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](https://en.wikipedia.org/wiki/Lebesgue%27s_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 $\varepsilon$.



Quantitative Lebesgue density theorem
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](https://terrytao.wordpress.com/2007/06/18/the-lebesgue-differentiation-theorem-and-the-szemeredi-regularity-lemma/). In the same notes, he does provide some interesting rate of convergence theorems.


