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Let $(X, \mathcal X, \mathbb P)$ be a probability space, let $\mathcal G$ be a $\sigma$-subfield, and let $\mathbb P^{\mathcal G}$ be a regular conditional probability for $\mathbb P$ given $\mathcal G$. That is, for all $x \in X$, $\mathbb P_x^{\mathcal E}$ is a probability measure, $x \mapsto \mathbb P_x(A)$ is $\mathcal G$ measurable for all $A \in \mathcal X$, and $$\mathbb P(A \cap G) = \int_G \mathbb P^{\mathcal G}_x(A)\mathbb P(dx)$$ holds for all $A \in \mathcal X$ and $G \in \mathcal G$.

Is it true that $P_x^{\mathcal G} \ll \mathbb P$ for almost every $x$?

I can confirm that this is true if $\mathcal X$ is separable, but I'm not sure it holds in general.

user435571
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    My guess is that this is false without separability of the sigma field. – Kavi Rama Murthy Dec 26 '19 at 05:56
  • @KaviRamaMurthy I was starting to think that too. For instance, separability is needed to ensure that the regular conditional probabilities are almost surely unique, so maybe we can get a counterexample when a.s. uniqueness fails. – user435571 Dec 26 '19 at 06:02
  • i don't know if this helps, but I was trying to disprove the absolute continuity of a joint probability measure $p_{X,Y}$ with respect to the product of its marginals $p_X\times p_Y$ and in the end it was enough to consider $X\sim\mathcal U([0,1])$, $Y=X$ and the set $A={(x,x):x\in[0,1]}$, then $p_{X,Y}(A)=1$ and $p_X\times p_Y(A)=0$. I feel like this could be adapted for your counter example since your conditional probability measure is just a Rado-Nykodym derivative of a joint. – P. Quinton Dec 26 '19 at 08:50

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Let $X=[0,1]^2$, $\mathcal X$ is its Borel algebra and $\mathcal G=\{[0,1]\times B : [0,1]\times B \in \mathcal X\}$. If $\mathbb P(A) = \lambda(\{a:(a,a)\in A\})$ for any $A\in \mathcal X$ (with $\lambda$ the Lebesgue measure on $[0,1]$) then for $(x,y)\in[0,1]^2$ we have $\mathbb P_{(x,y)}^{\mathcal G}(A)=1_{A}\{(y,y)\}$ and $\mathbb P(\{(0,0)\})=0\neq 1=\mathbb P_{(0,0)}^{\mathcal G}(\{(0,0)\})$.

P. Quinton
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  • Thanks, but there are some things I don't understand here. First, not every member of $\mathcal X$ is of the form $A \times B$ (where I assume you take $A$ and $B$ to be Borel subsets of $[0,1]$). Second, I don't get $\mathbb P^{\mathcal G}x(A \times B) = 1{A \cap B}(x)$. $\mathbb P^{\mathcal G}$ is a function on $X$, so $x$ must be in $[0,1]^2$, but then $1_{A \cap B}(x)$ does not seem well-formed. – user435571 Dec 26 '19 at 20:34
  • I tried too fix it but I'm not convinced. I think the idea may be good but I made stupid mistakes. – P. Quinton Dec 27 '19 at 09:39