Let $(X,d_X)$ and $(Y,d_Y)$ be two compact metric space with Hausdorff dimensions $\dim_H(X)=n$ and $\dim_H(Y)=m$ and Hausdorff measures $\mathcal{H}^{n}$ and $\mathcal{H}^{m}$. Assume that $\dim_H(X\times Y)=n+m$ for the cartesian product $(X\times Y, d)$ where $d=\sqrt{d_X^2+d_Y^2}$, then we have $(n+m)$-dimensional Hausdorff measure $\mathcal{H}^{n+m}$ on it.

Do we have $\mathcal{H}^{n+m}=\mathcal{H}^{n}\otimes\mathcal{H}^{m}$?

In particular, does the equation hold, if $Y$ is a smooth Riemannian manifold?

  • 1
    $\begingroup$ For a smooth manifold, the Hausdorff measure corresponds to the Riemannian volume, so that everything you want holds true. $\endgroup$ Dec 20, 2020 at 15:29
  • $\begingroup$ Does the equation hold, if $Y$ is a smooth Riemannian manifold and $X$ is a compact metric space? $\endgroup$ Dec 20, 2020 at 16:31
  • $\begingroup$ Sorry, I read for two manifolds. I am neither sure, not hopeless for the case of one manifold and one arbitrary space. The best I can suggest right now is to look at Falconer's book cited in answer, and possibly also Mattila's book. $\endgroup$ Dec 20, 2020 at 16:46

1 Answer 1


As noted in a comment, for Riemannian manifolds, the Hausdorff measures are equal to (up to a constant) the usual volumes. So this works.

The metric case you mention can fail. There are metric spaces of Hausdorff dimension $1$ that are not "rectifiable". Every subset has either $\mathcal H^1(E) = 0$ or $\mathcal H^1(E) = \infty$. Discussion is in Chapter 3 of

Falconer, K. J., The geometry of fractal sets, Cambridge Tracts in Mathematics, 85. Cambridge etc.: Cambridge University Press. XIV, 162 p. (1985). ZBL0587.28004.

And also note this is incorrect for general dimensions. Reference: third edition of

Falconer, Kenneth, Fractal geometry. Mathematical foundations and applications, Hoboken, NJ: John Wiley & Sons (ISBN 978-1-119-94239-9/hbk). xxx, 368 p. (2014). ZBL1285.28011.

Here are some things from Chapter 7.

Proposition 7.1
If $E \subset \mathbb R^n, F \subset \mathbb R^m$ are Borel sets with $\mathcal H^s(E), \mathcal H^t(F) < \infty$, then $$ \mathcal H^{s+t}(E \times F) \ge c \mathcal H^s(E)\;\mathcal H^t(F) \tag1$$ where $c > 0$ depends only on $s$ and $t$.

The opposite inequality $\le$ can fail.

Example 7.8
There exist sets $E, F \subset \mathbb R$ with $\dim_\mathcal H E = \dim_\mathcal H F = 0$ and $\dim_\mathcal H(F \times F) \ge 1$.

Falconer credits these results to:

Besicovitch, A. S.; Moran, P. A. P., The measure of product and cylinder sets, J. Lond. Math. Soc. 20, 110-120 (1945). ZBL0063.00354.


Marstrand, J. M., The dimension of Cartesian product sets, Proc. Camb. Philos. Soc. 50, 198-202 (1954). ZBL0055.05102.


Those two papers are among those reprinted in

Edgar, Gerald A. (ed.), Classics on fractals, Reading, MA: Addison-Wesley Publishing Company. x, 366 p. (1993). ZBL0795.28007.

More on this from a student of mine:

Mullins, Edmond N., Jr, Derivation bases, interval functions, and fractal measures. Thesis (Ph.D.)–The Ohio State University. 1996. 97 pp. ISBN: 978-0591-18087-9 (ProQuest LLC)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.