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Let $X$, $M$ be two metric spaces and $\nu:X\rightarrow \mathcal{M}_1(M)$, $x\mapsto \nu_x$ a map, where $\mathcal{M}_1(M)$ is the space of all probabilities over $M$ with the Boral $\sigma$-algebra, $\mathcal{M}_1(M)$ with the weak* topology. The following statements are equivalent

  1. $\nu$ is measurable with respsect to the Borel $\sigma$-algebras of $X$ and $\mathcal{M}_1(M)$;

  2. The map $X\rightarrow \mathbb{R}$, $x\mapsto \int \varphi d\nu_x$ is measurable, for all continuous bounded map $\varphi:M\rightarrow \mathbb{R}$;

  3. The map $X\rightarrow \mathbb{R}$, $x\mapsto \int \psi d\nu_x$ is measurable, for all measurable bounded map $\psi>M\rightarrow \mathbb{R}$;

  4. The map $X\rightarrow \mathbb{R}$, $x\mapsto \nu_x(E)$ is measurable, for all $E\subset M$ measurable.

I managed to prove some of them: $(2)\Rightarrow(1)$, $(4)\Rightarrow(3)$ and $(3)\Rightarrow(2)$.

I was trying to prove $(2)\Rightarrow(1)$ and $(2)\Rightarrow(4)$ because I think it's easier! Here are my ideas:

$(2)\Rightarrow (4):$ Given a measurable set $E\subset M$ and its characteristic map $\chi_E$, the natural approach is to approximate $\chi_E$ by a sequence of continuous maps. The problem is that it converges a.e. and I can't determine for which measure! I just just don't know how to solve this problem!

$(1)\Rightarrow (2):$ For this one I'm kind of lost! I tried to prove that the map $\mathcal{M}_1(M)\rightarrow \mathbb{R}$, $\mu\mapsto \int \varphi d\mu$, for a giver $\varphi:M\rightarrow\mathbb{R}$ continuous bounded map, is measurable but I couldn't manage to do that! (or at least measurable in $\operatorname{Im}(\nu)\cap \mathcal{M}_1(M)$)

Do you guys have any kind of idea? Any hint?

Thank you so much in advance, guys!

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$(1)\Rightarrow (2)$ is immediate, given the definition of the weak* topology on $\mathcal{M}_1(M)$. For $(2)\Rightarrow (4)$, use a $\pi$-$\lambda$ argument to prove that it's enough to prove that $(2)\Rightarrow (4)$ in the special case where $E$ is closed. Then use the fact that if $E$ is a closed set in a metric space $X$, $$ \lim_{n\to\infty} \max(1-n d(x,E),0)= \chi_E(x), \qquad \text{for all } x\in X. $$

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