If $\lim_{x\to\infty}\int_0^x a(t)\,dt=L$, is it true that $\lim_{x\to\infty}a(x)=0?$ Let $s(x)=\int_0^xa(t)\,dt$. I wonder:
If $\displaystyle \lim_{x\to\infty}s(x)=L$ is it true that $\displaystyle\lim_{x\to\infty}a(x)=0?$
 A: The statement is false even for continuous non-negative functions.
As an example, consider the function $a: [0, \infty) \to \mathbb R$ whose
graph consists of triangles with baseline $[n, n + 1/n^2]$ and height $2$
for each $n \in \mathbb N$ and is zero otherwise.
Then $\int_0^\infty a(x)dx = \sum_{n=1}^\infty 1/n^2 = \pi^2/6$,
but $\lim_{x \to \infty} a(x)$ does not exist.
A: No. $\lim_{x\to\infty} a(x)$ need not even exist, $a$ could be $1$ in a set of measure zero.
A: One hypothesis that does yield this result is that $a$ is absolutely integrable and uniformly continuous. Comparing to Martin R's answer, uniform continuity makes it so a triangle of a fixed height cannot be arbitrarily narrow.
Edit: a sketch of a proof. Suppose $f$ is absolutely integrable, uniformly continuous, and it is not true that $\lim_{x \to \infty} f(x) = 0$. Repeatedly apply the definition of the last statement to get $\{ x_n \}_{n=1}^\infty$ with $|f(x_n)| > \varepsilon$ and $|x_n-x_{n-1}| \geq 1$. Use uniform continuity to get intervals $I_n$ centered at $x_n$ with $|f(x)| > \varepsilon/2$ on $I_n$, where the $I_n$ all have length at least $\delta$. (When assuming only continuity, we cannot assume that the $I_n$ all have length at least $\delta$, which is why Martin R's example works.) Derive a contradiction of the absolutely integrable hypothesis from here.
This still works if you replace the absolute integrability with improper Riemann integrability, because in that case the integral would have to diverge either to $\pm \infty$ or by oscillation. This is because we guarantee that the integral over $I_n$ is at least a fixed number, namely $\varepsilon \delta/4$. We do not guarantee that it is at least, for example, $1/n$.
