Show $\lim\limits_{h\to0} \sup_{y\in B_h(0)} \int_{A_h} |f(x+y)-f(x)|^pdx=0$
where $A \subset\mathbb R^n$ is open. $f\in L^p(A)$. And for $h>0$ we set $A_h=\{x\in A: d(x, \partial A)>h\}$
How would one show the claim? I think that we want to use that $f\in L^p(A)$ to take $f$ out of the integral and use that the volume will go to zero but right now I don't see any way to go there or how to start at all. Any ideas? Thank you very much.
Let $F(x)=f(x)$ for $x \in A$ and $0$ for $x \notin A$. Then $F\in L^{p}=L^{p}(\mathbb R^{n})$ and $\int_{A_{h}} |f(x+y)-f(x)|^{p} dx \leq \int_{\mathbb R^{n}} |F(x+y)-F(x)|^{p} dx $ for $y \in B_h(0)$ and (it is well known that translation is continuous in $L^{p}$ in the sense) $\int_{\mathbb R^{n}} |F(x+y)-F(x)|^{p} dx \to 0$ as $y \to 0$.