# What can we say about the image of a measureable map in the support of its push forward measure

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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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 $\operatorname{supp}(\mu)$.
Ok, for some reason I was thinking that when $f$ is a measurable map, then $f$ does not have to be defined on all of $\Omega$, i.e. that we only have $f^{-1}(E)\subset\Omega$. But yes, in this sense you are right –  Andy Teich Oct 8 '12 at 13:02
@AndyTeich: I edited to make my discussion more exactly match your definition, and incidentally it now also handles the case of partial functions. You could also handle this case by restricting $f$ to the (measurable) set where it is defined, and treating this set as your new $\Omega$ (noting that its total measure may now be less than 1). –  Nate Eldredge Oct 8 '12 at 13:10
but in general, you can't say if $E\subseteq \text{supp}\mu$ or $\text{supp}\mu\subseteq E$ ? –  Andy Teich Oct 20 '12 at 14:51
@AndyTeich: No. Let $\Omega = [0,1) \cup \{2\}$ equipped with Lebesgue measure, and let $f : \Omega \to \mathbb{R}$ be the inclusion map. Then the image of $f$ is $E = [0,1) \cup \{2\}$, but you can check that the support of $\mu$ is $[0,1]$. Neither of these sets contains the other. –  Nate Eldredge Oct 21 '12 at 21:21
@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. –  Nate Eldredge Nov 8 '12 at 13:21