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\vspace*{.3in}
\noindent{\bf Beals, page 34 \#2:}
The assertion is false.  To see this, consider the function which
is zero except on closed intervals $[n-1/4^n,n+1/4^n]$, $n\geq 2$.
And to define $f$ on each of these intervals take $f(n) = 2^n$ and then
make $f$ linear on each of the remaining subintervals (i.e. the graph of
$f$ on the interval is a triangle of height $2^n$ and width $2/4^n$).
Then on the one hand $\lim_{n\to\infty} f(n) = \lim 2^n = \infty$, so $f$ is 
not bounded.  But on the other hand,
$$
\int f = \sum_{n=2}^\infty \frac12 \frac{2}{4^n} 2^n
       = \sum_{n=2}^\infty \frac{1}{2^n} = \frac12 <\infty.
$$
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