If $f$ is Lebesgue integrable on $ R$ then $\lim_{h\to 0} \int_R |f(x+h)-f(x)|dx=0? $

Question

If $f$ is Lebesgue integrable on $R$ then $\lim_{h\to 0} \int_R |f(x+h)-f(x)|dx=0$

My attempt: I am trying to use the definition of Lebesgue integration but I am stuck. Can anyone give me some hints? Thanks

Answer 1

Here's an outline:

1. Let $g\in C_c(\mathbb{R})$. Show that $g_h(x)=g(x+h)\in C_c(\mathbb{R})$ for sufficiently small $h$. Moreover, $\text{supp}\, g_h,\; \text{supp}\, g\subset K$ for some $K$ compact.
2. $\int|g_h-g|\leq ||g_h-g||m(K)\to 0$ as $h\to 0$, where $||\cdot||$ is the $\sup$ norm.
3. Note that $\int f_h=\int f$ for all $h\in\mathbb{R}$. Now use the fact that each $f\in L^1(\mathbb{R})$ can be approximated by some $g\in C_c(\mathbb{R})$ in the $L^1$ norm. Then $$\int |f_h-f|\leq \int |f_h-g_h|+\int |g_h-g|+\int |g-f|=\int |f-g|_h+\int |g_h-g|+\int |g-f|=\int |g_h-g|+2\int |g-f|$$

The first integral goes to $0$ by $2$ as $h\to 0$, and the second goes to $0$ by the discussion in $3$. Thus $\int |f_h-f|\to 0$.

Answer 2

Continuous compactly supported functions are dense in $L^1(\mathbb{R})$. For continuous functions the conclusion is obvious, and the remainder can be made arbitrarily small in $L^1$.