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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?

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  • $\begingroup$ 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. $\endgroup$
    – mlk
    Commented Apr 21 at 7:05

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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.

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