$\lim_{t \rightarrow 0} \frac{1}{2t} \int_{b-t}^{b+t} |f- \frac{1}{2t} \int_{b-t}^{b+t} f| = 0$ $\lim_{t \rightarrow 0} \frac{1}{2t} \int_{b-t}^{b+t} |f- \frac{1}{2t} \int_{b-t}^{b+t} f| = 0$
Can I use the Monotone convergence theorem besides the L'Hopital theorem to show this? with $t=1/n$
 A: Partial Solution:
Note that
\begin{align*}
&\dfrac{1}{2t}\int_{r-t}^{r+t}\left|f-\dfrac{1}{2t}\int_{r-t}^{r+t}f\right|\\
&\leq\dfrac{1}{2t}\int_{r-t}^{r+t}|f-f(r)|+\dfrac{1}{2t}\int_{r-t}^{r+t}\left|f(r)-\dfrac{1}{2t}\int_{r-t}^{r+t}f\right|\\
&=\dfrac{1}{2t}\int_{r-t}^{r+t}|f-f(r)|+\left|f(r)-\dfrac{1}{2t}\int_{r-t}^{r+t}f\right|.
\end{align*}
By Lebesgue's Differentiation Theorem one has for a.e. $r$ that
\begin{align*}
\dfrac{1}{2t}\int_{r-t}^{r+t}|f-f(r)|\rightarrow 0
\end{align*}
and hence the proposition is true for a.e. $r$.
For a fixed $b$, I have no solution.
A: I consider the case when $f$ is continuous, then general case is covered in the answer below..
We have
$$ \vert f(x)-\frac{1}{2t} \int_{b-t}^{b+t} f(y) dy \vert 
=\vert \frac{1}{2t} \int_{b-t}^{b+t} (f(x)-f(y)) dy\vert 
\leq \frac{1}{2t} \int_{b-t}^{b+t} \vert f(x)-f(y) \vert dy
\leq \sup_{y\in (b-t,b+t)} \vert f(x)-f(y)\vert.$$
Thus,
$$\frac{1}{2t} \int_{b-t}^{b+t}  \vert f(x)-\frac{1}{2t} \int_{b-t}^{b+t} f(y) dy \vert dx 
\leq \frac{1}{2t} \int_{b-t}^{b+t} \sup_{y\in (b-t,b+t)} \vert f(x)-f(y)\vert
\leq \sup_{x,y\in (b-t,b+t)} \vert f(x)-f(y)\vert.$$
However, if $f$ is continuous, then $f$ is uniformly continuous on compact sets and hence you obtain that the RHS of the last inequality goes to zero, which yields the claim.
