Let $A \subset [0, 1]$ be a measurable set, and $\mathbf 1_A$ its indicator function, viewed as a function on $\mathbb R$. Define for each $\delta > 0$, the function $f_{A, \varepsilon}: \mathbb R \to [0, 1]$ given by
$$f_{A, \varepsilon} (x) = \sup_{r \leq \varepsilon} \frac{1}{2r} \int_{x-r}^{x+r} |\mathbf 1_A (y) – \mathbf 1_A (x)| \, dy.$$
By the Lebesgue density theorem, $f_{A, \varepsilon}$ converges pointwise a.e. to the zero function as $\varepsilon \to 0$, and hence in $L^1$ by the dominated convergence theorem.
Question: Do there exist some absolute constants $C, \delta > 0$ such that for any measurable subset $A$ of $[0, 1]$, and all $\varepsilon < \delta$, we have
$$\|f_{A, \varepsilon}\|_{L^1} < C\varepsilon?$$
Remark: The role of $\delta$ is not quite necessary, but it saves us the trouble of having to deal with the trivial case of large $\varepsilon$.
Best Answer
Split $[0,1]$ into $n$-dyadic intervals and consider the set of alternating intervals:
$$A_{n}\triangleq\Big[0,\frac{1}{2^{n}}\Big]\cup \Big[\frac{2}{2^{n}},\frac{3}{2^{n}}\Big]\cup\cdots\cup \Big[1-\frac{2}{2^{n}},1-\frac{1}{2^{n}}\Big]$$
and the function $f_{n}(x)=1_{A_{n}}(x)$. Then we have that for $x\in A_{n}$ and $r>2^{-n}$ $$ \begin{split} \frac{1}{2r} &\int_{x-r}^{x+r} |\mathbf 1_A (y) - \mathbf 1_A (x)| \, dy\\ & = 1-\frac{|A_n \cap (x-r,x+r)|}{2r}=\frac{1}{2}. \end{split} $$ So for the $L^1$-norm and $\epsilon>2^{-n}$ we have the lower bound
$$\int_{A_{n}}f_{A,\epsilon}(x)dx=|A_{n}|\frac{1}{2}=\frac{1}{4}.$$
This example is from T.Tao's notes. In the same notes, he does provide some interesting rate of convergence theorems.