# The Unapologetic Mathematician

## The Radon-Nikodym Derivative

Okay, so the Radon-Nikodym theorem and its analogue for signed measures tell us that if we have two $\sigma$-finite signed measures $\mu$ and $\nu$ with $\nu\ll\mu$, then there’s some function $f$ so that

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

But we also know that by definition

$\displaystyle\nu(E)=\int\limits_E\,d\nu$

If both of these integrals were taken with respect to the same measure, we would know that the equality

$\displaystyle\int\limits_Ef\,d\mu=\int\limits_Eg\,d\mu$

for all measurable $E$ implies that $f=g$ $\mu$-almost everywhere. The same thing can’t quite be said here, but it motivates us to say that in some sense we have equality of “differential measures” $d\nu=f\,d\mu$. In and of itself this doesn’t really make sense, but we define the symbol

$\displaystyle\frac{d\nu}{d\mu}=f$

and call it the “Radon-Nikodym derivative” of $\nu$ by $\mu$. Now we can write

$\displaystyle\int\limits_E\,d\nu=\int\limits_Ef\,d\mu=\int\limits_E\frac{d\nu}{d\mu}\,d\mu$

The left equality is the Radon-Nikodym theorem, and the right equality is just the substitution of the new symbol for $f$. Of course, this function — and the symbol $\frac{d\nu}{d\mu}$ — is only defined uniquely $\mu$-almost everywhere.

The notation and name is obviously suggestive of differentiation, and indeed the usual laws of derivatives hold. We’ll start today by the easy property of linearity.

That is, if $\nu_1$ and $\nu_2$ are both $\sigma$-finite signed measures, and if $a_1$ and $a_2$, then $a_1\nu_1+a_2\nu_2$ is clearly another $\sigma$-finite signed measure. Further, it’s not hard to see if $\nu_i\ll\mu$ then $a_1\nu_1+a_2\nu_2\ll\mu$ as well. By the Radon-Nikodym theorem we have functions $f_1$ and $f_2$ so that

\displaystyle\begin{aligned}\nu_1(E)&=\int\limits_Ef_1\,d\mu\\\nu_2(E)&=\int\limits_Ef_2\,d\mu\end{aligned}

for all measurable sets $E$. Then it’s clear that

\displaystyle\begin{aligned}\left[a_1\nu_1+a_2\nu_2\right](E)&=a_1\nu_1(E)+a_2\nu_2(E)\\&=a_1\int\limits_Ef_1\,d\mu+a_2\int\limits_Ef_2\,d\mu\\&=\int\limits_Ea_1f_1+a_2f_2\,d\mu\end{aligned}

That is, $a_1f_1+a_2f_2$ can serve as the Radon-Nikodym derivative of $a_1\nu_1+a_2\nu_2$ with respect to $\mu$. We can also write this in our suggestive notation as

$\displaystyle\frac{d(a_1\nu_1+a_2\nu_2)}{d\mu}=a_1\frac{d\nu_1}{d\mu}+a_2\frac{d\nu_2}{d\mu}$

which equation holds $\mu$-almost everywhere.

July 9, 2010 - Posted by | Analysis, Measure Theory

## 3 Comments »

1. […] Radon-Nikodym Chain Rule Today we take the Radon-Nikodym derivative and prove that it satisfies an analogue of the chain […]

Pingback by The Radon-Nikodym Chain Rule « The Unapologetic Mathematician | July 12, 2010 | Reply

2. already is w fine…

Comment by juanmarqz | July 15, 2010 | Reply

3. all the way at the top, I think you mean v<<u. You have this reversed.

Comment by Bobby Brown | March 30, 2011 | Reply