Talk:Riesz representation theorem

"Combined with the Radon-Nikodým theorem, this amounts to the following..." — really? In order to apply the Radon-Nikodým theorem we must have two measures, one absolutely continuous w.r.t. the other. But here we have only one measure $\nu$. What is the point of $\mu$ if we cannot choose it once and serve all $L$ by it? --Boris Tsirelson 21:03, 21 July 2012 (CEST)