Let P be Lebesgue measure on the three-dimensional cube [0,1]3. Assume the Axiom of Choice. Then there will be P-nonmeasurable sets (e.g, as there is no finitely additive (much less countably additive, rigid-motion-invariant probability measure on all subsets of [0,1]3 by the Banach-Tarski paradox. Now let F be the Lebesgue measurable subsets of [0,1]3. One might then hope that one can define P(X|Y) for all X and Y in F, as long as Y is non-empty.
Turns out that one can't, at least if one expects finite additivity and rigid-motion invariance. The reason for that is that if we let Y be (the surface of) a sphere in [0,1]3, then all subsets of Y will be in F, but there is no finitely additive rotation invariant probability measure on all subsets of a sphere, by the Hausdorff paradox. This problem disappears if we restrict ourselves to the Borel subsets of [0,1]3, but the extension to the Lebesgue measurable ones is epistemologically very plausible—obviously any subset of a null set should be a null set.
But here's a funny thing. While I haven't checked all the details—it's grading time so I can only give this so much thought—it turns out that one can sensibly define P(X|Y) for all X in F and all non-measurable Y. The easiest case is where Y is maximally non-measurable, i.e., all its measurable subsets have null measure and all its measurable supersets have full measure. In that case, one can simply define P(X|Y)=P(X). Moreover, one can naturally extend this measure to all X's in FY, where FY is the smallest σ-field generated by F and Y (basically by setting P(X∩Y|Y)=P(X) and P(X∩Yc|Y)=0 for X in F).
This means that the Popper function approach to conditional probability on which P(X|Y) is defined for all X and Y in a single field or σ-field is not general enough, at least if we want P(X|[0,1]3) to be defined for all Lebesgue measurable X's. For in fact it seems we have more freedom as to what Y's we get to plug in and less as to what X's.