We start today with a possibly surprising result; pulling back a -ring puts significant restrictions on measurable functions. If is a function from a set into a measurable space , and if is measurable with respect to the -ring on , then whenever .
To see this fix a point , and let be a measurable set containing . Its preimage is then a measurable set containing . We can also define the level set , which is a measurable set since is a measurable function. Thus the intersection
is measurable. That is, it’s in , and so there exists some measurable so that is this intersection. Clearly , and so is as well, by assumption. But then , and we conclude that .
From this result follows another interesting property. If is a mapping from a set onto a measurable space , and if is a measurable function, then there is a unique measurable function so that . That is, any function that is measurable with respect to a measurable structure pulled back along a surjection factors uniquely through the surjection.
Indeed, since is surjective, for every we have some so that . Then we define , so that , as desired. There is no ambiguity about the choice of which preimage of to use, since the above result shows that any other choice would lead to the same value of . What’s not immediately apparent is that is itself measurable. But given a set we can consider its preimage , and the preimage of this set:
which is measurable since is a measurable function. But then this set must be the preimage of some measurable subset of , which shows that the preimage is measurable.
It should be noted that this doesn’t quite work out for functions that are not surjective, because we cannot uniquely determine if has no preimage under .