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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?

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    $\begingroup$ What about Haar measure on a compact, non-separable group? (By the way, don't you mean "smallest closed set with full measure"?). $\endgroup$
    – jbc
    Commented Jun 1, 2013 at 23:02
  • $\begingroup$ Good, concise answer, and thanks for the pointer on the typo. Thanks, @jbc. $\endgroup$
    – Tom LaGatta
    Commented Jun 3, 2013 at 3:08

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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.

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    $\begingroup$ 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? $\endgroup$
    – Nate Eldredge
    Commented Jun 2, 2013 at 12:37
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    $\begingroup$ 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. $\endgroup$
    – Nate Eldredge
    Commented Jun 2, 2013 at 12:59
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    $\begingroup$ Bogachev has other notes and references in Section 7.14 (viii). $\endgroup$
    – Nate Eldredge
    Commented Jun 2, 2013 at 13:02
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    $\begingroup$ 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. $\endgroup$
    – jbc
    Commented Jun 2, 2013 at 13:50
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    $\begingroup$ 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. $\endgroup$
    – Joseph Van Name
    Commented Jun 2, 2013 at 17:13

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