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 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 it is also linear and bounded because 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 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 () 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 $A\in\mathcal{F}$