I've been sitting on this question for a while now:
Let $f(x) \in L^1(\mathbb{R})$. Compute $$ \lim_{h\to \infty} \int_{\mathbb{R}} |f(x+h)-f(x)|dx. $$
I've managed to convince myself that the solution is $2 \int_{\mathbb{R}} |f|$. However, it alludes me of how to make this a quick and easy. Any way I can thing this approach this problem begins with taking a continuous functions with compact support and approximating the integral with this. Do anything have a more slick way to think about this problem?
Suppose first that $f$ has support in some $[-a,a].$ Then for $h> 2a,$
$$|f(x+h)-f(x)| = |f(x+h)|+ |f(x)|,$$ simply because these functions have disjoint supports. It follows that
$$(1)\int|f(x+h)-f(x)| = 2\int|f|$$
for $h>2a.$
Now let $\epsilon > 0.$ Then, by the DCT, we can choose $a>0$ such that $\int|f-f\chi_{[-a,a]}|< \epsilon.$ Set $g = f\chi_{[-a,a]}.$ Then
$$\int|f(x+h)-f(x)| \le \int|f(x+h)-g(x+h)| +\int|g(x+h)-g(x)| + \int|g(x)-f(x)|$$ $$ < \int|g(x+h)-g(x)| + 2\epsilon.$$
Therefore
$$\int|f(x+h)-f(x)| - 2\int|f| \le \int|g(x+h)-g(x)| - 2\int|g|+2\epsilon .$$
Now apply $(1)$ to $g$ to get
$$\limsup_{h\to \infty} \left( \int|f(x+h)-f(x)| - 2\int|f| \right) \le 2\epsilon.$$
Since $\epsilon$ is arbitrary, the $\limsup$ above is $0.$ A similar argument shows the $\liminf $ is $\ge -3\epsilon$ or so. Thus $\int|f(x+h)-f(x)| \to 2\int|f|$ as desired.