Talk:Rectifiable set

Definition 2: is it clear what is meant by "$k$-dimensional graphs of $\mathbb R^n$"? --Boris Tsirelson 09:43, 4 August 2012 (CEST)

As far as I remember, Hausdorff dimension k does not imply finite (or even σ-finite) k-dimensional Hausdorff measure. Really so? And then "might not be $\mathcal{H}^k$-measurable" looks strange. Is it OK? Do the definitions implicit here conform to definitions in Hausdorff measure and Hausdorff dimension? --Boris Tsirelson 09:50, 4 August 2012 (CEST)

Sorry, I was saving while constructing the page... I guess I should use a Sandbox instead, sorry again for wasting your time. I have to check still everything once again, since I just typed. Anyway:

• For Lipschitz graphs I can add an explanation if you think it is needed.
• For consistence with the pages Hausdorff measure and Hausdorff dimension I will check later (it is in my "to do list").
• You are right: Hausdorff dimension k does not imply finite (or even σ-finite). If you are referring to the decomposition of a general set in rectifiable and purely unrectifiable part you need the $\sigma$-finiteness hypothesis (which I just added). Rectifiable sets are automatically $\sigma$-finite because of the covering with countably many $C^1$ submanifolds (a $C^1$ submanifold is automatically $\sigma$-finite). The $\sigma$-finiteness assumption might be needed somewhere else and it is the typical thing on which I might slip inadvertently: I will check again everything with care.
• If I drop the Borel assumption in the definition of rectifiability, you might have the following example: Take a common $C^1$ injective curve $\gamma: [0,1]\to\mathbb R^2$ and take the typical Vitali non-Lebesgue measurable subset $V\subset [0,1]$. Now $\gamma (V)$ has Hausdorff dimension $1$ and it can be covered by a single one-dimensional submanifold, so it is rectifiable without Borel assumption. But it is not $\mathcal{H}^1$ measurable. Do you think it is worth to add this example?

Camillo 10:15, 4 August 2012 (CEST)

I see. Yes, sometimes sandbox is worth to use. Well, now I wonder, why Hausdorff dimension k is stipulated in these three definitions. It follows from the other requirements that it cannot exceed k, right? Thus, you just want to exclude the degenerate case, the dimension less than k, right? But then, did you exclude the case of dimension k but zero $\mathcal{H}^k$? --Boris Tsirelson 13:14, 4 August 2012 (CEST)