## Boolean Rings

A “Boolean ring” is a commutative ring with the additional property that each and every element is idempotent. That is, for any we have . An immediate consequence of this axiom is that , since we can calculate

The typical example we care about in the measure-theoretic context is a ring of subsets of some set , with the operation for addition and for multiplication. You should check that these operations satisfy the axioms of a Boolean ring. Since this is our main motivation, we will just consistently use and to denote addition and multiplication in Boolean rings, whether they arise from a measure theoretic context or not. From here it looks a lot like set theory, but keep in mind that the objects we’re looking at may have nothing to do with sets.

We can use these operations to define the other common set-theoretic operations. Indeed

and

and we can then define orders in the usual manner: .

As usual, the union of two elements is the “smallest” (with respect to this order) element above both of them, and the intersection of two elements is the “largest” element below both of them. The same goes for any finite number of elements, but if we try to move to an infinite number of elements there is no guarantee that there is *any* element above or below all of them, much less that such an element is unique. A “Boolean -ring” is a Boolean ring so that every countably infinite set of elements has a union. In this case, it is immediately true that any countably infinite set of elements has an intersection as well. The typical example, of course, is a -ring of subsets of a set .

A “Boolean algebra” is a Boolean ring for which there is some element so that for all elements . A “Boolean -algebra” is both a Boolean -ring and a Boolean algebra.

In the obvious way we have a full subcategory of the category of rings. It contains full subcategories of Boolean -rings, Boolean algebras, and Boolean -algebras.

## Measurable Functions on Pulled-Back Measurable Spaces

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 .

## Pulling Back and Pushing Forward Structure

Remember that we defined measurable functions in terms of inverse images, like we did for topological spaces. So it should be no surprise that we move a lot of measurable structure around between spaces by “pulling back” or “pushing forward”.

First of all, let’s say that is a measurable space and consider a function . We can always make into a measurable function by pulling back the -ring . For each measurable subset we define the preimage as usual, and define the pullback to be the collection of subsets of of the form for . Taking preimages commutes with arbitrary setwise unions and setwise differences, and , and so is itself a -ring. Every point gives us a point , and every point is contained in some measurable set . Thus is contained in the set , and so we find that is a measurable space. Clearly, contains the preimage of every measurable set , and so is measurable.

Measures, on the other hand, go the other way. Say that is a measure space and is a measurable function between measurable spaces, then we can define a new measure on by “pushing forward” the measure . Given a measurable set , we know that its preimage is also measurable, and so we can define . It should be clear that this satisfies the definition of a measure. We’ll write for this measure.

If is a measurable function, and if is a measure on , then we have the equality

in the sense that if either integral exists, then the other one does too, and their values are equal. As usual, it is sufficient to prove this for the case of for a measurable set . Linear combinations will extend it to simple functions, the monotone convergence theorem extends to non-negative measurable functions, and general functions can be decomposed into positive and negative parts.

Now, if is the characteristic function of , then if — that is, if — and otherwise. That is, . We can then calculate

As a particular case, applying the previous result to the function shows us that

We can go back and forth between either side of this equation by the formal substitution .

Finally, we can combine this with the Radon-Nikodym theorem. If is a measurable function from a measure space to a totally -finite measure space so that the pushed-forward measure is absolutely continuous with respect to . Then we can select a non-negative measurable function

so that

again, in the sense that if one of these integrals exists then so does the other, and their values are equal. The function plays the role of the absolute value of the Jacobian determinant.