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Given that the local dimension of a measure $\nu$ at a point $w$ is defined by $$d[\nu](w) = \limsup_{r \to 0} \frac{\log(B(w,r))}{\log r}$$ is it possible to find a measure which has local dimension $\alpha$ at each point in its support but which is not absolute continuous with respect to $m_\alpha$; the Hausdorff measure of dimension $\alpha$?

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I believe the answer is yes. Here is (the idea for) an example of a measure $\nu$ on a Cantor set $C$ in $[0,1]$, with $d[\nu]\equiv 1$ on its support, but $m_1(C)=0$. Start with $C_0=[0,1]$, remove the middle $1/3$ interval to get $C_1 = [0,1/3] \cup [2/3,1]$, remove the middle $1/4$ of both of those intervals to get $C_2$, etc. I.e., in the $n$-th step you remove the middle $1/(n+2)$-th of the $2^{n-1}$ intervals of $C_{n-1}$ to get $C_n$. Define $C = \bigcap_n C_n$, and define the measure $\nu$ so that each interval in the $2^n$-th step has measure $2^{-n}$.

The Lebesgue (or 1-dim Hausdorff measure) of $C_n$ is $(1-1/3)(1-1/4)\cdots(1-1/(n+2))$, and by standard estimates this goes to $0$. (You can probably modify the construction to get an easier limit argument here.) Now if you zoom in to small scales, the set $C$ contains a "fat" Cantor set of dimension arbitrarily close to 1, and the measure $\nu$ looks like the symmetric measure on such a Cantor set, so your $\limsup$ should be 1.

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Thanks! This example works perfectly :) –  malin Oct 8 '12 at 18:51
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