# About the sigma algebra generated by the Hausdorff measure on $\mathbb R^n$

Let $$\mathcal{H}^k$$ be the $$k-$$dimensional Hausdorff measure on $$\mathbb R^n$$, with $$k \in \{1, \ldots n\}$$. By Carathéodory's theorem we know that there exists a sigma algebra $$\mu(\mathcal{H}^k)$$ of $$\mathcal{H}^k-$$measurable sets and it's posible to show that if $$f:\mathbb R^k \to \mathbb R^n$$ is Lipschitz and $$A \subset \mathbb R^k$$ is a Lebesgue measurable set of $$\mathbb R^k$$ then $$f(A) \in \mu(\mathcal{H}^k)$$ my questions are:

Is the converse of the last assertion true?

There exists a characterization of the sets in $$\mu(\mathcal{H}^k)$$ ?

What relations exist between the Lebesgue sigma algebra on $$\mathbb R^n$$ and $$\mu(\mathcal{H}^k)$$?

For the third question using the identity function $$I:\mathbb R^n \to \mathbb R^n$$ which is Lipschitz, its easy to see that if $$A$$ is a Lebesgue measurable subset of $$\mathbb R^n$$ then $$A=I(A) \in \mu(\mathcal{H}^n)$$, then in this case the Lebesgue sigma algebra is contained in $$\mu(\mathcal{H}^n)$$ this contention is proper?

• The converse of your assertion is not true. Even if you exclude sets of zero and infinite measure, all the sets you construct are rectifiable and for $k<n$ there are measurable, purely unrectifiable sets of nonzero measure.
– mlk
Commented Apr 21 at 7:05

This is a partial answer. In $$\mathbb{R}^n$$, $$\mathcal{H}^n$$ coincides with the Lebesgue measure so $$\mu(\mathcal{H}^n)$$ coincides with the $$\sigma$$-algebra of Lebesgue measurable sets.