MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a topological space, equipped with its Borel $\sigma$-algebra $\mathcal B(X)$, and let $\mathbb P$ be a Radon probability measure on $(X, \mathcal B(X))$. Recall that the support of the measure $\mathbb P$ is the smallest closed set of full measure.

Is the support necessarily separable? If so, why? If not, what is a counterexample?

share|cite|improve this question
What about Haar measure on a compact, non-separable group? (By the way, don't you mean "smallest closed set with full measure"?). – jbc Jun 1 '13 at 23:02
Good, concise answer, and thanks for the pointer on the typo. Thanks, @jbc. – Tom LaGatta Jun 3 '13 at 3:08
up vote 3 down vote accepted

Let $I$ be a set of cardinality larger than the continuum. Then the product topology $[0,1]^{I}$ is compact but not separable. Give the interval $[0,1]$ the Lebesgue measure, then give $[0,1]^{I}$ the product measure $\mu$. If $U$ is a non-empty open subset of $[0,1]^{I}$, then $U$ contains a basic open set $\prod_{i\in I}U_{i}$ where $|\{i\in I|U_{i}\neq[0,1]\}|$ is finite. Therefore $0<\mu(\prod_{i\in I}U_{i})\leq \mu(U)$. Said differently, if $C$ is a closed subset of $[0,1]^{I}$ with $\mu(C)=1$, then $C=[0,1]^{I}$.

If you replace $[0,1]$ with the circle $S$, then $S^{I}$ is a compact non-separable group which does not have separable support as jbc mentioned.

share|cite|improve this answer
A priori, $\mu$ is only defined on the product $\sigma$-algebra on $[0,1]^I$, which is strictly smaller than the Borel $\sigma$-algebra. How do we extend $\mu$ to a Borel measure? And once this is done, how do we see that the measure is Radon? – Nate Eldredge Jun 2 '13 at 12:37
It appears that this follows from Theorems 7.14.3 and 7.2.2 (iii) of Bogachev's Measure Theory. It would be nice if there were a simpler example. – Nate Eldredge Jun 2 '13 at 12:59
Bogachev has other notes and references in Section 7.14 (viii). – Nate Eldredge Jun 2 '13 at 13:02
The reason I suggested using a group was to avoid such questions, since in this situation you can use the existence theorem for Haar measure (which, by the way, has a very short and transparent proof in the case of a compact group, using the weak star compactness of the dual ball of $C(K)$). A particularly simple solution to the OP is then provided by a large cardinal product of the two-point group. – jbc Jun 2 '13 at 13:50
The product $\sigma$-algebra on $[0,1]^{I}$ contains every Baire set, so since $[0,1]^{I}$ is compact we can extend this Baire measure to a Radon measure on the Borel $\sigma$-algebra in a unique way. – Joseph Van Name Jun 2 '13 at 17:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.