# The Unapologetic Mathematician

## Mathematics for the interested outsider

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