There is a remarkably nice proof of the Lebesgue decomposition theorem (described below) by von Neumann. This leads immediately to the Radon-Nikodym theorem.
If and are two finite measures on then there exists a non-negative (w.r.t. both measures) measurable function and a -null set such that
for each .
Let and consider the operator
It is obvious that the operator is linear and moreover for any we have
so that is a linear functional on . By the Reisz representation theorem for Hilbert spaces there exists a such that
Now consider the following sets;
First by (*)
which gives that .
Next we have that
so that .
For the last set consider , then by rearranging (*) we have,
Let then by applying monotone convergence and recalling that we have,
Thus putting this all together, for any we have
There is a rather obvious extension of this to -finite measures but the theorem does not hold for infinite measures.
We say that a measure is absolutely continuous with respect to (written ) if whenever then . The Radon-Nikodym theorem deals with this. Essentially it says that we has a density with respect to . For instance many people know the p.d.f. of a Gaussian distribution but the reason that the p.d.f. exists in because the Gaussian measure is absolutely continuous with respect to the Lebesgue measure.
A (sigma) finite measures is absolutely continuous with respect to an other (sigma) finite measure if and only if there exists a measurable function such that
Clearly if two measures are related by the given formula then .
The converse follows from the theorem above as .