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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?

Rudy the Reindeer
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Pacciu
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2 Answers2

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

t.b.
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  • PS: TeX lapses sometimes lately but at the moment it works. – t.b. Mar 30 '11 at 23:58
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    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 00:04
  • @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 00:06
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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 $\mathbb{R}$.

  • Great! I didn't know that. – t.b. Mar 31 '11 at 00:04
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    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 00:15
  • @George You beat me to it! –  Mar 31 '11 at 00:19
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    That the range of a nonatomic measure is an interval is a result of Sierpinski, and Wikipedia has a proof sketch. http://en.wikipedia.org/w/index.php?title=Atom_(measure_theory)&oldid=408858789#Non-atomic_measures – George Lowther Mar 31 '11 at 00:26
  • 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 00:33
  • @Theo: I don't think so. Krein-Milman tells you that the integral of an $L^\infty$ is in the closure of convex combinations of measures of sets, but that seems a bit circular. You still need to show that any such convex combination is the measure of a single set. – George Lowther Mar 31 '11 at 00:44
  • @Theo @George: David A. Ross has a note called An Elementary Proof of Lyapunov's Theorem in the August-Sept. 2005 issue (p. 651-653) of the American Mathematical Monthly. He notes that many proofs have been given over the years using tools such as Banach-Alaoglu and/or Krein Milman. His own proof uses only the intermediate value theorem. –  Mar 31 '11 at 00:51
  • @George: Right. But I think I see how to do it (I haven't used non-atomic so far). Let me represent the measure $\mu_i$ as $L^1$-functions $f_i$ with respect to $\mu = |\mu_1| + \cdots + |\mu_n|$. The subset of all points in $L^\infty(\mu)$ evaluating to the same point in $\mathbb{R}^n$ via $f \mapsto (\mu_i(f))$ is again compact and convex and contains an extremal point $g$, and I claim that such a point is in fact a characteristic function. – t.b. Mar 31 '11 at 01:13
  • If not, for $\varepsilon$ small the set $X = {x:\varepsilon \lt g(x) \lt \varepsilon}$ has positive $\mu$-measure. Then I can find $n+1$ disjoint subsets $X_0, \ldots, X_n$ of $X$ of positive measure. Then I solve $0 = \sum_{i=1}^{n} \lambda_j\int_{X_j} f_{i} \mu$ ($i=1,\ldots,n$) for the $\lambda_{i}$'s and make $|\lambda_{i}| \lt \varepsilon$ by using the free parameter. Put $h = \sum \lambda_{i} [X_]$ and observe that $g = \frac{1}{2} (g+h) + \frac{1}{2}(g-h)$ is not an extremal point. – t.b. Mar 31 '11 at 01:13
  • @Byron: Thanks a lot for the reference (and sorry for all the pings). - I meant $0 = \sum_{j} \lambda_j \int_{X_j} f_i,d\mu$ for $i = 1,\ldots n$. – t.b. Mar 31 '11 at 01:15
  • And of course it should have been $X = {x: \varepsilon \lt g(x) \lt 1-\varepsilon}$ – t.b. Mar 31 '11 at 01:23