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\vspace*{.3in}
\noindent{\bf Beals page 11, \#6:}
\begin{description}
\item[a)] Let $A_n = [0,1]$ if $n\in\N$ is even and $A_n = \emptyset$
if $n$ is odd.  Then $\limsup A_n = [0,1]$ and $\liminf A_n = \emptyset$.
\item[b)] I show only that $\limsup A_n$ is measurable.  The proof for
$\liminf A_n$ is similar.  For each $N\in\N$, let
$$
B_N = \bigcup_{n\geq N} A_n
$$
Then $B_N$ and $B \eqdef \cap_{N\in\N} B_N$ are measurable by assertion $I$ on 
page 9 of Beals notes.  

I claim that $B = \limsup A_n$.  To see that this
is so, let $x\in B$.  Then $x\in B_N$ for every $N\in\N$.  In other words,
for each $N\in\N$, there exists $n\geq N$ such that $x\in A_n$.  This
can only be the case if $x\in A_n$ for infinitely many $n$ (otherwise,
we could let $N$ be one larger than the maximum of those finitely many $n$
for which $x\in A_n$, and it would follow that $x\notin B_N$).  So
$x\in\limsup A_n$.  

In the other direction, suppose that $x\notin B$.  Then $x\notin B_N$ for
some $N\in\N$.  Then $x\in A_n$ for at most $N$ values of $n$.  It follows
that $x\notin\limsup A_n$.  This and the preceding paragraph show that
$x\in B$ if and only if $x\in\limsup A_n$, so the two sets are equal, and
I conclude that $\limsup A_n$ is measurable.
\qed
\end{description}

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