## A Counterexample

I’ve been thinking about yesterday’s post about the product of measurable spaces, and I’m not really satisfied. I’d like to present a concrete example of what can go wrong when one or the other factor isn’t a total measurable space — that is, when the underlying set of that factor is not a measurable subset of itself.

Let be the usual real line. However, instead of the Borel or Lebesgue measurable sets, let be the collection of all countable subsets of . We check that this is a -ring: it’s clearly closed under unions and differences, making it a ring, and the union of a countable collection of countable sets is still countable, so it’s monotone, and thus a -ring. And every point is in the measurable set , so is a measurable space. We’ll let be two more copies of the same measurable space.

Now we define the “product” space . A subset of is in if it can be written as the union of a countable number of measurable rectangles with and . But if and are both countable, then is also countable, and thus any measurable subset of is countable. On the other hand, if a subset of is countable, it’s the countable union of all of its singletons , each of which is a measurable rectangle. That is, if a subset of is countable, then it is measurable. That is, is the collection of all the countable subsets of .

Next we define the function by setting for any real number . We check that it’s measurable: given a measurable set we calculate

but if is measurable, then it’s countable, and it can contain only countably many points of the form . The preimage must then be countable, and thus measurable, and thus is a measurable function.

That said, the component function is *not* measurable. Indeed, we have for all real numbers . The set is countable — and thus measurable — but its preimage is the entire real line, which is uncountable and is *not* measurable. This is exactly the counterintuitive result we were worried about.

## Product Measurable Spaces

Now we return to the category of measurable spaces and measurable functions, and we discuss product spaces. Given two spaces and , we want to define a -ring of measurable sets on the product .

In fact, we’ve seen that given rings and we can define the product as the collection of finite disjoint unions of sets , where and . If and are -rings, then we define to be the smallest monotone class containing this collection, which will then be a -ring. If and are -algebras — that is, if and — then clearly , which is thus another -algebra.

Indeed, is a measurable space. The collection is a -algebra, and every point is in some one of these sets. If , then and , so . But is this really the -algebra we want?

Most approaches to measure theory simply define this to be the product of two measurable spaces, but we have a broader perspective here. Indeed, we should be asking if this is a product object in the category of measurable spaces! That is, the underlying space comes equipped with projection functions and . We must ask if these projections are measurable for our choice of -algebra, and also that they satisfy the universal property of a product object.

Checking measurability is pretty straightforward. It’s the same for either projector, so we’ll consider explicitly. Given a measurable set , the preimage must be measurable as well. And here we run into a snag: we know that is measurable, and so for any measurable the subset is measurable. In particular, if is a -algebra then is measurable right off. However, if is not itself measurable then is not the limit of any increasing countable sequence of measurable sets either (since -rings are monotonic classes). Thus is not the limit of any increasing sequence of measurable products , and so can’t be in the *smallest* monotonic class generated by such products, and thus can’t be measurable!

So in order for to be a product object, it’s necessary that be a -algebra. Similarly, must be a -algebra as well. What would happen if this condition fails? Consider a measurable function defined by . We can write , but since is not measurable we have no guarantee that will be measurable!

On the other hand, all is not lost; if and are both measurable, and if and , then we calculate the preimage

which is thus measurable. Monotone limits of finite disjoint unions of such sets are easily handled. Thus if both components of are measurable, then is measurable.

Okay, so back to the case where and are both -algebras, and so and are both measurable. We still need to show that the universal property holds. That is, given two measurable functions and , we must show that there exists a *unique* measurable function so that and . There’s obviously a unique function on the underlying set: . And the previous paragraph shows that this function must be measurable!

So, the uniqueness property always holds, but with the one caveat that the projectors may not themselves be measurable. That is, the full subcategory of *total* measurable spaces has product objects, but the category of measurable spaces overall does not. However, we’ll still talk about the “product” space with the -ring understood.

A couple of notes are in order. First of all, this is the first time that we’ve actually used the requirement that every point in a measurable space be a member of some measurable set or another. It will become more important as we go on. Secondly, we define a “measurable rectangle” in to be a set so that and — that is, one for which both “sides” are measurable. The class of all measurable sets is the -ring generated by all the measurable rectangles.

## Lebesgue Decomposition

As we’ve said before, singularity and absolute continuity are diametrically opposed. And so it’s not entirely surprising that if we have two totally -finite signed measures and , then we can break into two uniquely-defined pieces, and , so that , , and . We call such a pair the “Lebesgue decomposition” of with respect to .

Since a signed measure is absolutely continuous or singular with respect to if and only if it’s absolutely continuous or singular with respect to , we may as well assume that is a measure. And since in both cases is absolutely continuous or singular with respect to if and only if and both are, we may as well assume that is also a measure. And, as usual, we can break our measurable space down into the disjoint union of countably many subspaces on which both and are both totally finite. We can assemble the Lebesgue decompositions on a collection such subspaces into the Lebesgue decomposition on the whole, and so we can assume that and are totally finite.

Now that we can move on to the proof itself, the core is based on our very first result about absolute continuity: . Thus the Radon-Nikodym theorem tells us that there exists a function so that

for every measurable set . Since , we must have -a.e., and thus -a.e. as well. Since

Let us define and . Then we calculate

and thus (by the finiteness of ), . Defining and , then it’s clear that . We still must prove that .

If , then we calculate

and, therefore

But -a.e., which means that we must have , and thus .

Now, suppose and are two Lebesgue decompositions of with respect to . Then . We know that both singularity and absolute continuity pass to sums, so is singular with respect to , while is absolutely continuous with respect to . But the only way for this to happen is for them both to be zero, and thus and .

## Corollaries of the Chain Rule

Today we’ll look at a couple corollaries of the Radon-Nikodym chain rule.

First up, we have an analogue of the change of variables formula, which was closely tied to the chain rule in the first place. If and are totally -finite signed measures with , and if is a finite-valued -integrable function, then

which further justifies the the substitution of one “differential measure” for another.

So, define a signed measure as the indefinite integral of . Immediately we know that is totally -finite and that . And, obviously, is the Radon-Nikodym derivative of with respect to . Thus we can invoke the above chain rule to conclude that -a.e. we have

We then know that for every measurable

and the substitution formula follows by putting in for .

Secondly, if and are totally -finite signed measures so that — that is, and — then -a.e. we have

Indeed, , and by definition we have

so serves as the Radon-Nikodym derivative of with respect to itself. Putting this into the chain rule immediately gives us the desired result.

## The Radon-Nikodym Chain Rule

Today we take the Radon-Nikodym derivative and prove that it satisfies an analogue of the chain rule.

If , , and are totally -finite signed measures so that and , then -a.e. we have

By the linearity we showed last time, if this holds for the upper and lower variations of then it holds for itself, and so we may assume that is also a measure. We can further simplify by using Hahn decompositions with respect to both and , passing to subspaces on which each of our signed measures has a constant sign. We will from here on assume that and are (positive) measures, and the case where one (or the other, or both) has a constant negative sign has a similar proof.

Let’s also simplify things by writing

Since and are both non-negative there is also no loss of generality in assuming that and are everywhere non-negative.

So, let be an increasing sequence of non-negative simple functions converging pointwise to . Then monotone convergence tells us that

for every measurable . For every measurable set we find that

and so for all the simple we conclude that

Passing to the limit, we find that

and so the product serves as the Radon-Nikodym derivative of in terms of , and it’s uniquely defined -almost everywhere.

## The Radon-Nikodym Derivative

Okay, so the Radon-Nikodym theorem and its analogue for signed measures tell us that if we have two -finite signed measures and with , then there’s some function so that

But we also know that by definition

If both of these integrals were taken with respect to the same measure, we would know that the equality

for all measurable implies that -almost everywhere. The same thing can’t quite be said here, but it motivates us to say that in some sense we have equality of “differential measures” . In and of itself this doesn’t really make sense, but we define the symbol

and call it the “Radon-Nikodym derivative” of by . Now we can write

The left equality is the Radon-Nikodym theorem, and the right equality is just the substitution of the new symbol for . Of course, this function — and the symbol — is only defined uniquely -almost everywhere.

The notation and name is obviously suggestive of differentiation, and indeed the usual laws of derivatives hold. We’ll start today by the easy property of linearity.

That is, if and are both -finite signed measures, and if and , then is clearly another -finite signed measure. Further, it’s not hard to see if then as well. By the Radon-Nikodym theorem we have functions and so that

for all measurable sets . Then it’s clear that

That is, can serve as the Radon-Nikodym derivative of with respect to . We can also write this in our suggestive notation as

which equation holds -almost everywhere.

## The Radon-Nikodym Theorem for Signed Measures

Now that we’ve proven the Radon-Nikodym theorem, we can extend it to the case where is a -finite signed measures.

Indeed, let be a Hahn decomposition for . We find that is a -finite measure on , while is a -finite measure on .

As it turns out that on , while on . For the first case, let be a set for which . Since , we must have , and so . Then by absolute continuity, we conclude that , and thus on . The proof that on is similar.

So now we can use the Radon-Nikodym theorem to show that there must be functions on and on so that

We define a function on all of by for and for . Then we can calculate

which in exactly the conclusion of the Radon-Nikodym theorem for the signed measure .

## The Radon-Nikodym Theorem (Proof)

Today we set about the proof of the Radon-Nikodym theorem. We assumed that is a -finite measure space, and that is a -finite signed measure. Thus we can write as the countable union of subsets on which both and are finite, and so without loss of generality we may as well assume that they’re finite to begin with.

Now, if we assume for the moment that we’re correct and an does exist so that is its indefinite integral, then the fact that is finite means that is integrable, and then if is any other such function we can calculate

for every measurable . Now we know that this implies a.e., and thus the uniqueness condition we asserted will hold.

Back to the general case, we know that the absolute continuity is equivalent to the conjunction of and , and so we can reduce to the case where is a finite measure, not just a finite signed measure.

Now we define the collection of all nonnegative functions which are integrable with respect to , and for which we have

for every measurable . We define

Since is the supremum, we can find a sequence of functions in so that

For each we define

Now if is some measurable set we can break it into the finite disjoint union of sets so that on . Thus we have

and so .

We can write , which tells us that the sequence is increasing. We define to be the limit of the — is the maximum of all the — and use the monotone convergence theorem to tell us that

Since all of the integrals on the right are bounded above by , their limit is as well, and . Further, we can tell that the integral of over all of must be . Since is integrable, it must be equal -a.e. to some finite-valued function . What we must now show is that if we define

then is identically zero.

If it’s *not* identically zero, then by the lemma from yesterday there is a positive number and a set so that and so that

for every measurable set . If we define , then

for every measurable set , which means that . But

which contradicts the maximality of the integral of . Thus must be identically zero, and the proof is complete.

## The Radon-Nikodym Theorem (Statement)

Before the main business, a preliminary lemma: if and are totally finite measures so that is absolutely continuous with respect to , and is not identically zero, then there is a positive number and a measurable set so that and is a positive set for the signed measure . That is, we can subtract off a little bit (but not zero!) of from and still find a non--negligible set on which what remains is completely positive.

To show this, let be a Hahn decomposition with respect to the signed measure for each positive integer . Let be the union of all the and let be the intersection of all the . Then since and is negative for we find

for every positive integer . This shows that we must have . And then, since is not identically zero we must have . By absolute continuity we conclude that , which means that we must have for at least one value of . So we pick just such a value, set , and , and everything we asserted is true.

Now for the Radon-Nikodym Theorem: we let be a totally -finite measure space and let be a -finite signed measure on which is absolutely continuous with respect to . Then is an indefinite integral. That is, there is a finite-valued measurable function so that

for every measurable set . The function is unique in the sense that if any other function has as its indefinite integral, then -almost everywhere. It should be noted that we don’t assert that is integrable, which will only be true of is actually finite. However, either its positive or its negative integral must converge or we wouldn’t use the integral sign for a divergent integral.

Let’s take a moment and consider what this means. We know that if we take an integrable function , or a function whose integral diverges definitely, on a -finite measure space and define its indefinite integral , then is a -finite signed measure that is absolutely continuous with respect to the measure against which we integrate. What the Radon-Nikodym theorem tells us that *any* such signed measure arises as the indefinite integral of some such function . Further, it tells us that such a function is essentially unique, as much as any function is in measure theory land. In particular, we can tell that if we start with a function and get its indefinite integral, then any other function with the same indefinite integral must be a.e. equal to .

## Singularity

Another relation between signed measures besides absolute continuity — indeed, in a sense the opposite of absolute continuity — is singularity. We say that two signed measures and are “mutually singular” and write if there exists a partition of into two sets so that for every measurable set the intersections and are measurable, and

We sometimes just say that and are singular, or that (despite the symmetry of the definition) “ is singular with respect to “, or vice versa.

In a manner of speaking, if and are mutually singular then all of the sets that give a nonzero value are contained in , while all of the sets that give a nonzero value are contained in , and the two never touch. In contradistinction to absolute continuity, not only does the vanishing of not imply the vanishing of , but if we pare away portions of a set for which gives zero measure then what remains — essentially the only sets for which doesn’t automatically vanish — is necessarily a set for which *does* vanish. Another way to see this is to notice that if and are signed measures with both and , then we must necessarily have ; singularity says that must vanish on any set with , and absolute continuity says must vanish on any set with .

As a quick and easy example, let and be the Jordan decomposition of a signed measure . Then a Hahn decomposition for gives exactly such a partition showing that .

One interesting thing is that singular measures can be added. That is, if and are both singular with respect to , then . Indeed, let and be decompositions showing that and , respectively. That is, for any measurable set we have

Then we can write

It’s easy to check that must vanish on measurable subsets of , and that must vanish on measurable subsets of the remainder of .