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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 maximal inscribed and minimal 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?

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The answer to both questions is positive (assuming the definition of aspect ratio for sets is modified to refer to the maximal inscribed and minimal circumscribed balls). The family of all sets $V \subseteq \mathbb{R}^n$ with aspect ratio at most $a$ has bounded eccentricity in the sense of the Wikipedia article on this topic.

https://en.wikipedia.org/wiki/Lebesgue_differentiation_theorem

Any family of sets with bounded eccentricity satisfies the Lebesgue differentiation theorem.

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  • $\begingroup$ The minimal/maximal was a typo -- thanks for pointing out, fixed. Regarding the aspect ratio version, do you have a specific reference? The wiki link doesn't contain a proof of the general case. $\endgroup$ – Aryeh Kontorovich May 9 '19 at 18:06
  • $\begingroup$ Suppose $V \subseteq \mathbb{R}^n$ has aspect ratio at most $a$. Let $B$ be a ball of radius $R$ containing $V$ and let $C$ be a ball of radius $r$ contained in $V$ such that $R \leq ar$. Then we have $\mu(V) \mu(B)^{-1} \geq \mu(C) \mu(B)^{-1} = r^n R^{-n} \geq a^{-n}$. Therefore $a^{-n}$ works as the constant $c$ in the definition of bounded eccentricity on the Wikipedia page. $\endgroup$ – burtonpeterj May 9 '19 at 21:23

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