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2017/10/07 23:39 · douc


In this post, I introduce the beautiful proof of Von-Neumann. To get the idea, we start with a very simple example that gathers all the different ingredients of this wonderful meal.

Assume that there exist two measures $\mu$ and $\nu$ on $(\Omega, \mcf)$ such that for all $A \in \Omega$, $\nu(A) \leq \mu(A)\leq \mu(\Omega)<\infty$. Then, $\nu(f) \leq \mu(f)$ for any nonnegative function $f$ and the Cauchy-Schwarz inequality yields $$ |\nu(f)| \leq [\nu(\Omega)]^{1/2} [\nu(f^2)]^{1/2} \leq [\nu(\Omega)]^{1/2} [\mu(f^2)]^{1/2} $$ Therefore, the linear mapping: $f \mapsto \nu(f)$ is continuous on $\Ltwo(\mu)$ and this allows to apply the Riez-Theorem: there exists $g \in \Ltwo(\mu)$ such that $\nu(f)=\mu(fg)$ for all $f\in \Ltwo(\mu)$. We can show that $\mu(g\notin[0,1])=0$ by combining the previous equality with $\nu(f) \leq \mu(f)$ and by taking the specific functions $f=\indi{g<0}$ or $f=\indi{g>1}$.

All the ingredients are here: the linear mapping $f \mapsto \nu(f)$ where $f$ in taken in a $\Ltwo$ space associated to another measure, the Radon-Nikodym theorem and some particular choices of $f$ to get properties on $g$. But since the assumption “for all $A \in \Omega$, $\nu(A) \leq \mu(A)\leq \mu(\Omega)<\infty$” is too restrictive, we will use $\nu(A) \leq \pi(A)$ where $\pi=\nu+\mu$, which is direct since $\mu$ is a (nonnegative) measure!!!!!

The Radon-Nikodym theorem

Theorem If $\mu$ and $\nu$ are two finite measures on $(\Omega,\mathcal{F})$ then there exist a nonnegative measurable function $g$ and a $\mu$-null set B such that $$ \nu(A)=\int_A g \, d\mu+ \nu(A \cap B) $$ for each $A \in \mathcal{F}$.

$\bproof$ Define $\pi:=\mu+\nu$ and consider the linear operator $T(f):=\nu(f)$. Note that $T$ is continuous on $L^2(\pi)$. Indeed, by the Cauchy-Schwarz inequality, for all $f \in L^2(\pi)$, $$ |T(f)| \leq [\nu(\Omega)]^{1/2} \|f\|_{L^2(\nu)}\leq [\nu(\Omega)]^{1/2} \|f\|_{L^2(\pi)} $$ According to the Riesz Theorem, there exists a $h \in L^2(\pi)$ such that for all $f\in \Ltwo(\pi)$, $T(f)=\pi(f h)$ so that \begin{equation} \label{eq:fond} \forall f \in \Ltwo(\pi),\quad \nu(f)=\mu(f h)+\nu( f h)\eqsp. \end{equation} From this equation, we would like to deduce that $\nu(f(1-h))=\mu(fh)$ and then $\nu(f)=\mu(fh/(1-h))$ but we have to be rigorous since we don't know yet if $h$ takes values only between $0$ and $1$. Consider the following sets $$ N \eqdef \{h <0\}, \quad M\eqdef \{0 \leq h < 1\}, \quad B\eqdef \{h \geq 1\}. $$ Let $A \in \mcf$. Define $M_n:=\{0 \leq h \leq 1-1/n\}$ and note that $f\eqdef\frac{\indi{A} \indi{M_n}}{1-h} \in \Ltwo(\pi)$ since this function is bounded by $n$. Since \eqref{eq:fond} can be rewritten as $ \nu(f(1-h))=\mu(fh)$, the monotone convergence then yields $$ \nu(A \cap M)=\lim_{n \to \infty} \nu\lr{\indi{A \cap M_n}}=\lim_{n \to \infty} \nu\lr{\frac{\indi{A\cap M_n}}{1-h} (1-h)}=\lim_{n \to \infty} \mu\lr{\frac{\indi{A\cap M_n}}{1-h} h}=\mu\lr{\indi{A} \underbrace{\frac{\indi{M}h}{1-h}}_{g}}=\mu(\indi{A}g) $$ Plugging $f=\indi{N}$ into \eqref{eq:fond} implies $0\leq \nu(N)=\mu(\indi{N} h)+\nu(\indi{N} h) \leq 0$ so that $\nu(N)=0$ and $$ \nu(A)=\underbrace{\nu(A \cap N)}_{0} + \nu(A \cap M) + \nu(A \cap B) = \int_A g \, d\mu + \nu(A \cap B) $$ To complete the proof, it remains to show that $B$ is a $\mu$-null set, that is $\mu(B)=0$. But plugging $f=\indi{B}$ into \eqref{eq:fond}, we get $0 \geq \int_B (1-h) d\nu=\mu(\indi{B} h)\geq \mu(\indi{B}) \geq 0$ so that $\mu(B)=0$. $\eproof$

From this theorem, we can show that if $\mu$ and $\nu$ are $\sigma$-finite measures on $(\Omega,\mcf)$, then there exist a nonnegative measurable function $g$ and a $\mu$-null set B such that $$ \nu(A)=\int_A g \, d\mu+ \nu(A \cap B) $$ for each $A \in \mathcal{F}$. As a direct consequence, if $\mu$ is $\sigma$-finite and if $\nu \preceq \mu$ in the sense that $\mu$-null sets are $\nu$-null sets, then there exists a nonnegative, measurable function $g$ such that $\rmd \nu= g \ \rmd \mu$ (which is an “infinitesimal” notation for saying that $\nu= g \cdot \mu$).

An easy exercise on Radon Nikodym derivatives


Let $\mu,\nu$ be two measures on $(\Xset,\Xsigma)$. Assume that $\nu \preceq \mu$ and that $\frac{\rmd \nu}{\rmd \mu}(x)>0$, for $\mu$-almost all $x\in \Xset$. Show that $\mu \preceq \nu$ and $\frac{\rmd \mu}{\rmd \nu}(x)=\lr{\frac{\rmd \nu}{\rmd \mu}(x)}^{-1}$ , for $\nu$-almost all $x\in \Xset$.


By assumption, for any non-negative measurable function $f$ on $\Xset$,

\begin{equation} \label{eq:rel} \nu(f)=\int_\Xset \mu(\rmd x) \frac{\rmd \nu}{\rmd \mu}(x) f(x) \end{equation}

Taking $f=\indi{\Xset}$, we get $\mu(\frac{\rmd \nu}{\rmd \mu})=\nu(\Xset)=1<\infty$ and therefore $\mu(\frac{\rmd \nu}{\rmd \mu}<\infty)=1$. Since by assumption $\mu(\frac{\rmd \nu}{\rmd \mu}=0)=1$, we obtain $\mu(A)=1$ where $$ A=\set{x \in \Xset}{\frac{\rmd \nu}{\rmd \mu}(x)\in (0,\infty)} $$ This, combined with \eqref{eq:rel}, in turn implies $$ \nu(A)=\int_\Xset \mu(\rmd x) \frac{\rmd \nu}{\rmd \mu}(x) \indi{A}(x)=\int_\Xset \mu(\rmd x) \frac{\rmd \nu}{\rmd \mu}(x)= \nu(\Xset)=1 $$ For all $B\in \Tsigma$, using successively $\mu(A)=1$, Eq. \eqref{eq:rel} and $\nu(A)=1$, \begin{align*} \mu(B)&=\mu(B \cap A)=\int_\Xset \mu(\rmd x)\lrb{ \frac{\rmd \nu}{\rmd \mu}(x) \times \lr{\frac{\rmd \nu}{\rmd \mu}(x)}^{-1}}\indi{B}(x)\indi{A}(x)\\ &=\int_\Xset \nu(\rmd x) \lr{\frac{\rmd \nu}{\rmd \mu}(x)}^{-1} \indi{B}(x)\indi{A}(x)= \int_\Xset \nu(\rmd x) \lr{\frac{\rmd \nu}{\rmd \mu}(x)}^{-1} \indi{B}(x) \end{align*} This completes the proof.

world/radon.txt · Last modified: 2022/03/16 07:40 (external edit)