TITLE: A measure theory question
QUESTION [11 upvotes]: 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.

REPLY [10 votes]: 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 ${\bf R}^n$.