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I can prove the following result. Here $\operatorname{dim} X$ stands for the topological dimension and $\mathcal{H}^n$ denotes the Hausdorff measure.

Theorem. Suppose that $f:\mathbb{R}^n\supset\Omega\to X$, $\Omega$ open, is a Lipschitz continuous map onto a metric space $X$, $f(\Omega)=X$. Then $\operatorname{dim} X=n$ if and only if $\mathcal{H}^n(X)>0$.

Question. Is it a known result?

The following result is due to Szpilrain: If $\mathcal{H}^{n+1}(X)=0$, then $\operatorname{dim} X\leq n$, see (see Unknown work of Nöbeling on topological/Hausdorff dimension). Hence, the implication $\operatorname{dim} X=n$ $\implies$ $\mathcal{H}^n(X)>0$ follows. Also $\mathcal{H}^{n+1}(X)=0$ (Lipschitz image of a subset of $\mathbb{R}^n$) so $\operatorname{dim} X\leq n$ and it remains to show that if $\mathcal{H}^n(X)>0$, then $\operatorname{dim} X\geq n$.

My proof of this fact is not particularly difficult (2 pages in LaTeX), but at the same time it is not elementary as it is based on the Kirchheim-Rademacher theorem (https://mathoverflow.net/a/324877/121665) and the Kirchheim area formula. On the other hand the theorem stated above sounds rather classical so I wonder if the result has already been proved in the literature.

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    $\begingroup$ It reminds the following simpler problem (page 67 here: arxiv.org/pdf/0906.0290v14.pdf): Let X be a contractible metric space with zero (n+1)-dimensional Hausdorff measure. Assume that ∆ and ∆' are two embedded n-disks in X having the same boundary. Show that ∆=∆'. $\endgroup$
    – Anton Petrunin
    Commented May 4, 2021 at 18:06
  • $\begingroup$ @AntonPetrunin I bought your book. How did you manage to make it so cheap? I am very interested in publishing a cheap book. I just looked at your problem. Perhaps it is a stupid idea, but can one apply the Mayer-Vietoris sequence? $\endgroup$
    – Piotr Hajlasz
    Commented May 4, 2021 at 19:33
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    $\begingroup$ The fact that the topological dimension does not exceed the Hausdorff dimension is a classical result of L. Pontrjagin, L. Schnirelmann, Sur une propriété métrique de la dimension, Ann. Math. (2) 33 (1) (1932) 156–162. So, $dim(X)\le dim_H(X)$. The fact that $X$ is a Lipschitz image of $\mathbb R^n$ implies $dim_H(X)\le\dim_H(\mathbb R^n)=n$. If $H^n(X)>0$, then $dim_H(X)\ge n$ and hence $dim(X)\le dim_H(X)=n$. So, the problem why the first inequality is in fact, the equality. $\endgroup$
    – Taras Banakh
    Commented May 5, 2021 at 4:56
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    $\begingroup$ @TarasBanakh Actually the result you mentioned was proved by Nobeling in 1931, see mathoverflow.net/q/360384/121665. $\endgroup$
    – Piotr Hajlasz
    Commented May 5, 2021 at 15:29

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