Is the Lebesgue integral of a continuous function necessarily continuous? Suppose that $f:\mathbb R \rightarrow [0,\infty]$ is measurable and that $\int_{\mathbb R}f \, dx<\infty$.
(a) Prove that $F(x)=\int^x_{-\infty}f(y) \, dy$ is a continuous function.
(b) Is $F$ necessarily uniformly continuous? Justify your answer.
I've managed to get the first part by considering a sequence of $\{z_n\}\subset (-\infty,x]$ with $z_n\rightarrow z$. Then setting $f_n=f_{\chi(-\infty,z_n]}$ and using the Dominated Convergence Theorem, I get sequentially continuity and hence continuity.
How do I go about trying to prove (b)? I'm guessing that the answer is no.
I was trying to come up with a sequence of continuous functions that eventually has a discontinuity, but I don't know if this is possible?
My other thoughts were along the lines making defining $f(y)$ such that $F(x)=x^2$ and we know already that $x^2$ is not uniformly continuous. Can that work?
 A: This addresses (b). Since $f$ is integrable, $F$ has a finite limit at $+\infty$. Furthermore, the limit of $F$ at $-\infty$ is zero. By (a), $F$ is continuous. To sum up, one is given a continuous function with finite limits at $\pm\infty$. Then:

Every continuous function with finite limits at $\pm\infty$ is uniformly continuous.

To prove this, call $G$ such a function, and pick $\varepsilon\gt0$. There exists $K$ such that $|G(x)-G(+\infty)|\leqslant\frac12\varepsilon$ for every $x\geqslant K$ and $|G(x)-G(-\infty)|\leqslant\frac12\varepsilon$ for every $x\leqslant-K$. On $[-K-1,K+1]$, $G$ is uniformly continuous hence there exists $\delta\gt0$, say $\delta\leqslant1$, such that, if $|x-y|\leqslant\delta$, $|x|\leqslant K+1$, $|y|\leqslant K+1$, then $|G(x)-G(y)|\leqslant\varepsilon$. One sees that if $|x-y|\leqslant\delta$ (without the conditions on $|x|$ and $|y|$), then either $x\leqslant-K$ and $y\leqslant-K$, or $-K-1\leqslant x\leqslant K+1$ and $-K-1\leqslant y\leqslant K+1$, or $x\geqslant K$ and $y\geqslant K$. In all three cases, $|G(x)-G(y)|\leqslant\varepsilon$. QED.
A: Since $f\ge 0$, we have $F$ weakly increasing and $F(x)\to0$ as $x\to-\infty$.  Since $f$ is integrable, we have $F(x)\to c<\infty$ as $x\to\infty$.  Let $\varepsilon>0$.  Choose $A>0$ so large that if $x>A$ then $c-F(x)<\varepsilon$ and $F(-x)<\varepsilon$.  Then try to show that $F$ is uniformly continuous on $[-A,A]$.  Then $\delta$ you get should work outside that interval as well.
