proof of Radon-Nikodym theorem

The following proof of Radon-Nikodym theorem is based on the original argument by John von Neumann. We suppose that $\mu$ and $\nu$ are real, nonnegative, and finite. The extension to the $\sigma$-finite case is a standard exercise, as is $\mu$-a.e. uniqueness of Radon-Nikodym derivative. Having done this, the thesis also holds for signed and complex-valued measures.

Let $(X,ℱ)$ be a measurable space and let $\mu ,\nu :ℱ\to [0,R]$ two finite measures on $X$ such that $\nu (A)=0$ for every $A\in ℱ$ such that $\mu (A)=0$. Then $\sigma =\mu +\nu$ is a finite measure on $X$ such that $\sigma (A)=0$ if and only if $\mu (A)=0$.

Consider the linear functional $T:L^2(X,ℱ,\sigma )\to ℝ$ defined by

$$Tu={\int }_{X}u𝑑\mu \forall u\in L^2(X,ℱ,\sigma ).$$ (1)

$T$ is well-defined because $\mu$ is finite and dominated by $\sigma$, so that $L^2(X,ℱ,\sigma )\subseteq L^2(X,ℱ,\mu )\subseteq L^1(X,ℱ,\mu );$ it is also linear and bounded because $|Tu|\le {\parallel u\parallel }_{L^2(X,ℱ,\sigma )}\cdot \sqrt{\sigma (X)}.$ By Riesz representation theorem, there exists $g\in L^2(X,ℱ,\sigma )$ such that

$$Tu={\int }_{X}u𝑑\mu ={\int }_{X}u\cdot g𝑑\sigma$$ (2)

for every $u\in L^2(X,ℱ,\sigma )$. Then $\mu (A)={\int }_{A}g𝑑\sigma$ for every $A\in ℱ$, so that $\mu$- and $\sigma$-a.e. (Consider the former with $A=\{x\mid g(x)\le 0\}$ or $A=\{x\mid g(x)>1\}$.) Moreover, the second equality in (LABEL:eq:q) holds when $u={\chi }_A$ for $A\in ℱ$, thus also when $u$ is a simple measurable function by linearity of integral, and finally when $u$ is a ($\mu$- and $\sigma$-a.e.) nonnegative $ℱ$-measurable function because of the monotone convergence theorem.

Now, $1/g$ is $ℱ$-measurable and nonnegative $\mu$- and $\sigma$-a.e.; moreover, ${\displaystyle \frac{1}{g}}\cdot g=1$ $\sigma$- and $\mu$-a.e. Thus, for every $A\in ℱ$,

$${\int }_A\frac{1}{g}𝑑\mu ={\int }_A𝑑\sigma =\sigma (A)$$ (3)

Since $\sigma$ is finite, $1/g\in L^1(X,ℱ,\mu )$, and so is $f={\displaystyle \frac{1}{g}}-1$. Then for every $A\in ℱ$

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

Title	proof of Radon-Nikodym theorem
Canonical name	ProofOfRadonNikodymTheorem
Date of creation	2013-03-22 18:58:03
Last modified on	2013-03-22 18:58:03
Owner	Ziosilvio (18733)
Last modified by	Ziosilvio (18733)
Numerical id	5
Author	Ziosilvio (18733)
Entry type	Proof
Classification	msc 28A15
Synonym	Hilbert spaces proof of Radon-Nikodym’s theorem
Synonym	measure- theoretic proof of Radon-Nikodym theorem