Radon-Nikodym theorem

Let $(X,\mathcal{A})$ be a measurable space, $\lambda$ and $\mu$ two measure on this space.

Def A measure $\mu$ algebra is called absolutely continuous w.r.t. another measure $\lambda$ if

$$\lambda (A)=0⟹\mu (A)=0,\forall A\in \mathcal{B}(\mathbb{R})$$

this is written as $\mu <<\lambda$.

Def A measure $\mu$ is called singular if there is a $N\in \mathcal{A}$ with

$$\lambda (N)=0\text{ and }\mu (N^c)=0$$

this is written as $\mu \perp \lambda$.

Usually one works with measures on the reals, so that the measurable space is $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ and $\lambda$ is the Lebesgue measure.

Some examples:

• The zero measure is absolutely continuous w.r.t any measure.
• Every measure is absolutely continuous w.r.t itself.
• The Dirac measure ${\delta }_0$ is singular w.r.t. the Lebesgue measure, since:

$${\delta }_0(\{0{\}}^c)=0\text{ and }\mathcal{L}(\{0\})=0$$

Theorem (Radon-Nikodym and Lebesgue decomposition) Let $\mu :\mathcal{B}\to [0,\infty ]$ any $\sigma$-finite measure on the borel sigma algebra of the reals. Then

1. we can decompose it in two uniquely determined measures

$$\mu ={\mu }_{ac}+{\mu }_s,{\mu }_{ac}<<\mathcal{L},{\mu }_s\perp \mathcal{L}$$

2. There is a measurable map $g:\mathbb{R}\to [0,\infty )$ such that

$${\mu }_{ac}(A)={\int }_{A}gd\mathcal{L},\forall A\in \mathcal{B}(\mathbb{R})$$

$g$ is called the Radon-Nikodym derivative.

Proof (Von-Neumann) First we prove a result for finite measure spaces, and then extend it to $\sigma$-finite spaces.

Let $(X,\mathcal{A})$ be a measurable space, and $\mu ,\nu :\mathcal{A}\to [0,R]$ two finite measure such that $\nu <<\mu$. Then $\sigma :=\mu +\nu$ is also a finite measure on $X$ and $\sigma (A)=0$ iff $\mu (A)=0$.

Consider the linear functional $T:L^2(X,\mathcal{A},\sigma )\to \mathbb{R}$ defined by:

$$T[u]:={\int }_{X}ud\mu$$

$T$ is well defined because $\mu$ is finite and dominated by $\sigma$, so that

$$L^2(\sigma )\subseteq L^2(\mu )\subseteq L^1(\mu )$$

it’s also bounded since

$$|T[u]|\le \Vert \mu {\Vert }_1\le \Vert \mu {\Vert }_2\sqrt{\mu (X)}$$

where we have used the Holder inequality. Then by Riesz representation theorem there exists a $g\in L^2(X,\mathcal{A},\sigma )$ such that

$$T[u]={\int }_{X}ud\mu ={\int }_{u}ugd\sigma ,\forall u\in L^2(\sigma )$$

Then $\mu (A)={\int }_{A}gd\sigma$ for any $A\in \mathcal{A}$, so that $0<g\le 1$ $\mu$ and $\sigma$ a.e. Consider the measurable set $Z:=g^{-1}(0)$, if this has positive $\mu$ measure we get a contradiction. The same for the set $P:=g^{-1}([1,\infty ))$.

The second equality works also with any non-negative measurable functions (using the standard machinery, simple functions and then the Monotone Convergence Theorem]).

Since $\frac{1}{g}$ is non-negative and measurable, then

$${\int }_A\frac{1}{g}d\mu ={\int }_{A}d\sigma =\sigma (A),\forall A\in \mathcal{A}$$

Since $\sigma$ is finite, $\frac{1}{g}\in L^1(\mu )$ and so is $f:=\frac{1}{g}-1$. Then for evert $A\in \mathcal{A}$:

$$\nu (A)=\sigma (A)-\mu (A)={\int }_A\frac{1}{g}d\mu -{\int }_{A}d\mu ={\int }_{A}fd\mu$$

It’s easy to extend this result to $\sigma$-finite measures:

$$\nu (A)=\nu (\bigcup _{n\ge 1}{A}_n)=\sum _{n\ge 1}\nu (A_n)=\sum _{n\ge 1}{\int }_{A_n}{f}_{n}d\mu ={\int }_{A}fd\mu$$

where $f:={\sum }_{n\ge 1}{f}_{n}1(A_n)$ (the limits can be exchanged by MON). $\square$