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 and on such that for all , . Then, for any nonnegative function and the Cauchy-Schwarz inequality yields Therefore, the linear mapping: is continuous on and this allows to apply the Riez-Theorem: there exists such that for all . We can show that by combining the previous equality with and by taking the specific functions or .

Define and consider the linear operator . Note that is continuous on . Indeed, by the Cauchy-Schwarz inequality, for all , According to the Riesz Theorem, there exists a such that for all , so that From this equation, we would like to deduce that and then but we have to be rigorous since we don't know yet if takes values only between and . Consider the following sets Let . Define and note that since this function is bounded by . Since \eqref{eq:fond} can be rewritten as , the monotone convergence then yields Plugging into \eqref{eq:fond} implies so that and To complete the proof, it remains to show that is a -null set, that is . But plugging into \eqref{eq:fond}, we get $0 \geq \int_B (1-h) d\nu=\mu(\indi{B} h)\geq \mu(\indi{B}) \geq 0\mu(B)=0$.

Let be two measures on . Assume that and that , for -almost all . Show that and , for -almost all .

By assumption, for any non-negative measurable function on ,

Taking , we get and therefore . Since by assumption , we obtain where This, combined with \eqref{eq:rel}, in turn implies For all , using successively , Eq. \eqref{eq:rel} and , This completes the proof.