Let $A \subset [0,1]^n$ with $A$ measurable and such that $\mathcal{L}^n (A)= \delta >0$, and consider a partition of $[0,1]^n$ in $\epsilon$-cubes (i.e. cubes of side $\epsilon)$.  
 For $\epsilon \to 0$ (say, $\epsilon= 1/n$ with $n \in \mathbb{N}$ so that $1/\epsilon$ is always an integer) we should have that, at the limit, it holds the Lebesgue differentiation theorem, so you have that points in $A$ have density 1 and those in $A^c$ have density 0. What I mean is that given every (in fact $\mathcal{L}^n$ almost every, but whatever) point $x$ in $[0,1]^n$ I think that if you take the sequence of the $\epsilon$ -cubes $Q\_\epsilon (x)$ such that they contain $x$, one should have that
$$
\frac{| A \cap Q\_\epsilon (x)|}{\epsilon ^n} \to I\_{A} (x),
$$ with $I\_A$ the indicator function of $A$.  
 In fact the Lebesgue theorem is about shrinking balls centered at the point $x$ you consider, but at the limit I think this shouldn't matter if you consider finer and finer partitions.


What I would like to know is if there is a sort of quantitative version of this result for fixed $\epsilon$. What I mean is, given a partition $\{Q\_\epsilon (i)\}\_{i=1, \dots, \epsilon^{-n}}$ , having some result which gives you a sort of balance between how big $f\_{A, \epsilon } (i) =\frac{| A \cap Q\_\epsilon (i)|}{\epsilon ^n}$ is and in how many cubes you have this.  
 So for example, if I fix $\epsilon$ we can't have $ f\_{A, \epsilon} (i)= 1$ for more than $\delta/ \epsilon ^n$ indices, but it's easy to give an example of an $A$ such that $f\_{A, \epsilon} (i)= \delta$ for every $i$.


I was thinking that since the convergence from the Lebesgue theorem is pointwise I can play a bit with Egoroff's theorem or something like that, but maybe these kind of estimates are already considered and have a name in the literature (although I couldn't find anything).



Quantitative version of Lebesgue points theorem
$\newcommand\ep\epsilon\newcommand\de\delta$Let $\ep=1/m$ for a natural $m$, so that
$$N\_\ep:=\ep^{-n},$$
the number of the $\ep$-cubes, is an integer.


Take any $c\in(0,1]$ and let
$$N\_{A,\ep}(c):=\#\{i\in[N\_\ep]\colon f\_{A,\ep}(i)\ge c\},$$
the number of the $\ep$-cubes with relative $A$-content $\ge c$; as usual, $[k]:=\{1,\dots,k\}$. Then, by Markov's inequality,
$$N\_{A,\ep}(c):=\sum\_{i\in[N\_\ep]}\,1(f\_{A,\ep}(i)\ge c)
\le\sum\_{i\in[N\_\ep]}\,\frac{f\_{A,\ep}(i)}c \\
=\frac1c\,\sum\_{i\in[N\_\ep]}\,|A\cap Q\_\ep(i)|\,N\_\ep
=\frac1c\,|A|\,N\_\ep=\frac\de c\,N\_\ep=:M\_{\de,\ep}(c).$$


So, $M\_{\de,\ep}(c)$ is an upper bound on the number $N\_{A,\ep}(c)$ of the $\ep$-cubes with relative $A$-content $\ge c$. This bound is nontrivial only if $c>\de$.
Also, the bound $M\_{\de,\ep}(c)$ is exact in (say) the following sense:


Suppose that $c\in[\de,1]$ is such that $M\_{\de,\ep}(c)$ is an integer (which will be $\le N\_\ep$, since $c\ge\de$). Clearly, then there is a measurable subset $A$ of $[0,1]^n$ such that $f\_{A,\ep}(i)=c\,1(i\le M\_{\de,\ep}(c))$ for all $i\in[N\_\ep]$. Then $N\_{A,\ep}(c)=M\_{\de,\ep}(c)$, so that the upper bound $M\_{\de,\ep}(c)$ on $N\_{A,\ep}(c)$ is attained.


On the other hand, whenever the exact upper bound $M\_{\de,\ep}(c)$ on $N\_{A,\ep}(c)$ is nontrivial (that is, for any $c\in(\de,1]$, the only lower bound on $N\_{A,\ep}(c)$ is the trivial bound $0$ -- which will be attained if e.g. $f\_{A,\ep}(i)=\de$ for all $i\in[N\_\ep]$.


