Let $f,f' \in L^1$, $f$ is differentiable everywhere, show $\int_\mathbb{R} f'=0$
I have no clue how to do this problem. I had some ideas but they all require extra assumptions like:
$f(x)$ is uniformly continuous on $\mathbb{R}$ or $f'$ is bounded on each compact set. Any hints would be appreciated.
I was trying to mess around with $\int_\mathbb{R} \frac{f(x+1/n)-f(x)}{1/n}$ and try to move the limit around but could not justify doing that. If I could the result would easily follow from translation invariance. I was thinking MVT could help us find a dominating function ($f'$) but I do not think that is the case.
In the following I am assuming that $f'$ exists everywhere.
We have $f(b)-f(a) = \int_a^b f'(t)dt$. Since $f'$ is integrable we see that $\lim_{x \to -\infty} f(x)$ and $\lim_{x \to \infty} f(x)$ exist, call them $f(-\infty)$ and $f(\infty)$ respectively.
Suppose $f(-\infty) >0$, then for some $M$ if $x<M$, we have $f(x)>{1 \over 2} f(-\infty)$ and hence if $a,b <M$ we have $\int_a^b f(s)ds > {1 \over 2}f(\infty)(b-a)$ in which case $f$ is not integrable. Similarly if $f(-\infty) <0$.
Repeat the process to get $f(\infty) = 0$.
Hence $f(\infty)-f(\infty) = 0 -0 = \int f'$.
Aside:
If we drop the requirement that $f'$ exists everywhere then the result is not true. For example, $f(x)=x \cdot 1_{[0,1]}$ is differentiable everywhere except $0,1$ and both $f,f'$ are integrable, but $\int f' = {1 \over 2}$.