$\quad$ Let $f$ be a differentiable function in $\mathbb R$, and $f$ is also a Lebesgue integrable function in $\mathbb R$. Let $f'$ be a Lebesgue integrable function in $\mathbb R$, too. How to prove that:
$$\int_{\mathbb R}f'(x)\mathrm dx=0$$
$\quad$ My intuition tells me that when a function is integrable over the whole space, that is, when its integral is finite, the value of the function always starts at $0$ and ends at $0$. The way up is the way down, so the derivative's integration is $0$. But I know that even if the function is integrable on the whole space, it doesn't have to be $$\lim_{x\to\infty}f(x)=0$$ This function is not necessarily an absolutely continuous function, so there seems to be no direct relationship between the integral of its derivative and the original function in my mind. I can't think any more. Is there any way to deal with this problem?
Any discussion and suggestions are welcome!
Edit:
According to PhoemueX, I learned that $f$ is absolutely continuous over any interval $[a,b]$.
Based on Kurt G's answers and Stefan's hints, I understand the significance of limits there. So it seems that we can get the result without doing an odd-even decomposition.
My idea is that already known $$\int_a^bf'(x)\mathrm dx=f(b)- f(a)$$ and $f'(x)$ is integrable, so $$\lim_{a\to+\infty}\int_{-a}^0f'(x)\mathrm dx,\quad\lim_{b\to+\infty}\int_0^bf'(x)\mathrm dx$$ both exist, so $$\lim_{a\to+\infty}(f(0)- f(-a)),\quad\lim_{b\to+\infty}(f(b)- f(0))$$ both exist, therefore $f(\pm\infty)$ both exist. Then according to $f$ being integrable, $f(\pm\infty)$ can only be $0$, so we get the final conclusion.
Best Answer
Setting $$ g(x):=\frac{f(x)+f(-x)}{2}\,,\quad h(x):=\frac{f(x)-f(-x)}{2} $$ we can decompose $f$ into an even function $g$ plus an odd function $h\,:$ $$ f(x)=g(x)+h(x)\,. $$ The derivative of $g$ is odd and the derivative of $h$ is even: $$ g'(x)=-g'(-x)\,,\quad h'(x)=h'(-x)\,. $$ Since the integral of an odd function is zero we get $$ \int_\mathbb Rf'(x)\,dx=\int_\mathbb Rh'(x)\,dx\,. $$ Since $h$ is even we get $$ \int_\mathbb Rh'(x)\,dx=\lim_{a\to\infty}\big(h(a)-h(-a)\big)=\lim_{a\to\infty}2h(a)\,. $$ If this limit were different from zero this would imply that $h$ is not integrable which is a contradiction.