Let $P$ and $Q$ be probabilities on $(\Omega, \mathcal{A})$, and let $\mathcal{F}$ be a sigma-subalgebra of $\mathcal{A}$. Assume $P \ll Q$. Assume that $P(\cdot \mid \mathcal{F})$ and $Q(\cdot \mid \mathcal{F})$ are regular.
Is it true that
$$P\big(\big\{\omega: P(\cdot \mid \mathcal{F})(\omega) \ll Q(\cdot \mid \mathcal{F})(\omega)\big\}\big)=1?$$
Is the set in question even measurable in general?
This seems like a very natural question to me, but I have not been able to find a single reference addressing it.
My first attempt was as follows. Suppose for contradiction that there were some $A \in \mathcal{A}$ such that
$$0< P\big(\big\{\omega: 0=Q(A \mid \mathcal{F})(\omega) < P(A \mid \mathcal{F})(\omega)\big\}\big).$$
Since $F=\{\omega: 0=Q(A \mid \mathcal{F})(\omega) < P(A \mid \mathcal{F})(\omega)\} \in \mathcal{F}$, we can conclude that $Q(A \cap F) = 0 < P(A \cap F)$, which contradicts $P \ll Q$.
But that observation doesn't use regularity at all, and, indeed, all we have shown is that for each $A$, there's a $P$-probability $1$ set on which $Q(A \mid \mathcal{F})=0 \implies P(A \mid \mathcal{F})=0$. But I am asking about something stronger.
I should note that the claim I ask about seems plausible to me because it holds for elementary conditional probabilities: If $P(F)>0$ and $P \ll Q$, then $P(\cdot \mid F) \ll Q(\cdot \mid F)$. Using this basic fact, one can show that the claim is true when $\mathcal{F}$ is generated by a countable partition. So all that is left really is to account for general $\mathcal{F}$.
Here is an attempt to prove the claim for the case where $\mathcal{A}$ is generated by a countable algebra $\mathcal{B}$. For arbitrary probability measure $\mu_1$, $\mu_2$ defined on an algebra $\mathcal{G}$, write $\mu_1 \ll_s \mu_2$ iff
$$\forall m \in \mathbb{N}, \ \exists r \in \mathbb{N}, \ \forall G \in \mathcal{G}: \mu_2(G) < 1/r \implies \mu_1(G)<1/m.$$
I make use of the following
Claim: If $\mu_1$ and $\mu_2$ are probability measures on $(\Omega, \mathcal{A})$, a measurable space, and $\mathcal{G}$ is an algebra that generates $\mathcal{A}$, then $\mu_1 \ll \mu_2$ if and only if $\mu_1 \ll_s \mu_2$.
So, in our case
$$\{\omega: P(\cdot \mid \mathcal{F})(\omega) \ll Q(\cdot \mid \mathcal{F})(\omega)\} = \{\omega: P^{\mathcal{B}}(\cdot \mid \mathcal{F})(\omega) \ll_s Q^{\mathcal{B}}(\cdot \mid \mathcal{F})(\omega)\},\tag{1}$$
where $P^{\mathcal{B}}(\cdot \mid \mathcal{F})(\omega)$ is the restriction of $P(\cdot \mid \mathcal{F})(\omega)$ to $\mathcal{B}$, and similarly for $Q^{\mathcal{B}}(\cdot \mid \mathcal{F})(\omega)$. So it suffices to show that the set on the right-hand side of (1) has $P$-probability $1$.
First, note that
\begin{align}
\{\omega: P^{\mathcal{B}}(\cdot \mid \mathcal{F})(\omega) \ll_s Q^{\mathcal{B}}(\cdot \mid \mathcal{F})(\omega)\} &= \bigcap_{m \in \mathbb{N}} \bigcup_{r \in \mathbb{N}} \bigcap_{B \in \mathcal{B}} \{Q(B \mid \mathcal{F}) < 1/r \implies P(B \mid \mathcal{F}) < 1/m\}\\ &=:C.
\end{align}
So, $P(C) < 1$ implies
$$\exists m, \ \forall r, \ \exists B \in \mathcal{B}: P\big(Q(B \mid \mathcal{F}) < 1/r \implies P(B \mid \mathcal{F}) < 1/m \big) < 1$$
iff
$$\exists m, \ \forall r, \ \exists B \in \mathcal{B}: P\big(\{Q(B \mid \mathcal{F})<1/r\} \cap \{P(B \mid \mathcal{F}) \geq 1/m\} \big)>0.\tag{2}$$
Now,
$$P\big(\{Q(B \mid \mathcal{F})<1/r\} \cap \{P(B \mid \mathcal{F}) \geq 1/m\} \big)>0.$$
implies
$$Q\big(\{Q(B \mid \mathcal{F})<1/r\} \cap \{P(B \mid \mathcal{F}) \geq 1/m\} \big)>0$$
because $P \ll Q$, in which case
$$P(B \cap \{P(B \mid \mathcal{F}) \geq 1/m\}) = \int_{\{P(B \mid \mathcal{F}) \geq 1/m\}} P(B \mid \mathcal{F})dP \geq 1/m$$
and
$$Q(B \cap \{Q(B \mid \mathcal{F}) < 1/r\}) = \int_{\{Q(B \mid \mathcal{F}) < 1/r\}} Q(B \mid \mathcal{F})dQ <1/r.$$
But then, by (2),
$$\exists m, \ \forall r, \ \exists B \in \mathcal{B}: Q(B \cap \{Q(B \mid \mathcal{F}) < 1/r \ \text{and} \ P(B \cap \{P(B \mid \mathcal{F}) \geq 1/m,$$
which implies
$$\exists m, \ \forall r, \ \exists A \in \mathcal{\mathcal{A}}: Q(A) < 1/r \ \text{and} \ P(A) \geq 1/m.$$
This contradicts $P \ll Q$.
Best Answer
Here is another approach. Let $\varphi$ denote the Radon-Nikodym derivative of $P$ with respect to $Q$. Define $$ K(\omega, A):={ \int_A \varphi(\omega')Q(d\omega'|\mathcal F)(\omega)\over \int_\Omega\varphi(\omega')Q(d\omega'|\mathcal F)(\omega)},\qquad A\in\mathcal A, $$ with the understanding that the ratio vanishes when the denominator is zero. You can check that (i) $(\omega,A)\mapsto K(\omega,A)$ has the properties required of the regular conditional distribution $P(\cdot|\mathcal F)(\omega)$, and (ii) clearly $K(\omega,\cdot)\ll Q(\cdot|\mathcal F)(\omega)$ for a.e. $\omega$.