## The Radon-Nikodym Theorem (Statement)

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

To show this, let be a Hahn decomposition with respect to the signed measure for each positive integer . Let be the union of all the and let be the intersection of all the . Then since and is negative for we find

for every positive integer . This shows that we must have . And then, since is not identically zero we must have . By absolute continuity we conclude that , which means that we must have for at least one value of . So we pick just such a value, set , and , and everything we asserted is true.

Now for the Radon-Nikodym Theorem: we let be a totally -finite measure space and let be a -finite signed measure on which is absolutely continuous with respect to . Then is an indefinite integral. That is, there is a finite-valued measurable function so that

for every measurable set . The function is unique in the sense that if any other function has as its indefinite integral, then -almost everywhere. It should be noted that we don’t assert that is integrable, which will only be true of 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 , or a function whose integral diverges definitely, on a -finite measure space and define its indefinite integral , then is a -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 . 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 and get its indefinite integral, then any other function with the same indefinite integral must be a.e. equal to .

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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 |

The measures and are not signed measures here.

Comment by John Armstrong | December 29, 2012 |

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 |

Go back to the definition of a measure: “an extended real-valued, non-negative, countably additive set function defined on an algebra , and satisfying .” By definition, measures are nonnegative.

Comment by John Armstrong | January 6, 2013 |

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 |

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 |