# The product of two Hausdorff measures

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?

• For a smooth manifold, the Hausdorff measure corresponds to the Riemannian volume, so that everything you want holds true. Dec 20, 2020 at 15:29
• Does the equation hold, if $Y$ is a smooth Riemannian manifold and $X$ is a compact metric space? Dec 20, 2020 at 16:31
• 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. Dec 20, 2020 at 16:46

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.

and

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

plugs:

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)