# If $P \ll Q$, are the regular conditional probabilities a.s. absolutely continuous?

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 $$P \ll Q$$.

• Your contradiction is odd: $A$ depends on $\omega.$ – Will M. Nov 8 '18 at 22:33
• Set $E$ the event $P(\cdot \mid \mathcal{F}) \not\ll Q(\cdot \mid \mathcal{F}).$ Then, if $\omega \in E$ there exists a set $A = A(\omega) \in \mathcal{A}$ such that $P(A\mid \mathcal{F}) > 0$ while $Q(A\mid \mathcal{F}) = 0$. – Will M. Nov 8 '18 at 22:39
• @WillM. Ah, I see. Thanks. – aduh Nov 8 '18 at 22:42

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. $$\omega$$.