So given a probability space $(\Omega, \mathcal{A}, P)$ and sub-$\sigma$-algebra $\mathcal{G}\subseteq\mathcal{A}$, consider the conditional probability $P(A|\mathcal{G})$ for any $A\in\mathcal{A}$. Now $P(A|\mathcal{G})$ has to satify
(i) that it is $\mathcal{G}$-measurable
(ii) That for all $G\in\mathcal{G}$ it holds: $$\int_G P(A|\mathcal{G})dP=P(A\cap G). $$
It is an import fact that $P(A|\mathcal{G})(\omega)$ need not be a probability measure for every $\omega\in\Omega$. My question is now if it even can fail to be a measure?
Billingsley in his book Probability and Measure seems to claim this in Problem 33.11, since he there writes: "The following construction shows that conditional probabilities may not give measures". However, then he goes to assume that $P(A|\mathcal{G})(\omega)$ is a probabiality measure for all $\omega\in\Omega$ and then creates a contradiction. So it seems that he only shows that $P(A|\mathcal{G})(\omega)$ need not be a probability measure for every $\omega\in\Omega$ (which is well known), but does not show that a conditional probability can even fail to be a measure. So do there exist conditional probabilities that even fail to be measures?
