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Let $\Theta$ and $X$ be two (Hausdorff) topological spaces. Let $\mathbb P : \Theta \to \Delta(X)$ be a "statistical model", i.e., a continuous function from parameter space $\Theta$ to the space of probability measures $\Delta(X)$ (equipped with the topology of weak convergence of measures).

Let $S_\theta \subseteq X$ denote the support of the measure $\mathbb P_\theta$ (i.e., the smallest closed set of full measure). Define the relation $$\Theta' := \{ (\theta, x) : x \in S_\theta \}.$$ Is $\Theta'$ a closed subset of the product space $\Theta \times X$? If not, is $\Theta'$ at least measurable?

Edit: fedja points out that $\Theta'$ need not be closed. In that example, $\Theta' \subseteq [0,1] \times \mathbb R$, and $\Theta' = \big\{(0,0)\big\} \cup \big( (0,1] \times \mathbb R \big)$.

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  • $\begingroup$ Here's the naïve approach, which doesn't work. Suppose $(\theta^s, x^s) \to (\theta,x)$. Let $U$ be an open neighborhood of $x$ in $X$. We want to show that $\mathbb P_\theta(U) > 0$. By weak convergence of measures, $\mathbb P_\theta(U) \le \liminf_s \mathbb P_{\theta^s}(U)$. Unfortunately, this is the wrong direction for the inequality, and doesn't show that $\mathbb P_\theta(U)$ is positive. $\endgroup$
    – Tom LaGatta
    Commented Oct 21, 2013 at 0:25
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    $\begingroup$ Certainly not closed: take $\Theta=[0,1]$, $X=\mathbb R$, $P_0=\delta_0$ and $P_\theta=\frac{\theta}{\pi(x^2+\theta^2)}\,dx$ for $\theta>0$ (or any other approximate identity). I'm not sure about Borel measurability: need to think more. $\endgroup$
    – fedja
    Commented Oct 21, 2013 at 0:50
  • $\begingroup$ Are your probability measures on $X$'s Borel sets? $\:$ What if there is no smallest closed set of full measure? $\endgroup$
    – user5810
    Commented Oct 21, 2013 at 1:25
  • $\begingroup$ @RickyDemer: yes, I made the implicit assumption that the probability measures are all Borel, and maybe even Radon. If the support doesn't exist (or is empty), then my question is vacuous as stated. However, Vakhania introduced an alternate notion of support, which is always well-defined. When the topological support exists, the two notions agree, and the Vakhania support has full measure. See Vakhania's classic article here: projecteuclid.org/DPubS/Repository/1.0/… $\endgroup$
    – Tom LaGatta
    Commented Oct 21, 2013 at 1:33

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For $\mu\in \Delta(X)$ and $x\in X$ (which I assume to be a metric space with distance $d$) we have $x\in \operatorname{supp}(\mu)$ if and only if for any positive integer $n$ the function $\delta_{x,n}:=\big(1/n - d(\cdot,x)\big)_+$ has positive integral wrto the measure $\mu$. Therefore $$\{(x,\mu)\in X\times\Delta(X)\,:\, x\in \operatorname{supp}(\mu)\}=\bigcap_{n\in\mathbb{N}_+}\bigcup_{k\in\mathbb{N}_+} \{(x,\mu)\,:\, \langle\mu,\delta_{x,n }\rangle\ge 1/k\}\, . $$ For any $n$ and $k$ the set $\{(x,\mu)\,:\, \langle\mu,\delta_{x,n }\rangle\ge 1/k\} $ is closed, just because both the map $X\ni x\mapsto \delta_{x,n }\in \big(C_b(X),\|\cdot\|_\infty\big)$ and the pairing $C_b(X)\times\Delta(X)\ni(f,\mu) \mapsto \langle \mu,f \rangle$ are continuous ($\Delta(X)$ being endowed with the weak topology). This shows that the above set (the graph of the support map) is a Borel set; note that by consequence $\Theta'$ is Borel even for measurable $\mathbb{P}:\Theta\to\Delta(X)\,.$

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  • $\begingroup$ I assumed $X$ is a metric space. The argument can be extended to more general $X$, but there could be problems (starting with the definition of support) if $X$ is not paracompact, or not first-countable. $\endgroup$
    – Pietro Majer
    Commented Oct 21, 2013 at 8:52

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