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Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space and $f:\Omega\rightarrow\mathbb{R}^d$ a measurable function. Let $\mu$ be the probability measure defined by $\mu(B):=\mathbb P(f^{-1}(B))$ for any Borel set $B\in\mathbb{R}^d$. What can we say about the set $\{ f(\omega)\mid\omega\in\Omega\}\cap\text{supp}(\mu)$? Is it dense in $\text{supp}(\mu)$? Here, we define $\text{supp}(\mu)=\bigcap_{A\text{ closed}, \mu(A^c)=0} A.$

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Yes.

Let $E = \{f(\omega) : \omega \in \Omega\}$ be the image of $f$, and $\bar{E}$ its closure. Since $\bar{E}^c$ is disjoint from the image of $f$, we have $f^{-1}(\bar{E}^c) = \emptyset$. Hence $\mu(\bar{E}^c) = P(\emptyset) = 0$, i.e. $\bar{E}$ is a closed set whose complement has $\mu$-measure 0. Since $\operatorname{supp}(\mu)$ is by definition the smallest such set, we must have $\operatorname{supp}(\mu) \subset \bar{E}$, which is to say that $E$ is dense in $\operatorname{supp}(\mu)$.

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    @AndyTeich: Take $\Omega' = f^{-1}(\operatorname{supp}\mu)$; this is measurable because $\operatorname{supp}\mu$ is closed, and $f(\Omega') = f(\Omega) \cap \operatorname{supp}\mu$, which is a dense subset of $\operatorname{supp}\mu$ as argued above.2012-11-08