There is a remarkably nice proof of the Lebesgue decomposition theorem (described below) by von Neumann. This leads immediately to the Radon-Nikodym theorem.

**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 .*

**Proof:**

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,

in particular

.

Let then by applying monotone convergence and recalling that we have,

.

Thus putting this all together, for any we have

as claimed.

Q.E.D.

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.

**Corollary:**

*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*

*.*

**Proof:**

Clearly if two measures are related by the given formula then .

The converse follows from the theorem above as .

Q.E.D.

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## 3 comments

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02/05/2011 at 19:38

KostasIf only you posted something about random partitions…

07/11/2011 at 08:34

beni22sofVery nice proof.

16/05/2013 at 13:16

F. DragosWow, this is a very nice and easy to understand proof. I think you should include it in “For Students” category also.