Construction of "pathological" measures A selfposed but never solved problem:

Is it possible to find a measure space
  $(X, \mathcal{M} ,\mu )$  such that
  the range of $\mu$ is something like
  the Cantor set (i.e. a bounded,
  perfect, uncountable, totally
  disconnected set)?

I was thinking about this problem some time ago and now, reading some old MT post, it came back to my mind.
I remember I solved a similar selfposed problem, showing that one can construct a measure over an interval whose range is the union of a finite number of disjoint intervals, but the one listed on top resisted my efforts.
Any ideas?
P.S.: It seems TeX tags don't work, isn't it?
 A: Added: After posting this, I realized that the $n=1$ case is pretty straightforward
and doesn't require a big theorem. In fact, it is a nice exercise to show that any non-atomic probability space supports a uniform(0,1) random variable $U$. For $0\leq \alpha\leq 1$, the set $\lbrace\omega: U(\omega)\leq \alpha\rbrace$ has measure $\alpha$.  
This result is Corollary 1.12.10 (page 56) in Bogachev's Measure Theory Volume 1. 

Liapunov's convexity theorem implies (take $n=1$) that the range of any finite non-atomic measure is a compact, convex set of $\mathbb{R}$. 
A: Recall that the Cantor set $C$ can be identified with the set of numbers in $[0,1]$ admitting a ternary expansion consisting entirely of $0$'s and $2$'s. Take $\mathbb{N}$ with the measure $\mu(n) = \frac{2}{3^{n}}$. Then for every subset $A \subset \mathbb{N}$ we have $\mu(A) \in C$ and for every $x = \sum_{n=1}^{\infty} a_{n} \frac{2}{3^n} \in C$ with $a_{n} \in \{0,1\}$ we find $A$ with $\mu(A) = x$ by taking $A = \{n \in \mathbb{N}\,:\,a_{n} = 1\}$.
