I'll state the classic result in its density (rather than the more general differentiation) version. Let $\mu$ be a measure on the Borel $\sigma$-algebra of $\mathbb{R}^n$ and $A\subset \mathbb{R}^n$ a measurable set.

For $x\in\mathbb{R}^n$, define $$ \chi_{A,r}(x)=\frac{\mu(A\cap B_r(x))}{\mu(B_r(x))},$$ where $B_r(x)$ 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 $A$.

My question is: what if we use other "nice" test sets instead of balls? Rudin (Real and Complex Analysis, Sec. 7.9) defines classes of "nicely shrinking sets", but I'm thinking in a different direction.

Let us fix a compact $K\subset\mathbb{R}^n$ and consider some measurable $A\subset K$. Let $\mathcal{V}=\{V_1,\ldots,V_N\}$ be a measurable partition of $K$, and denote, for $x\in K$, the partition element to which it belongs by $V(x)$. Define $\rho(\mathcal{V})$ and $\alpha(\mathcal{V})$, respectively, to be the maximal diameter and aspect ratio of any partition element. (The aspect ratio of $V$ is the the ratio of the radii of the minimal inscribed and maximal circumscribed balls of $V$.)

Now define, for $x\in K$, $$ \chi_{A,r,a}(x)=\sup_{\mathcal{V}}\frac{\mu(A\cap V(x))}{\mu(V(x))},$$ where the supremum is over all measurable partitions $\mathcal{V}$ with $\rho(\mathcal{V})\le r$ and $\alpha(\mathcal{V})\le a$.

I think that for fixed $a<\infty$, it's still true that $\lim_{r\to0}\chi_{A,r,a}(x)$ converges almost everywhere to the characteristic function of $A$. Before I reinvent the wheel (or worse, spend time trying to prove a falsehood), two questions. 1. Is this true? 2. Is the result known, or perhaps a simple consequence of a known result?


Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.