If $f\in L^1(\mathbb R)$ and $f'\in L^1(\mathbb R)$, then $\lim_{x\to \infty }f(x)=0$. Let $f,f'\in L^1(\mathbb R)$. Prove that $$\lim_{x\to \infty }f(x)=0.$$

First of all, is $f'$ defined a.e. ? Because there are no assumption on the fact that $f$ is derivable. So, is $$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$
for almost every $x$ ?

My attempt for the statement : Suppose WLOG $f'&gt;0$. I'm not sure if it's true, but I would say that : Let $\varepsilon &gt;0$. Since $f'$ is $L^1$, there is a ball $[a,b]$ s.t. $$\int_{]-\infty ,a[\cup]b,+\infty [}f'(x)dx&lt;\varepsilon .$$
In particular, if $x&gt;y&gt;b$, then $$f(x)-f(y)=\int_y^x f'(t)dt&lt;\varepsilon,$$
but I just can conclude that $\lim_{x\to \infty }f(x)-f(y)&lt;\varepsilon .$
I think the question is not properly posed. Normally the statement is interpreted as follows: if $f$ is integrable and differentiable almost everywhere with $f'$ also integrable then $f(x) \to 0$ as $ x \to \infty$. This statement is false. $f(x)=1$ when $x$ is an integer and $0$ otherwise gives a counterexample. Some additional assumptions on $f$ are necessary to prove that $f(x) \to 0$ as $ x \to \infty$.
The idea is to show that $f(x)-f(0)=\int_0^x f'$ hence $f$ has a limit in $\infty$, which must be $0$ because $f\in L^1$.