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I have a (probability) measure $\mu$ on space $X$, a function $f: X \rightarrow Y$, and a pushforward measure $\nu$ on $Y$ induced by applying $f$ to $\mu$.

Suppose we have $\text{supp}(\nu)$. Can we say anything about the set $f^{-1}(\text{supp}(\nu))$ in relation to a set involving $\mu$? My guess is that it needn't equal $\text{supp}(\mu)$. [I have read this post.]

If additionally $f$ is continuous, then it seems like we should have $\text{supp}(\mu) = f^{-1}(\text{supp}(\nu))$? Is this right? (edit: it is not! see comments)

Alp Uzman
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housed_off_space
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    For the second question, let $f$ be a constant function from $\mathbb{R}$ to $\mathbb{R}$ with value $1$ and let $\mu$ be the Dirac measure concentrated on $0$. – Michael Greinecker Mar 15 '23 at 23:00
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    So as Michael's example shows, when $f$ is continuous, the preimage $f^{-1}(\operatorname{supp} \nu)$ is a closed set of measure 1, but there is no reason why it must be the smallest one - unless $f$ is in fact a homeomorphism, but then everything is trivial. – Nate Eldredge Mar 16 '23 at 02:18
  • Good, that makes sense. So yes, with continuity, $supp(\mu) \subseteq f^{-1}(supp(\nu))$. I suppose also trivially, if I restricted the domain of $f$ such that we had $f: supp(\mu) \rightarrow Y$, then we would have equality of the sets. – housed_off_space Mar 16 '23 at 04:35

1 Answers1

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At least when $X$ and $Y$ are separable and metrizable, one can relate the support of a measure and the pullback of the support of the pushforward. We'll use the following (see Parthasarathy's Probability Measures on Metric Spaces, pp.27-28, Thm.2.1):

Theorem: Let $X$ be a separable metrizable space. Then for any Borel probability measure $\mu$ on $X$, its support $\text{supp}(\mu)$ is defined by any one of the following statements:

  • $\text{supp}(\mu)=\bigcap\{F\subseteq X \,|\, F\text{ is closed and } \mu(X\setminus F)=0\}$.
  • $\text{supp}(\mu)= \{x\in X\,|\, \forall U\in\text{Nbhd}(x):\mu(U)>0\}$.
  • $\mu(\text{supp}(\mu))=1$ and for any closed subset $F\subseteq X$ with $\mu(F)=1$, $\text{supp}(\mu)\subseteq F$.

Next let $X,Y$ be separable and metrizable, $f:X\to Y$ be Borel measurable. Let $\mu$ be a Borel probability measure on $X$ and denote by $f_\ast(\mu)$ the pushforward of $\mu$ by $f$. $f^{-1}(\text{supp}(f_\ast(\mu)))$ is Borel measurable and has $\mu$-measure $1$; consequently we have

$$\text{supp}(\mu)\subseteq \overline{f^{-1}\left(\text{supp}(f_\ast(\mu))\right)}$$

(so that $f^{-1}\left(\text{supp}(f_\ast(\mu))\right)$ is dense in $\text{supp}(\mu)$, similar to the linked discussion). Interpreted differently, we also have that all three of the sets

$$\text{supp}(\mu),\quad f^{-1}\left(\text{supp}(f_\ast(\mu))\right),\quad \overline{f^{-1}\left(\text{supp}(f_\ast(\mu))\right)}$$

represent the same element of the measure algebra of $X$ w/r/t $\mu$ (see my answer at Why do we write $P\ll Q$ and why is $P$ called absolutely continuous w.r.t. $Q$?).

Alp Uzman
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