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.