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Talk:Riesz representation theorem

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"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)

You derive the last statement from the previous as follows. Let $L\in C(X)'$. By the first statement there is a $\mathbb R$-valued measure $\nu$ with finite total variation such that $L (f) = \int f d\nu$. You denote by $\mu$ the total variation measure of $\nu$, which is then a usual measure (i.e. taking values in $\mathbb R^+$) and hence it is Radon in the sense of Radon measure. The Radon-Nykodim theorem is applied to get the existence of a measurable $g$ such that $\nu = g\mu$, from which you get the last statement. The point of $\mu$ is that we got rid of the sign changes, which are instead now pushed to the $g$ (which in fact takes only the values $\pm 1$). It has certain advantages. Of course in this context we could also phrase it as: Any $L\in C(X)'$ can be written as the difference of integrals against two Radon measures. But the way I stated easily generalizes to characterize the dual of $C(X, V)$ where $V$ is a finite vector space (or even a general Banach space). Camillo 22:49, 21 July 2012 (CEST)
Maybe it is just my upbringing: in geometric measure theory it is customary to take into account vector valued measures and the Radon-Nykodim is phrased in general terms, so to derive from it that any vector-valued measure can be written as a measurable vector function times its total variation. In this case the $g$ is less silly: it has $|g|=1$, so you factored the direction from the size of the measure. Just to give you an example why we are so fond about it. Take a smooth open set $A$ in $\mathbb R^n$ and consider its indicator function. The distributional derivative is a vector-valued measure. The above $g\mu$ decompoisition gives you the surface measure on the $\partial A$ as $\mu$ and the outward unit normal to the boundary as $g$. In this way you got a purely functional-analytic object encoding the geometry of a surface. It is easy to generalize it to much rougher objects, but retaining some functional analytic structure. When I will deal with the Radon-Nykodim and with finer entries in Geometric measure theory I will then get to this somewhere. Camillo 22:49, 21 July 2012 (CEST)
Yes, I see, you are right. I just did not note the turn from positive to signed; sorry. --Boris Tsirelson 08:09, 22 July 2012 (CEST)
How to Cite This Entry:
Riesz representation theorem. Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Riesz_representation_theorem&oldid=27172