Is Lebesgue density independent of the choice of neighborhoods?

Question

Let $E$ be a Borel subset of $\mathbb R^d$ and $m$ be the Lebesgue measure on $\mathbb R^d$ and by definition of density (Folland real analysis 2nd edition ,Exercise 25 in page 100):

$$D_E(x):= \lim_{r\to 0} \frac{m(E\cap B(x,r))}{m(B(x,r))}$$

I wonder if we can replace the open balls $B(x,r)$'s (centered at the origin with radius $r$) by any neighborhoods of $x$ and still get the same "density". Namely, let $N_n(x)$ be a sequence of neighborhoods of $x$ (we don't even assume they are open, as long as each of them contains an open neighborhood of $x$) such that

$$\lim_{n\to \infty} m(N_n(x))=0.$$

Do we necessarily have

$$D_E(x):= \lim_{n\to \infty} \frac{m(E\cap N_n(x))}{m(N_n(x))}?$$

For a seemingly endless rabbit hole that your question leads to, see the survey paper Differentiation of Integrals by Andrew M. Bruckner (1971) and the last paragraph of my answer to Is there a garden of derivatives? Also, a google search for Saks rarity theorem will lead to some interesting results. — Dave L. Renfro, Jun 28 '21 at 17:13

score 2 · Accepted Answer · answered Jun 28 '21 at 17:46

As stated, no. If $a < b$ are two points in $\mathbb R$, then $$ N_n := B_{1/n}(a) \cup B_{1/n}(b) $$ is a neighborhood of $a$. Now take the interval $E= (-\infty,\frac{a+b}{2})$ and compute $$ \lim_{n\to \infty} \frac{m(E\cap N_n)}{m(N_n)} $$

A classic text on derivations is:

Hayes, C. A.; Pauc, C. Y., Derivation and martingales, Ergebnisse der Mathematik und ihrer Grenzgebiete. 49. Berlin-Heidelberg-New York: Springer-Verlag. VII, 203 p. (1970). ZBL0192.40604.

score 1 · Answer 2 · answered Jun 28 '21 at 17:40

Unfortunately, the geometry of your neighborhoods can have quite a bit of influence. As one simple example, take the subset of $\mathbb{R}^2$ given by $S = \{(x, y) \mid y \geq 0\}.$ Then, if you take Folland's ball based definition, the density at $(0, 0)$ will be $1/2.$ But if we take less symmetric neighborhoods, you can see how to force more than $1/2$ the area of the neighborhood to be in the upper half plane, even in the limit.

Is Lebesgue density independent of the choice of neighborhoods?

2 Answers2