I have the problem
Prove that if $f$ and $f'$ are Lebesgue integrable over $\mathbb{R}$, then $\int_\mathbb{R}f' = 0$, where $f'$ is defined everywhere.
Honestly, not sure where to start. I had the idea to perhaps use the fundamental theorem of calculus $f(x) = \int_0^x f' + f(0)$ and take limits as $x$ goes to $0$ and trying to work from there. However, I'm not sure that the theorem holds in this case. Any help getting started is appreciated!
On
$$f(x) = f(0) + \int_0^x f'$$
Taking the appropriate limits...
$$0=f(\infty) = f(0) + \int_0^\infty f'$$
$$0=f(-\infty) = f(0) + \int_0^{-\infty} f'=f(0) - \int_{-\infty}^0 f'$$
$$\therefore \int_{-\infty}^\infty f'= \int_{-\infty}^0 f' + \int_0^\infty f'=f(0)-f(0)=0$$
Where we needed $f$ to be Lebesgue integrable to know it tended to $0$ on the tails, and $f'$ to be Lebesgue integrable to rearrange those integrals.
As a counter-example you can take a compactly supported function, say $f(x)=x^2$ over $[0, 1]$.
Edit: As pointed out in the comments, the answer to this question really depends on how you interpret the question. This answer assumes that $f'$ means the derivative of the $L^1$ function $f$ which is defined almost everywhere.