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)