2
$\begingroup$

The Lebesgue density theorem in $\mathbb{R}^n$ may be stated as follows. For a Lebesgue-measurable $A\subseteq\mathbb{R}$ and $r>0, x\in\mathbb{R}^n$, define $$ \chi_{A,r}(x)=\frac{\mu(A\cap B_r(x))}{\mu(B_r(x))},$$ where $\mu(\cdot)$ is the Lebesgue measure and $B_r(x)$ is the closed ball about $x$ with radius $r$. Then the limit $\chi_A(x)=\lim_{r\to0}\chi_{A,r}(x)$ exists for $\mu$-almost all $x\in\mathbb{R}^n$ and is either 0 or 1, depending on $x$'s membership in $A$.

I would like to prove a uniform version of this claim, something along the following lines: for all $R>0$, $$ \lim_{r\to0} \mathrm{ess}\sup_{x\in B_R(0)} |\chi_{A,r}(x)-\chi_A(x)|=0.$$

If the above is known to be false (example?), I'd be happy with a "high probability version", something like: for all $\epsilon>0$ there is a $B'\subset B_R(0)$ with $\mu(B_R(0)\setminus B')<\epsilon$ such that $$ \lim_{r\to0} \mathrm{ess}\sup_{x\in B'} |\chi_{A,r}(x)-\chi_A(x)|=0.$$ The latter looks like it's provable using the techniques suggested here: https://terrytao.wordpress.com/2007/06/18/the-lebesgue-differentiation-theorem-and-the-szemeredi-regularity-lemma/ but if it's already known, I'd be grateful for a reference.

Improve this question
$\endgroup$
4
  • $\begingroup$ @RobertIsrael: I assume the suprema are meant to be essential suprema, since $\chi_A(x)$ is only defined almost everywhere. $\endgroup$
    – Nik Weaver
    Commented Jul 18, 2016 at 18:28
  • 5
    $\begingroup$ Convergence almost everywhere implies convergence in probability, which in turn is uniform on a set of almost full probability. This is Egorov theorem $\endgroup$
    – Fedor Petrov
    Commented Jul 18, 2016 at 18:55
  • $\begingroup$ @FedorPetrov ack, of course! Many thanks. And yes, those sups should be ess-sups; will edit. $\endgroup$
    – Aryeh Kontorovich
    Commented Jul 18, 2016 at 19:20
  • 3
    $\begingroup$ Sorry, intermediate step is not needed. Convergence in probability does not imply uniform convergence on a large set, but convergence almost everywhere does, and this already is Egorov theorem. $\endgroup$
    – Fedor Petrov
    Commented Jul 18, 2016 at 19:58

1 Answer 1

Reset to default
4
$\begingroup$

For the first version of your claim, consider $n=1$, $A = [0,1]$. For any $r \in (0,1)$, $\chi_{A,r}(x)$ is near $1/2$ in a neighbourhood of $0$, so $$\text{ess-sup}_{x \in B_R(0)} |\chi_{A,r}(x) - \chi_A(x)| \ge 1/2$$

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .