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\begin{document}







\vspace*{.3in}
\noindent{\bf Beals, page 25 \#3:}
Note that since $f$ is integrable and $F(y) - F(x) \leq \int_x^y |f|$,
there is no loss of generality in what follows if we assume that $f\geq 0$.

\begin{description}
\item[(a)]
Let $\{x_n\}\subset\R$ be any sequence converging to $x$.  
Let $I_n$ be the closed interval of points between $x_n$ and $x$.  Then
$$
F(x_n) - F(x) = \int_{I_n} f = \int \mathbf{1}_{I_n} f.
$$
Now the function $g_n := \mathbf{1}_{I_n} f$ satisfies $|g_n| \leq g := |f|$ 
for all
$n\in\N$.  Moreover, $\lim_{n\to\infty} g_n(y)=0$ for all $y\neq x$.  
Therefore,
$$
\lim_{n\to\infty}  F(x_n) - F(x) = \lim_{n\to\infty} \int g_n 
= \int\lim_{n\to\infty} g_n = \int 0 = 0.
$$
In other words $\lim_{n\to\infty}  F(x_n) = F(x)$, which means that $F$
is continuous at $x$.
\qed
\item[(b)]
$F$ must be uniformly continuous, too.  To see this, note that (since
$f\geq 0$) $F$ increases from $\lim_{x\to-\infty} F(x) = 0$ to 
$$
M := \lim_{x\to\infty} F(x) = \int f < \infty.
$$ 
So if $\epsilon > 0$ is given, then there is some number $T$ such that
$M - \epsilon/2 < F(x) < M$ for all $x\geq T$ and similarly
$0<F(x)<\epsilon/2$
for all $x<-T$.  In particular, if $x,y>T$ or $x,y<-T$, we have
$|F(x) - F(y)| < \epsilon/2$.


Moreover, $F$ is uniformly continuous on the compact set $[-T,T]$, so 
there is $\delta>0$ such that $x,y\in [0,T]$ and $|x-y|<\delta$ implies that
$$
|F(x) - F(y)| <\epsilon/2.
$$
Now if by chance $|x-y|<\delta$ and, say, $|x|<T$ while $|y|>T$, we have
either $-T$ or $T$ between $x$ and $y$---say, for argument's sake it's
$T$.  Then 
$$
|F(x) - F(y)| \leq |F(x)-F(T)| + |F(T) - F(y)| < \epsilon/2 + \epsilon/2
=\epsilon.
$$
So in all cases, $|x-y|<\delta$ implies that $|F(x)-F(y)|<\epsilon$,
and $F$ is uniformly continuous.
\qed
\end{document}
