Here's an intuitive answer, it's up to you to convert it into a rigorous argument!

If f is integrable then it tends to 0 at infinity. So there is a bounded region of , call it K, outside which |f| is very small. Choose h to be larger than the diameter of K. Inside K, |f(x+h)| will be very small, so |f(x+h)+f(x)| is approximately |f(x)|. Let L = {k–h : k ∈ K} (so L is the translate of K through –h). Inside L, |f(x)| will be very small, so |f(x+h)+f(x)| will be approximately |f(x+h)|. Outside K∪L, both |f(x)| and |f(x+h| are small, hence so is |f(x+h)+f(x)|.

Thus the integral of |f(x+h)+f(x)| will be approximately the sum of the integrals over K and L, each of which is approximately equal to the integral of |f|. In the limit, it follows that .