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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?

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up vote 5 down vote accepted

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 $\mathbb{R}$.

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Great! I didn't know that. – t.b. Mar 31 '11 at 0:04
Alternatively, every finite nonatomic measurable has a measure-preserving map to an interval $[0,a]$ equipped with the Borel sigma algebra and the Lebesgue measure. So the range of the measure will also be $[0,a]$. – George Lowther Mar 31 '11 at 0:15
@George You beat me to it! – Byron Schmuland Mar 31 '11 at 0:19
That the range of a nonatomic measure is an interval is a result of Sierpinski, and Wikipedia has a proof sketch.… – George Lowther Mar 31 '11 at 0:26
Doesn't Liapunov's theorem immediately follow from the fact that the positive part of the unit ball in the space of $L^\infty$-functions is weak$^\ast$-compact and that the characteristic functions are its extremal points (which are dense by Krein-Milman)? – t.b. Mar 31 '11 at 0:33

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 $A = \{n \in \mathbb{N}\,:\,a_{n} = 1\}$.

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PS: TeX lapses sometimes lately but at the moment it works. – t.b. Mar 30 '11 at 23:58
Nice example. And, the fact that you constructed a purely atomic measure is no coincidence. By another answer (now deleted, I guess he misread the question), this is not possible for a non-atomic measure. In fact, the range of the nonatomic part would be a closed interval, negating the possibility of getting a perfect set. – George Lowther Mar 31 '11 at 0:04
@George: Thanks! Luckily @Byron decided to undelete his answer, so it's back again (it's a great complement to the answer). – t.b. Mar 31 '11 at 0:06

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