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
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.
After all of yesterday’s definitions, we continue examining the countably infinite product of a sequence of totally finite measure spaces such that each .
If and are positive integers with , it may happen that a non-empty subset is both a -cylinder and a -cylinder. By yesterday’s result, we find that we can write both and for some subsets and . But we can rewrite the first of these equations as , and thus we conclude that .
If is measurable, then both and are measurable. We calculate
using the normalization of all the measures . It follows that we can define a set function unambiguously on each measurable -cylinder by the equation
Since every measurable rectangle is the product of a sequence with all but finitely many , we can choose a large enough so that for all . Then is a -cylinder, and is defined on all measurable rectangles. It’s straightforward to see that is finite, non-negative, and finitely additive where it’s defined.
We define the analogue of in by . If is defined on a cylinder then will be defined on each section , and we find that
So now, if is a sequence of totally finite measure spaces with for all , then there exists a unique measure defined on the countably infinite product of the measure spaces so that for every measurable of the form we have
This measure is called the product of the measures .
From what we know about continuity, we just have to show that is continuous from above at to show that it’s a measure. That is, if is a decreasing sequence of cylinders on which is defined so that for all , then the countable intersection of the is non-empty.
For each we define the set
We then find that
and so . Therefore, .
Now is a decreasing sequence of measurable subsets of , which is bounded in measure away from zero. And so by the continuity of the measure , we conclude that there is at least one point in their intersection. That is, for all .
But now everything we’ve said about is true as well for . We can thus replace the sequence with the sequence , and the bound with the bound . We then find a point so that for all . We can repeat this process to find a sequence with , such that
for all .
I say that this sequence belongs to all the . Indeed, given any of them, choose the so that is a -cylinder. We know that , and so there must be at least one point with for all from to . But then, since is a -cylinder, it must contain the point .
Thus, as asserted, the intersection of the is nonempty, and so is continuous from above at , and is thus a measure.
Because we know that product spaces are product objects in the category of measurable spaces — at least for totally measurable spaces — we know that the product functor is monoidal. That is, we can define -ary products unambiguously as iterated binary products. But things start to get more complicated as we pass to infinite products.
If is a countably infinite collection of sets, the product is the collection of all sequences with for all . If each is equipped with a -ring and a measure , it’s not immediately clear how we should equip the product space with a -ring and a measure. However, we can give meaning to these concepts if we specialize. First we shall insist that each be a -algebra, and second we shall insist that be totally finite. Not just totally finite, though; we will normalize each measure so that .
This normalization is, incidentally, always possible for a totally finite measure space. Indeed, if is a totally finite measure on a space , we can define
It is easily verified that is another totally finite measure, and .
Now we will define a rectangle as a product of the form
where for all , and where for all but finitely many . We define a measurable rectangle to be one for which each is measurable as a subset of . We then define a subset of the countably infinite product to be measurable if it’s in the -algebra generated by the measurable rectangles. This defines the product of a countably infinite number of measurable spaces.
Now if is any set of positive integers, we say that two sequences and agree on if for all . A set is called a -cylinder if any two points which agree on are either both in or both out of . That is, membership of a sequence in is determined only by the coordinates for .
We also define the sets
so that we always have Each is itself a countably infinite product space. For every set and each point , we define the subset as the section of determined by . It should be clear that every section of a measurable rectangle in is a measurable rectangle in
Now if , and if E\subseteq X$ is a (measurable) -cylinder, then , where is a (measurable) subset of . Indeed, let be an arbitrary point of , and let be the -section of determined by this point. The sets (by assumption) and (by construction) are both -cylinders, so if belongs to either of them, then so does the point .
It should now be clear that if such a point belongs to either or , then it belongs to the other as well. Again using the fact that both and are -cylinders, if belongs to either or then so does the point . We can conclude that and consist of the same points. The parenthetical assertion on measurability follows from the fact that every section of a measurable set is measurable.
We continue our assumptions that and are both -finite measure spaces, and we consider the product space .
The first step towards the measure-theoretic version of Fubini’s theorem is a characterization of sets of measure zero. Given a subset , a necessary and sufficient condition for to have measure zero is that the -section have for almost all . Another one is that the -section have for almost all . Indeed, the definition of the product measure tells us that
Since the function is integrable and nonnegative, our condition for an integral to vanish says that the integral is zero if and only if -almost everywhere. Similarly, we see that the integral of is zero if and only if -almost everywhere.
Now if is a non-negative measurable function on , then we have the following equalities between the double integral and the two iterated integrals:
If is the characteristic function of a measurable set , then we find that
Next we assume that is a simple function. Then is a finite linear combination of characteristic functions of measurable sets. But clearly all parts of the asserted equalities are linear in the function , and so since they hold for characteristic functions of measurable sets they must hold for any simple function as well.
Finally, given any non-negative measurable function , we can find an increasing sequence of simple functions converging pointwise to . The monotone convergence theorem tells us that
We define the functions
and conclude that since is an increasing sequence, must me an increasing sequence of non-negative measurable functions as well. For every the monotone convergence theorem tells us that
As a limit of a sequence of non-negative measurable functions, must also be a non-negative measurable function. One last invocation of the monotone convergence theorem tells us that
which proves the equality of the double integral and one of the iterated integrals. The other equality follows similarly.
And now we come to Fubini’s theorem itself: if is an integrable function on , then almost every section of is integrable. If we define the functions
wherever these symbols are defined, then and are both integrable, and
Since a real-valued function is integrable if and only if both its positive and negative parts are, it suffices to consider non-negative functions . The latter equalities follow, then, from the above discussion. Since the measurable functions and have finite integrals, they must be integrable. And since they’re integrable, they must be finite-valued a.e., which implies the assertions about the integrability of sections of .
Let and be two -finite measure spaces, and let be the product measure on .
If is a function on so that its integral is defined — either is integrable, or its integral diverges definitely — then we write it as any of
and call it the “double integral” of over . We can also consider the sections and . For any given , we set
if the integral exists. If the resulting function is integrable, then we write
The latter notation, with the measure before the integrand is less common, but it can be seen in older texts. I’ll usually stick to the other order.
Similarly, we can define the function as the integral of the -section if it exists. If is integrable, we write
where, again, the latter notation is deprecated. These integrals are called the “iterated integrals” of . We can also define double and iterated integrals over a measurable subset , as usual, and write
If is a -algebra and is a Borel-measurable function we defined the upper and lower ordinate sets and to be measurable subsets of . Now if we have a measure on and Lebesgue measure on the Borel sets, we can define the product measure on . Since we know and are both measurable, we can investigate their measures. I assert that
It will be sufficient to establish this for simple functions, since for either the upper or the lower ordinate set we can approximate any measurable by a monotone sequence of simple so that or . Then the limit will commute with (since measures are continuous), and it will commute with the integral as well.
So, we can assume that is simple, and write
with the a pairwise-disjoint collection of measurable sets. But now if the equality holds for each of the summands then it holds for the whole function. That is, we can assume — without loss of generality — that for some real number and some measurable subset .
And now the result should be obvious! Indeed, is the measurable rectangle , while is the measurable rectangle . Since the product measure on a measurable rectangle is the product of the measures of the two sides, these both have measure . On the other hand, we calculate the integral as
and so the equality holds for such rectangles, and thus for simple functions, and thus for all measurable functions.
Of course, it should now be clear that the graph of has measure zero. Indeed, we find that
These results put a precise definition to the concept of the integral as the “area under the graph”, which was the motivation behind our definition of the Riemann integral, way back when we introduced it.
Last time we too a measurable set and defined the functions and . We also showed that
That is, for every measurable we can define the real number
I say that this function is itself a -finite measure, and that for any measurable rectangle we have . Since measurable rectangles generate the -ring , this latter condition specifies uniquely.
To see that is a measure, we must show that it is countably additive. If is a sequence of disjoint sets then we calculate
where we have used the monotone convergence theorem to exchange the sum and the integral.
We verify the -finiteness of by covering each measurable set by countably many measurable rectangles with finite-measure sides. Since the sides’ measures are finite, the measure of the rectangle itself is the product of two finite numbers, and is thus finite.
We call the measure the “product” of the measures and , and we write . We thus have a -finite measure space that we call the “cartesian product” of the spaces and .
After considering the product of measurable spaces, let’s consider what happens if our spaces are actually equipped with measures. That is, let and be -finite measure spaces, and consider the product space .
Now, if is any measurable subset of , then we can define two functions — and — by the formulæ and , where and are the –section determined by and -section determined by , respectively. I say that both and are non-negative measurable functions, and that
where we take the first integral over and the second over .
Let be the collection of subsets of for which the assertions we’ve made are true. It’s straightforward to see that is closed under countable disjoint unions. Indeed, if is a sequence of disjoint sets for which the assertions hold — call the functions and respectively — then if is the disjoint union of the we can calculate
Since each of the are measurable, so is , and is similarly measurable. The equality of the integrals should be clear.
Now, since and are -finite, every measurable subset of can be covered by a countable disjoint union of measurable rectangles, each side of each of which has finite measure. Thus we just have to verify that the result holds for such measurable rectangles with finite-measure sides, and it will hold for all measurable subsets of , and that is a monotone class. That is a monotone class follows from the dominated convergence theorem and the monotone convergence theorem, and so we have only to show that the assertions hold for measurable rectangles with finite-measure sides.
Okay, so let be a measurable rectangle with and . We can simplify our functions to write
These are clearly measurable, and we find that
so the assertions hold for measurable rectangles with finite-measure sides, and thus for all measurable subsets of .
Yesterday, we showed that the graph of a non-negative measurable function is measurable. Today we’ll explore this further. We continue all the same notation as we used yesterday: is a measurable space and is the measurable space of real numbers and Borel sets.
First, if is any measurable subset, and if and are real numbers, then the set is also measurable. The affine transformation sends any measurable set in to another such set, so if is a measurable rectangle, then the transformed set will also be a measurable rectangle. It’s also straightforward to show that the transformation commutes with setwise unions and intersections, and that the assertion is true for . Thus the assertion holds on some -ring , which contains all measurable rectangles. Since measurable rectangles generate the -algebra of all measurable sets on , must contain all measurable sets, and thus the assertion holds for all measurable .
Now — as a partial converse to yesterday’s final result — if is a non-negative function so that (or ) is a measurable set, then is a measurable function. We will show this using an equivalent definition of measurability — that will be measurable if we can show that for every real number the set is measurable. This is clearly true for every nonpositive , and so we must show it for the positive .
For each positive integer we define the set
Our above lemma shows that the first set in the intersection is measurable — as a transformation of , which we assumed to be measurable — and the second set is a measurable rectangle. Thus the intersection is measurable. We take the union of these sets for all :
This is the union of a sequence of measurable sets, and so it is measurable. Taking any we find the -section determined by is the set . And this is thus measurable, since all sections of measurable sets are measurable.
This result is the basis of an alternative characterization of measurable functions. We could have defined a non-negative function to be measurable if its upper (or lower) ordinate set ( is measurable, and extended to general functions by insisting that this hold for both positive and negative parts.
Finally, we can extend our result from last time. If is any measurable function, then its graph is measurable. Indeed, we can take the positive and negative parts and , which are both measurable. Thus all four sets , , , and are measurable. Choosing and we reflect and to and . We then form the unions
The difference between these two sets is the graph of , which is thus measurable.
Let be a measurable space so that itself is measurable — that is, so that is a -algebra — and let be the real line with the -algebra of Borel sets.
If is a real-valued, non-negative function on , then we define the “upper ordinate set” to be the subset such that
We also define the “lower ordinate set” to be the subset such that
We will explore some basic properties of these sets.
First, if is the characteristic function of a measurable subset , then is the measurable rectangle , while is the measurable rectangle .
Next, if is a non-negative simple function, then we can write it as the linear combination of characteristic functions of disjoint subsets. If , then for each the upper ordinate set is the measurable rectangle while the lower ordinate set is the measurable rectangle . Since the are all disjoint, the upper ordinate set is the disjoint union of all the , and similarly for the lower ordinate sets. Thus the upper and lower ordinate sets of a simple function are both measurable.
Next we have some monotonicity properties: if and are non-negative functions so that for all , then and . Indeed, if then , so as well, and similarly for the lower ordinate sets.
If is an increasing sequence of non-negative functions converging pointwise to a function , then is an increasing sequence of sets whose union is . That is increasing is clear from the above monotonicity property, and just as clearly they’re all contained in . On the other hand, if , then But since increases to , this means that for some , and so . Thus is contained in the union of the . Similarly, if is decreasing to , then decreases to .
Finally (for now), if is a non-negative measurable function, then and are both measurable. The lower ordinate set is easier, since we know that we can pick an increasing sequence of non-negative measurable simple functions converging pointwise to . Their lower ordinate sets are all measurable, and they form an increasing sequence of measurable sets whose union is . Since this is a countable union of measurable sets, it must be itself measurable.
For we have to be a little trickier. First, if is bounded above by , then is non-negative and also bounded above by , and we can find an increasing sequence of non-negative measurable simple functions converging pointwise to . Then is a decreasing sequence of non-negative simple functions converging pointwise to . The catch is that the measurability of a simple function only asks that all the nonzero sets on which it is defined be measurable. That is, in principle the zero set of may be non-measurable. However, the zero set of is the complement of , and since this set is measurable we can use the assumed measurability of to see that is measurable as well. And so we see that is measurable as well. Thus is a decreasing sequence of measurable sets, converging to , which must thus be measurable.
Now, for a general , we can consider the sequence which replaces any value with . Each of these functions is still measurable (again using the measurability of ), and is now bounded. Thus is an increasing sequence of measurable sets, and I say that now their union is . Indeed, each is contained in , so the union must be. On the other hand, if , then . But since , there is some so that . Thus , and so , and is in the union as well. Since is the union of a countable sequence of measurable sets, it is itself measurable.
Incidentally, this implies that if is a non-negative measurable function, then the difference is measurable. But we can calculate this difference as
That is, is exactly the graph of the function , and so we see that the graph of a non-negative measurable function is measurable.