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Let $I$ be an index set, possibly uncountable, and $U_\iota\subseteq\mathbb{R}^d$ be closed for $\iota\in I$. Does there always exist a Borel probability measure $\mu$ such that $\text{supp}(\mu)=\bigcup_{\iota\in I} U_\iota$, where $\text{supp}(\mu)$ is the support of $\mu$, i.e. the smallest closed set $A$ such that $\mu(A^c)=0$?

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up vote 7 down vote accepted

Of course not. The infinite union of closed sets is generally not closed. It might even fail to be Borel. Let $N\subseteq\mathbb{R}^d$ be non-Borel, let $I=N$ and let $U_i=\{i\}$. Then $\bigcup_{i\in I}U_i=N$.

But for every closed set $C\subseteq\mathbb{R}^d$, there is a probability measure $\mu$ with support $C$. Just pick a countable subset $D\subseteq C$ such that $C$ is the closure of $D$. If $D$ is finite, let $\mu$ be the uniform distribution on $D$. If $D$ is infinite, write it as $\{d_1,d_2,d_3,\ldots\}$ and let $\mu$ be the discrete probability measure that puts probability $1/2^n$ on the point $d_n$.

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