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Here's an interesting problem one can formulate for a student. This problem arises when considering special ergodic theorems:

On a finite dimensional manifold $M$ with a Lebesgue measure $\mu$, does every measure zero set equal a countable union of the sets of less than full Hausdorff dimension?

For a diffeomorphism $f$ of $M$ and a continuous function $\varphi$ on $M$, define $$\overline \varphi = \lim_{n \rightarrow \infty} \frac{1}{n} \sum_0^{n-1} \varphi \circ f^k(x).$$ Then the Birkhoff theorem asserts that for almost all $x$, $\overline \varphi \rightarrow \int_M \varphi, n \rightarrow \infty$. But consider the set $K_{\alpha}$ of $x$ where $$\alpha \leq |\overline \varphi - \int \varphi|.$$ So Birkhoff says $\mu(K_\alpha)=0$, but what about the Hausdorff dimension of $\mu(K_\alpha)=0$? For some diffeomorphisms, for example hyperbolic maps, it was proven that $\dim_H K_\alpha < \dim X$. That fact gives a rise to my question. I expect a negative answer, but I can not find a counterexample.

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  • $\begingroup$ It's not clear what "decomposed" means. Do you mean that expressing the set of measure zero as union of the sets of a Hausdorff dimension less than full? Then, do you want to allow "uncountable" union? If so, any set of measure zero is union of points. $\endgroup$
    – Sungjin Kim
    Commented Mar 29, 2012 at 8:00
  • $\begingroup$ I mean expresing the set of measure zero as a countable (or less) union of the sets of a Hausdorff dimension less than full. In the remark, $\alpha$ could be taken $\frac{1}{n}$. So I do not accept such an easy solution. I hope, now it's more precise. $\endgroup$
    – Olga
    Commented Mar 29, 2012 at 8:49

1 Answer 1

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Consider a function $h$ defined on the unit interval $[0,1]$ which is monotone nondecreasing and for which $h(0)=0$, $h$ continuous at $0$. We may define a Hausdorff measure $H_h$ associated to $h$ (see Donoghue, Distributions and Fourier Transforms Academic Press New York 1969 p. 30--35, or C. A. Rogers, Hausdorff Measures, Cambridge University Press, 1970). When $h(x)=x^\alpha$ you get the ordinary Hausdorff measures $H_\alpha$. Consider also $f(x)=x\log(e/x)$.

Then a set $A\subset{\bf R}$ with $0< H_f(A)<1$ has measure of Lebesgue $0$ but it is not union of a numerable set $A=\bigcup A_n$ with $H_{\alpha_n}(A_n)=0$ and $0<\alpha_n<1$, because this implies $H_f(A_n)=0$ and so $H_f(A)=0$.

As similar construction applies to each ${\bf R}^n$.

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    $\begingroup$ Perhaps I must explain that the existence of $A\subset{\bf R}$ with $0<H_f(A)<1$ is well known and due to Dvoretzky: Dvoretzky, A. A note on Hausdorff dimension functions. Proc. Cambridge Philos. Soc. 44, (1948). 13–16. $\endgroup$
    – juan
    Commented Mar 29, 2012 at 19:49
  • $\begingroup$ Thank you so much for your answer, as far as I understand, for $\mathbb{R}^n$ we have to consider $f_n(x)=x^n \log (e/x)$ and everything will work as it has to. And the number e in the definition of f doesn't value much - we can take any positive number we want. $\endgroup$
    – Olga
    Commented Apr 3, 2012 at 19:54
  • $\begingroup$ Yes, the constant $e$ is only to make $x \log(e/x)$ monotone on $[0,1]$. In fact $H_f = H_g$ if $f$ and $g$ coincide on an interval $[0,\varepsilon]$, therefore there is some liberty in choosing $f$. $\endgroup$
    – juan
    Commented Apr 5, 2012 at 11:09

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