# lebesgue density theorem for smooth manifold (revised)

My question is about lebesgue density theorem:

Let $\mathcal{H}^s$ be $s-$dimensional Hausdorff measure.

If $A\subset \mathbb{R}^{n}$ with $0<\mathcal{H}^s(A)<\infty,$ then for $\mathcal{H}^{s}$ almost all $x\in A,$

$$\limsup_{r \rightarrow 0}\frac{\mathcal{H}^{s}(A\cap B(x,r))}{\beta_s r^s}\leq 1,$$ where $\beta_s$ is the $s-$dimensional Hausdorff measure of $s-$dimensional unit ball.

Do we have the above inequality for all $x\in A$, if we assume that $A$ is a subset of a $C^{1}-$manifold?

Thank you so much

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There is something wrong with this question. If $s<n$, then the $s$-dimensional Hausdorff measure of any open set (like $B(x,r)$) is always $\infty$, so your quotient is always $0$. If $s=n$, then the limit is a.e. equal to 1 (this is the Lebesgue density theorem), and if $s>n$, then you always have $0/0$ there... – Lukas Geyer Oct 10 '12 at 5:09
Hello Lukas Geyer, you are right. Here is my revised question. – user44174 Oct 13 '12 at 18:46
OK, this looks like an interesting question. I vaguely remember that Falconer has some results like this in his "Fractal Geometry" book. I am not home right now, I'll try to look it up when I get there. – Lukas Geyer Oct 14 '12 at 19:29