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 $T:L^{2}(X,\mathcal{F},\sigma)\to\mathbb{R}$ defined by

 $Tu=\int_{X}u\;d\mu\;\;\forall u\in L^{2}(X,\mathcal{F},\sigma)\;.$ (1)

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

 $Tu=\int_{X}u\;d\mu=\int_{X}u\cdot g\,d\sigma$ (2)

for every $u\in L^{2}(X,\mathcal{F},\sigma)$. Then $\mu(A)=\int_{A}g\,d\sigma$ for every $A\in\mathcal{F}$, so that $0 $\mu$- 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 (LABEL: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 $A\in\mathcal{F}$,

 $\int_{A}\frac{1}{g}\,d\mu=\int_{A}d\sigma=\sigma(A)$ (3)

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

 $\nu(A)=\sigma(A)-\mu(A)=\int_{A}\left(\frac{1}{g}-1\right)d\mu=\int_{A}f\,d\mu\,.$
Title proof of Radon-Nikodym theorem ProofOfRadonNikodymTheorem 2013-03-22 18:58:03 2013-03-22 18:58:03 Ziosilvio (18733) Ziosilvio (18733) 5 Ziosilvio (18733) Proof msc 28A15 Hilbert spaces proof of Radon-Nikodym’s theorem measure- theoretic proof of Radon-Nikodym theorem