# The Unapologetic Mathematician

## 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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Posted by John Armstrong | 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 | 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 | 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 | 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 | 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 | 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 | 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 | December 29, 2012 | Reply

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

Comment by | 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 | 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 | 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 | January 7, 2013 | Reply

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