If $f \in L^1(\mathbb{R})$, does $\lim_{s \rightarrow \infty}\int |f(x-s) - f(x)| dm$ exist?

Question

So if it were to exist, clearly it is nonnegative. Then for $s\in \mathbb{R}$, $$\int |f(x-s) -f(x)|dm(x) \leq \int|f(x-s)|dm(x) + \int|f(x)|dm(x)$$ because $f \in L^1(\mathbb{R})$. My intuition is that since the Lebesgue measure is translation invariant, these integrals should be the same. However, I'm not sure how to prove that, or even if my answer is correct.

edit: fixed my inequality.

Answer 1

$\int|f(x-s)|dm=\int|f(x)|dm$, so it is independent of $s$. Therefore $\lim_{s\to \infty}\int|f(x-s)|dm$ exists.

Answer 2

Hint: What is the answer if $f$ is supported in some interval?