\mu(E)<\infty.$  Since $E\cap\emptyset=\emptyset,$ by Axiom ({\em iii}) (countable additivity) we have
$$\mu(E)=\mu(E\cup \emptyset)=\mu(E)+\mu(\emptyset)$$ from which it follows that $\mu(\emptyset)=0$.

\item We have $\lim\sup E_j:=\bigcap_{n=1}^{\infty} \bigcup_{j=n}^{\infty} E_j $ by definition.  Suppose that $\sum_{j=1}^{\infty}\mu(E_j)<\infty$.
Observe, for each $n$ we have $\lim\sup E_j\subset \bigcup_{j=n}^{\infty} E_j$ so that, by the monotonicity of $\mu$, we have
$$\mu(\lim\sup E_j)\leq \mu(\bigcup_{j=n}^{\infty} E_j)=\sum_{j=n}^{\infty} \mu(E_j).$$
Now since $\sum_{j=1}^{\infty}\mu(E_j)<\infty$,
for any $\epsilon >0$ there exists some $N>0$ such that $\sum_{j=N}^{\infty}\mu(E_j)<\epsilon.$  It follows that for any $\epsilon >0$ we have
an $N>0$ with
$$\mu(\lim\sup E_j)\leq \mu(\bigcup_{j=N}^{\infty} E_j)< \epsilon.$$
Since this holds for any $\epsilon >0$ we have $\mu(\lim\sup E_j)=0$.

\item Let $\mu$ be a finitely additive outer measure and $\{E_j\}$ and countable collection of pairwise disjoint measurable sets.
Since $\bigcup_{j=1}^{N}E_j\subset \bigcup_{j=1}^{\infty} E_j$ for every $N>0$, by monotonicity of $\mu$, we have
$\mu(\bigcup_{j=1}^{N}E_j)\leq \mu(\bigcup_{j=1}^{\infty} E_j)$.  Thus, by finite additivity and pairwise disjointness of the $E_j$, we have
$$\sum_{j=1}^N \mu(E_j)\leq \mu(\bigcup_{j=1}^{\infty} E_j).$$  Since this inequality is independent of $N$, it holds in the limit, i.e.
$$\sum_{j=1}^{\infty} \mu(E_j)\leq \mu(\bigcup_{j=1}^{\infty} E_j).$$
Countable subadditivity of $\mu$ gives the reverse inequality, so we must have
$$\sum_{j=1}^{\infty} \mu(E_j)= \mu(\bigcup_{j=1}^{\infty} E_j).$$  Thus, $\mu$ is countably additive.

\item Recall that
$$\H_{\a,\e}(E)=\inf\left\{\sum \diam(Q_j)^{\a}:E\subset \bigcup Q_j,\ \diam(Q_j)<\epsilon\right\},$$ and that without loss of generality
we may suppose that $Q_j\subset E$ for all $j$ (which we assume from now on).
Observe that if $E\subset \bigcup Q_j$ then $f(E)\subset \bigcup f(Q_j)$, and that since
$$||f(x)-f(y)||\leq C||x-y||^{\eta}$$ we have $\diam(f(Q_j))^{\a}\leq C^{\a}\diam(Q_j)^{\a\eta}.$  It follows that for every
covering $\{Q_j\}$ of $E$ with $$\diam(Q_j)<\e $$ we have the covering $\{f(Q_j)\}$ of $f(E)$ with $$\diam(f(Q_j))<C\e^{\eta}$$ such that
$$\sum \diam(f(Q_j))^{\a}\leq \sum C^{\a}\diam(Q_j)^{\a\eta}.$$
Thus, for every element $x$ of the set
$$\left\{\sum \diam(Q_j)^{\a\eta}:E\subset \bigcup Q_j,\ \diam(Q_j)<\epsilon  \right\}$$
we have an element $y$ of the set
$$\left\{\sum \diam(f(Q_j'))^{\a}: f(E)\subset \bigcup  Q_j',\ \diam(Q_j')< C\epsilon^{\eta}  \right\}$$
with $y\leq C^{\a} x$ (just let $Q_j'=f(Q_j)$).
Therefore, by the properties of the infimum, we have
$$\H_{\a,C\e^{\eta}}(f(E))\leq C^{\a}\H_{\eta\a,\e}(E).$$
Since this holds for all positive $\epsilon,$ we have (since $C,\eta>0$)
$$\H_{\a}(f(E))=\lim_{\e\rightarrow 0}\H_{\a,\e}(f(E))=\lim_{\e\rightarrow 0}\H_{\a,C\e^{\eta}}(f(E))\leq
C^{\a}\lim_{\e\rightarrow 0}\H_{\eta\a,\e}(E)=C^{\a}\H_{\eta\a}(E),$$ so that
$$\H_{\a}(f(E))\leq C^{\a}\H_{\eta\a}(E).$$

\end{enumerate}
\end{document}