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. 



A measure theory question
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 ${\bf R}^n$. 


