Suppose $f$ is an integrable function on $\mathbb R$. Suppose further that $f$ is absolutely continuous on each closed and bounded interval $[a, b]$, and that its derivative $f'$ also is integrable over $\mathbb R$. Prove that $$\lim_{x\to \infty} f(x) = 0.$$
As $f$ is absolutely continuous on each closed and bounded interval then it is also bounded variation on there. So we can write $f$ as a difference of two funtions. However, I could not see where to use $f$ and $f'$ integrable conditions to get our result.
Hint:
By considering $|f|$ if necessary, it suffices to assume that $f\ge 0$.
We argue by contradiction. Assume the contrary that $f$ does not tend to $0$ when $x\to \infty$. Then there is a strictly increasing sequence so that $$\lim_{n\to \infty} a_n = + \infty\ \ \ \text{and } \ \ f(a_n) \ge \epsilon_0.$$
Now, using the fact that $f$ is integrable, show that there is a sequence $b_n$ so that $b_n \in (a_n, a_{n+1})$ so that $0\le f(b_n) \le \epsilon_0/2$ for $n$ large enough.
Now we use the fundamental theorem of calculus $$f(y) - f(x) = \int_x^y f'(s) ds$$ (which holds when $f$ is absolutely continuous) on the those interval $[a_n, b_n]$ for $n$ large. This will give you a contradiction that $f'$ is integrable.