ProofOfRadonNikodymTheorem 2009-07-02 07:19:47 2009-07-28 15:37:22 Proof Radon-Nikodym theorem 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here 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 $(X,\mathcal{F})$ be a measurable space and let $\mu,\nu:\mathcal{F}\to[0,R]$ two finite measures on $X$ such that $\nu(A)=0$ for every $A\in\mathcal{F}$ 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 \begin{math} T:L^2(X,\mathcal{F},\sigma)\to\mathbb{R} \end{math} defined by \begin{equation} \label{eq:T} Tu=\int_Xu\;d\mu\;\;\forall u\in L^2(X,\mathcal{F},\sigma)\;. \end{equation} $T$ is well-defined because $\mu$ is finite and dominated by $\sigma$, so that \begin{math} L^2(X,\mathcal{F},\sigma) \subseteq L^2(X,\mathcal{F},\mu) \subseteq L^1(X,\mathcal{F},\mu); \end{math} it is also linear and bounded because \begin{math} |Tu|\leq\|u\|_{L^2(X,\mathcal{F},\sigma)}\cdot\sqrt{\sigma(X)}. \end{math} By Riesz representation theorem, there exists $g\in L^2(X,\mathcal{F},\sigma)$ such that \begin{equation} \label{eq:g} Tu=\int_Xu\;d\mu=\int_Xu\cdot g\,d\sigma \end{equation} for every $u\in L^2(X,\mathcal{F},\sigma)$. Then \begin{math} \mu(A)=\int_Ag\,d\sigma \end{math} for every $A\in\mathcal{F}$, so that $0<g\leq 1$ $\mu$- and $\sigma$-a.e. (Consider the former with $A=\{x\mid g(x)\leq 0\}$ or $A=\{x\mid g(x)>1\}$.) Moreover, the second equality in (\ref{eq:q}) holds when $u=\chi_A$ for $A\in\mathcal{F}$, 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 $\mathcal{F}$-measurable function because of the monotone convergence theorem. Now, $1/g$ is $\mathcal{F}$-measurable and nonnegative $\mu$- and $\sigma$-a.e.; moreover, $\dfrac{1}{g}\cdot g=1$ $\sigma$- and $\mu$-a.e. Thus, for every $A\in\mathcal{F}$, \begin{equation} \label{eq:1/g} \int_A\frac{1}{g}\,d\mu =\int_Ad\sigma =\sigma(A) \end{equation} Since $\sigma$ is finite, $1/g\in L^1(X,\mathcal{F},\mu)$, and so is $f=\dfrac{1}{g}-1$. Then for every $A\in\mathcal{F}$ \begin{displaymath} \nu(A)=\sigma(A)-\mu(A) =\int_A\left(\frac{1}{g}-1\right)d\mu =\int_Af\,d\mu\,. \end{displaymath}