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.

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July 6, 2010 - Posted by | Analysis, Measure Theory

12 Comments »

  1. [...] Radon-Nikodym Theorem (Proof) Today we set about the proof of the Radon-Nikodym theorem. We assumed that is a -finite measure space, and that is a -finite signed measure. Thus we can [...]

    Pingback by The Radon-Nikodym Theorem (Proof) « The Unapologetic Mathematician | July 7, 2010 | Reply

  2. [...] Radon-Nikodym Theorem for Signed Measures Now that we’ve proven the Radon-Nikodym theorem, we can extend it to the case where is a -finite signed [...]

    Pingback by The Radon-Nikodym Theorem for Signed Measures « The Unapologetic Mathematician | July 8, 2010 | Reply

  3. [...] Radon-Nikodym Derivative Okay, so the Radon-Nikodym theorem and its analogue for signed measures tell us that if we have two -finite signed measures and with [...]

    Pingback by The Radon-Nikodym Derivative « The Unapologetic Mathematician | July 9, 2010 | Reply

  4. [...] the proof itself, the core is based on our very first result about absolute continuity: . Thus the Radon-Nikodym theorem tells us that there exists a function so [...]

    Pingback by Lebesgue Decomposition « The Unapologetic Mathematician | July 14, 2010 | Reply

  5. [...] we can combine this with the Radon-Nikodym theorem. If is a measurable function from a measure space to a totally -finite measure space so that the [...]

    Pingback by Pulling Back and Pushing Forward Structure « The Unapologetic Mathematician | August 2, 2010 | Reply

  6. [...] that , and thus that is a (signed) measure. It should also be clear that implies , and so . The Radon-Nikodym theorem now tells us that there exists an integrable function so [...]

    Pingback by Some Continuous Duals « The Unapologetic Mathematician | September 3, 2010 | Reply

  7. In the argument, when you said 0<=nu(Bo)<=1/n*mu(Bo). It is not clear to me why, 0<=nu(Bo). I tried to search through many books, but not found any where.

    Comment by suneel | December 29, 2012 | Reply

  8. The measures \mu and \nu are not signed measures here.

    Comment by John Armstrong | December 29, 2012 | Reply

  9. I’m also confused. Following suneel’s comment, though I understand that mu and nu are not signed measures, but why would that specifically mean that 0<=nu(Bo)?

    Comment by Alice | January 6, 2013 | Reply

  10. Go back to the definition of a measure: “an extended real-valued, non-negative, countably additive set function \mu defined on an algebra \mathcal{A}, and satisfying \mu(\emptyset)=0.” By definition, measures are nonnegative.

    Comment by John Armstrong | January 6, 2013 | Reply

  11. Now my question is different, Radon-Nikodym theorem is for signed measure and we are using this lemma(which is only for positive measure) in the proof of Radon-Nikodym, I am loosing some connection here? Thanks for clarification.

    Comment by suneel | January 7, 2013 | Reply

  12. There’s a separate version of Radon-Nikodym for signed measures. Look up at comment #2, which is the pingback from a later post called “The Radon-Nikodym Theorem for Signed Measures”.

    Comment by John Armstrong | January 7, 2013 | Reply


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