The Unapologetic Mathematician

Mathematics for the interested outsider

The Radon-Nikodym Theorem (Statement)

Before the main business, a preliminary lemma: if \mu and \nu are totally finite measures so that \nu is absolutely continuous with respect to \mu, and \nu is not identically zero, then there is a positive number \epsilon and a measurable set A so that \mu(A)>0 and A is a positive set for the signed measure \nu-\epsilon\mu. That is, we can subtract off a little bit (but not zero!) of \mu from \nu and still find a non-\mu-negligible set on which what remains is completely positive.

To show this, let X=A_n\uplus B_n be a Hahn decomposition with respect to the signed measure \nu-\frac{1}{n}\mu for each positive integer n. Let A_0 be the union of all the A_n and let B_0 be the intersection of all the B_n. Then since B_0\subseteq B_n and B_n is negative for \nu-\frac{1}{n}\mu we find

\displaystyle0\leq\nu(B_0)\leq\frac{1}{n}\mu(B_0)

for every positive integer n. This shows that we must have \nu(B_0)=0. And then, since \nu is not identically zero we must have \nu(A_0)=\nu(X\setminus B_0)>0. By absolute continuity we conclude that \mu(A_0)>0, which means that we must have \mu(A_n)>0 for at least one value of n. So we pick just such a value, set A=A_n, and \epsilon=\frac{1}{n}, and everything we asserted is true.

Now for the Radon-Nikodym Theorem: we let (X,\mathcal{S},\mu) be a totally \sigma-finite measure space and let \nu be a \sigma-finite signed measure on \mathcal{S} which is absolutely continuous with respect to \mu. Then \nu is an indefinite integral. That is, there is a finite-valued measurable function f:X\to\mathcal{R} so that

\displaystyle\nu(E)=\int\limits_Ef\,d\mu

for every measurable set E. The function f is unique in the sense that if any other function g has \nu as its indefinite integral, then f=g \mu-almost everywhere. It should be noted that we don’t assert that f is integrable, which will only be true of \nu is actually finite. However, either its positive or its negative integral must converge or we wouldn’t use the integral sign for a divergent integral.

Let’s take a moment and consider what this means. We know that if we take an integrable function f, or a function whose integral diverges definitely, on a \sigma-finite measure space and define its indefinite integral \nu, then \nu is a \sigma-finite signed measure that is absolutely continuous with respect to the measure against which we integrate. What the Radon-Nikodym theorem tells us that any such signed measure arises as the indefinite integral of some such function f. Further, it tells us that such a function is essentially unique, as much as any function is in measure theory land. In particular, we can tell that if we start with a function f and get its indefinite integral, then any other function with the same indefinite integral must be a.e. equal to f.

July 6, 2010 Posted by | Analysis, Measure Theory | 12 Comments

   

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