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 \mu and \nu are two finite measures on (\Omega,\mathcal{F}) then there exists a non-negative (w.r.t. both measures) measurable function f and a \mu-null set B such that

\nu(A)=\int_A f \, d\mu+ \nu(A \cap B)

for each A \in \mathcal{F}.

Proof:

Let \pi:=\mu+\nu and consider the operator

T(f):=\int f\, d\nu.

It is obvious that the operator is linear and moreover for any f \in L^2(\pi) we have

|T(f)| \leq ||f||_{L^2}

so that T is a linear functional on L^2(\pi). By the Reisz representation theorem for Hilbert spaces there exists a h \in L^2(\pi) such that

T(f)=\int f d\nu=\int f h \, d\pi=\int f h \, d\mu+\int f h \, d\nu. (*)

Now consider the following sets;

N:=\{h <0\}, \quad M:=\{0 \leq h < 1\}, \quad B:=\{h \geq 1\}.

First by (*)

0 \geq \int_N h \,d\pi=\int \mathbf{1}_N h\, d\pi=\nu(N)=\int_N h \, d\mu+\int_N h \, d\nu

which gives that \nu(N)=\mu(N)=0.

Next we have that

\nu(B) = T(\mathbf{1}_B) = \int_B h \, d\mu + \int_B h \, d\nu \geq \nu(B) + \mu(B)

so that \mu(B)=0.

For the last set consider M_n:=\{0 \leq h \leq 1-1/n\}, then by rearranging (*) we have,

\int (1-h)f \, d\nu = \int hf \, d\mu in particular

\nu(M_n)=\int \frac{\mathbf{1}_{M_n}}{1-h}(1-h)\, d \nu=\int h \frac{\mathbf{1}_{M_n}}{1-h}\, d\mu.

Let f=\frac{h}{1-h} then by applying monotone convergence and recalling that \mu(B)=\mu(N)=0 we have,

\nu(M \cap A)=\int_A f \, d\mu.

Thus putting this all together, for any A \in \mathcal{F} we have

\nu(A)=\nu(A \cap N) + \nu(A \cap M) + \nu(A \cap B) = \int_A f \, d\mu + \nu(A \cap B)

as claimed.

Q.E.D.

There is a rather obvious extension of this to \sigma-finite measures but the theorem does not hold for infinite measures.

We say that a measure \nu is absolutely continuous with respect to \mu (written \mu >> \nu) if whenever \mu(A)=0 then \nu(A)=0. The Radon-Nikodym theorem deals with this. Essentially it says that we \nu has a density with respect to \mu. 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 \nu is absolutely continuous with respect to an other (sigma) finite measure if and only if there exists a measurable function f such that

\nu(A)=\int_A f\, d\mu.

Proof:

Clearly if two measures are related by the given formula then \nu << \mu.

The converse follows from the theorem above as \nu(A \cap B)=0.

Q.E.D.

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