If is a signed measure then we know that the total variation is a measure. It then makes sense to discuss whether or not a measurable function is integrable with respect to . In this case, will be integrable with respect to both and . Indeed, since this is obviously true for simple , and general integrable functions are limits of simple integrable functions.
This being the case, we can define both integrals
and, since , it makes sense to define
This integral shares some properties with “positive” integrals. For instance, it’s clearly linear:
Unfortunately, it doesn’t play well with order. Indeed, if is a measurable -negative set, then everywhere, but
This throws off most of our basic properties. However, some can be salvaged. It’s no longer necessary that a.e. for the integral to be zero, but it’s sufficient. And, thus, if a.e. then their integrals are equal, although the converse doesn’t hold.
One interesting fact is that for every measurable set we find
where we take the supremum over all measurable functions with everywhere. Indeed, if we take a Hahn decomposition for , then since is measurable so are and . If we take and , then we find
Thus we can actually attain this value. Can we get any larger? No. We can’t achieve anything by adding to the value of outside , since the integral is only taken over anyway. And within we could only increase the positive component of the integral by increasing the value of in , or increase the negative component by decreasing the value of in . Either way, we’d make some , which isn’t allows. Thus the total variation over is indeed this supremum.
Now, as we’ve pointed out, this will be a measure so long as is a.e. non-negative. But now if is any integrable function at all, the indefinite integral is a finite signed measure.
Let’s give a Hahn decomposition corresponding to . I say that if we set
then is positive, is negative, and is a Hahn decomposition. Indeed, we know that and are measurable. Thus if is measurable, then is measurable, and we find
since everywhere. Similarly, we verify that is negative.
Now we can use this to find the Jordan decomposition of . We define
That is, the upper variation of is the indefinite integral of the positive part of , while the lower variation of is the indefinite integral of the negative part of . And then we can calculate
The total variation of is the indefinite integral of the absolute value of .
Today we consider what happens when we’re working over a -algebra — so the whole space is measurable — and we restrict our attention to totally finite signed measures. These form a vector space, since the sum of two finite signed measures is again a finite signed measure, as is any scalar multiple (positive or negative) of a finite signed measure.
Now, it so happens that we can define a norm on this space. Indeed, taking the Jordan decomposition, we must have both and , and thus . We define , and use this as our norm. It’s straightforward to verify that , and that implies that is the zero measure. The triangle inequality takes a bit more work. We take a Hahn decomposition for and write
So we know that this defines a norm on our space.
But is this space, as asserted, a Banach space? Well, let’s say that is a Cauchy sequence of finite signed measures so that given any we have for all sufficiently large and . But this is larger than any , which itself is greater than . If is a positive measurable set then this shows that is kept small, and we find similar control over the measures of negative measurable sets. And so the sequence is always Cauchy, and hence convergent. It’s straightforward to show that the limiting set function will be a signed measure, and that we will have control over . And so the space of totally finite signed measures is indeed a Banach space.
It’s not too hard to construct examples showing that Hahn decompositions for a signed measure, though they exist, are not unique. But if we have two of them — — there’s something we can show to be unique. For every measurable set we have and .
Indeed, it’s easy to see that , so (since is positive) . But we can also see that , so (since is negative) . And so , and similarly . We then see that
The proof that is similar.
We can thus unambiguously define two set functions on the class of all measurable sets
for any Hahn decomposition . We call these the “upper variation” and “lower variation”, respectively, of . We can also define a set function
called the “total variation” of . It should be noted that and have almost nothing to do with each other.
Now, I say that each of these set functions — , , and — is a measure, and that . If is (totally) finite or -finite, then so are and , and at least one of them will always be finite.
Each of these variations is clearly non-negative, and countable additivity is also clear from the definitions. For example, given a pairwise-disjoint sequence we find
and similarly for and . Thus each one is a measure. The equation is clear from the definitions. The fact that takes at most one of implies that one of is finite. Finally, if every measurable set (say, ) is a countable union of -finite sets (say, ), then is the countable union of the , and so as well, and similarly for . Thus and are -finite.
We see that every signed measure can be written as the difference of two measures, one of which is finite. The representation of a signed measure as the difference between its upper and lower variations is called the “Jordan decomposition” of .
Given a signed measure on a measurable space , we can use it to break up the space into two pieces. One of them will contribute positive measure, while the other will contribute negative measure. First: some preliminary definitions.
We call a set “positive” (with respect to ) if for every measurable the intersection is measurable, and . That is, it’s not just that has positive measure, but every measurable part of has positive measure. Similarly, we say that is “negative” if for every measurable the intersection is measurable, and . For example, the empty set is both positive and negative. It should be clear from these definitions that the difference of two negative sets is negative, and any disjoint countable union of negative sets is negative, and (thus) any countable union at all of negative sets is negative.
Now, for every signed measure there is a “Hahn decomposition” of . That is, there are two disjoint sets and , with positive and negative with respect to , and whose union is all of . We’ll assume that , but if takes the value (and not ) the modifications aren’t difficult.
We write , taking the infimum over all measurable negative sets . We must be able to find a sequence of measurable negative sets so that the limit of the is — just pick so that — and we can pick the sequence to be monotonic, with . If we define as the union — the limit — of this sequence, then we must have . The measurable negative set has minimal measure .
Now we pick , and we must show that is positive. If it wasn’t, there would be a measurable subset with . This cannot itself be negative, or else would be negative and we’d have , contradicting the minimality of .
So must contain some subsets of positive measure. We let be the smallest positive integer so that contains a subset with . Then observe that
So everything we just said about holds as well for . We let be the smallest positive integer so that contains a subset with . And so on we go until in the limit we’re left with
after taking out all the sets .
Since , the measure of is finite, and so the measure of any subset of must be finite as well. Thus the limits of the must be zero, so that the measure of the countable disjoint union of all the can converge. And so any remaining measurable set that can fit into must have . That is, must be a measurable negative set disjoint from . But we must have
which contradicts the minimality of just like would have if it had been a negative set. And thus the assumption that is untenable, and so every measurable subset of has positive measure.
We have a couple results about signed measures and certain sequences of sets.
If is a signed measure and is a disjoint sequence of measurable sets so that the measure of their disjoint union is finite:
then the series
is absolutely convergent. We already know it converges since the measure of the union is finite, but absolute convergence will give us all sorts of flexibility to reassociate and rearrange our series.
We want to separate out the positive and the negative terms in this series. We write if and otherwise. Similarly, we write if and otherwise. Then we write the two series
The terms of each series have a constant sign — positive for the first and negative for the second — and so if they diverge they can only diverge definitely — to in the first case and to in the second. But at least one must converge or else we’d have obtaining both infinite values. But the sum of all the converges, and so both series must converge — if the series of diverged to and the seris of converged, then there wouldn’t be enough negative terms in the series of for the whole thing to converge. But then since the positive terms and the negative terms both converge, the whole series is absolutely convergent.
Now we turn to some continuity properties. If is a monotone sequence — if it’s decreasing we also ask that at least one — then
The proofs of both of these facts are exactly the same as for measures, except we need the monotonicity result from the end of yesterday’s post to be sure that once we hit one finite , all the later will stay finite.
And so we introduce a “signed measure”. This is essentially just like a measure, except we allow negative values as well. That is, is an extended real-valued, countably additive set function. But we want to prune the concept slightly.
First off, we insist that . Additivity tells us that ; if there is any measurable set at all with finite, then follows, so our condition just rules out the degenerate cases where or for all measurable .
Secondly, we insist that can take only one of the values . That is, we can’t have one measurable with and another measurable with . Indeed, if this were the case then we’d have to deal with some indeterminate sums. We can’t quite be sure of this just considering since the two might not be disjoint, but we can consider the following three equations that follow from additivity:
Either is finite or it’s not. If it’s finite, then the first two equations tell us that and , and so the sum in the third equation is indeterminate. On the other hand, if then the sum in the second equation must be indeterminate to satisfy , and similarly the sum in the first equation would have to be indeterminate if . To avoid these indeterminate sums we make our restriction.
On the other hand, there are certain pathologies we don’t have to worry about. For instance, if is a pairwise disjoint sequence of measurable sets, then countable additivity tells us that
and so the sum either converges (if the measure of the union is finite) or it definitely diverges to . That is, even though we may have negative terms we don’t have to worry about an oscillating sum that fails to converge because the sequence of partial sums jumps around and never settles down. So it always makes sense to write such a sum down, even if its value may be infinite.
All the language about a measure being finite, -finite, totally finite, or totally -finite carries over. The only modification is that we have to ask for or (equivalently) instead of in the definitions.
Of course, just as for measures, signed measures are finitely additive (which we used above) and thus subtractive. It won’t be monotone, in general, though; Given a set and a subset we can write
If then , even though is the subset. However, we can at least say that if then as well. Indeed, using the same equation if either one of the summands on the right is infinite then is as well. If both are, then they’re both infinite in the same direction (since can only assume one of ) and so is again infinite. The only possibility for a finite is for both and to be finite.
Given an integrable function , we’ve defined the indefinite integral to be the set function
This is clearly real-valued, and we’ve seen that it’s countably additive. If is a.e. non-negative, then will also be non-negative, and so the indefinite integral is a measure. Since is integrable we see that
and so is a totally finite measure.
But this situation feels a bit artificially restrictive in a couple ways. First of all, measures can be extended real-valued — why do we never find ? Well, it makes sense to extend the definition of at least the symbol of integration a bit. If is not integrable, but a.e., there is really only one possibility: there is no upper bound on the integrals of simple functions smaller than . And so in this situation it makes sense to define
Similarly, if a.e. and fails to be integrable, it makes sense to define
In general, we can break a function into its positive and negative parts and , and then define
for all functions for which at most one of and fails to be integrable. That is, if the positive part is integrable but the negative part is not, then the integral can be defined to be . If the negative part is integrable but the positive part isn’t, we can define the integral to be . If both positive and negative parts are integrable then the whole function is integrable, while if neither part is integrable we still leave the integral undefined. We don’t know in general how to deal with the indeterminate form .
And so now we find that any a.e. non-negative function — integrable or not — defines a measure by its indefinite integral. If isn’t integrable, then we get an extended real-valued set function, but this doesn’t prevent it from being a measure. As a matter of terminology, we should point out that we don’t call a function whose integral is now defined to be positive or negative to be “integrable”. That term is still reserved for those functions whose indefinite integrals are totally finite, as above.
We can wrap up this introduction to the Lebesgue integral by outlining the alternate approach that commenter Cristi was referring to. We’ll do this from the perspective of our current track, and it should be clear how the alternative definitions would lead us to the same place.
If is a non-negative integrable function on a measure space . For every measurable set , we define
Also, for every finite, pairwise disjoint collection of measurable sets we define
We then assert that the supremum of all numbers for all finite, pairwise disjoint collections is equal to the integral of :
If is simple, this is obvious. Indeed, if is the collection of sets used to write as a finite linear combination of characteristic functions, then is exactly the integral of by definition. Any set that extends outside one of these sets will have , and so we can’t get any larger than the integral of .
On the other hand, for a general integrable function we consider a non-negative simple with , and we let be the sets used to express as a finite linear combination of characteristic functions:
We see that
If is an increasing sequence of non-negative simple functions converging pointwise a.e. to , then
where we use the definition of integrability, and we take the supremum over finite, pairwise disjoint collections . But it’s also clear that for every we have
for some non-negative simple .
So the alternate approach proceeds by defining the integral of a simple function as before, and defining general integrals of non-negative functions by the supremum above. General integrable functions overall are handled by using their positive and negative parts. Then you can prove the monotone convergence theorem, followed by Fatou’s lemma, and then the Fatou-Lebesgue theorem, which leads to dominated convergence theorem, and we’re pretty much back where we started.
Now we turn to the Fatou-Lebesgue theorem. Let be a sequence of integrable functions (this time we do not assume they are non-negative) and be some other function which dominates this sequence in absolute value. That is, we have a.e. for all . We define the functions
These two functions are integrable, and we have the sequence of inequalities
Again, this is often stated for a sequence of measurable functions, but the dominated convergence theorem allows us to immediately move to the integrable case. In fact, if the sequence converges pointwise a.e., then a.e. and the inequality collapses and gives us exactly the dominated convergence theorem back again.
Since dominates the sequence , the sequence will be non-negative. Fatou’s lemma then tells us that
Cancelling the integral of we find the first asserted inequality. The second one is true by the definition of limits inferior and superior. The third one is essentially the same as the first, only using the non-negative sequence .