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.