Prove that $\int_{\mathbb{R}}\left|f_{r}-f\right| d m \rightarrow 0 \text { as } r \rightarrow 0^{+} $

Question

Let $f \in L^{1}(\mathbb{R})$. For $r>0$ define $$ f_{r}(x):=\frac{1}{2 r} \int_{x-r}^{x+r} f d m \text { for } x \in \mathbb{R} . $$ Prove that $\int_{\mathbb{R}}\left|f_{r}-f\right| d m \rightarrow 0 \text { as } r \rightarrow 0^{+} .$

Here, I have showed that $\int_{\mathbb{R}}\left|f_{r}\right| d m \leq \int_{\mathbb{R}}|f| d m \text { for each } r>0.$ And by using convolution map we can obtain, $\left|f_{r}-f\right| = \left|\frac{1}{2r}\int_{\mathbb{R}}{(f{(x-t)}-f(x)})\chi_{[-r,r]} (t) dm(t) \right |$. What should I do next?

Answer 1

By Lebesgue's Differentiation Theorem $f_r \to f$ a.s. $L^{1}$ convergence follows immediately from DCT if $f$ is a continuous function with compact support. [Note that if $f$ vanishes outside $[-N,n]$ and $0<r<1$ then $f_r$ vanishes outside $[-N-1,N+1]$]. Now approximate $f$ in $L^{1}$ by a continuous function $g$ with compact support. By the inequality $\|f_r\| \leq \|f\|$ that you already have applied to $f-g$ (and triangle ineqaality for $L^{1}$ norm) you can easily get $L^{1}$ convergence of $f_r$ to $f$.

[$\|f_r-f\| \leq \|(f-g)_r\|+\|g-g_r\|+\|f-g\| \leq \|g-g_r\|+2\|f-g\|$]