The Unapologetic Mathematician

The Radon-Nikodym Theorem for Signed Measures

Now that we’ve proven the Radon-Nikodym theorem, we can extend it to the case where $\mu$ is a $\sigma$-finite signed measures.

Indeed, let $X=A\uplus B$ be a Hahn decomposition for $\mu$. We find that $\mu^+$ is a $\sigma$-finite measure on $A$, while $\mu^-$ is a $\sigma$-finite measure on $B$.

As it turns out that $\nu\ll\mu^+$ on $A$, while $\nu\ll\mu^-$ on $B$. For the first case, let $E\subseteq A$ be a set for which $\mu^+(E)=0$. Since $E\cap B=\emptyset$, we must have $\mu^-(E)=0$, and so $\lvert\mu\rvert(E)=\mu^+(E)+\mu^-(E)=0$. Then by absolute continuity, we conclude that $\nu(E)=0$, and thus $\nu\ll\mu^+$ on $A$. The proof that $\nu\ll\mu^-$ on $B$ is similar.

So now we can use the Radon-Nikodym theorem to show that there must be functions $f_A$ on $A$ and $f_B$ on $B$ so that

\displaystyle\begin{aligned}\nu(E\cap A)=&\int\limits_{E\cap A}f_A\,d\mu^+\\\nu(E\cap B)=&\int\limits_{E\cap B}f_B\,d\mu^-=-\int\limits_{E\cap B}-f_B\,d\mu^-\end{aligned}

We define a function $f$ on all of $X$ by $f(x)=f^+(x)$ for $x\in A$ and $f(x)=f^-(x)$ for $x\in B$. Then we can calculate

\displaystyle\begin{aligned}\nu(E)&=\nu((E\cap A)\uplus(E\cap B))\\&=\nu(E\cap A)+\nu(E\cap B)\\&=\int\limits_{E\cap A}f_A\,d\mu^+-\int\limits_{E\cap B}-f_B\,d\mu^-\\&=\int\limits_{E\cap A}f\,d\mu^+-\int\limits_{E\cap B}f\,d\mu^-\\&=\int\limits_Ef\,d\mu\end{aligned}

which in exactly the conclusion of the Radon-Nikodym theorem for the signed measure $\mu$.

July 8, 2010 - Posted by | Analysis, Measure Theory

1. […] 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 , then there’s some function […]

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

2. here, there are several formulas which does not display right

Comment by juanmarqz | July 13, 2010 | Reply

3. They’re all displaying correctly for me, at least for now. WordPress’ $\LaTeX$ support has been extremely buggy the last week or so. Is it saying something like “latex path not specified”?

Comment by John Armstrong | July 13, 2010 | Reply

4. yes “latex path not specified” in red and yellow background…

Comment by juanmarqz | July 13, 2010 | Reply

If nothing else, mouseover should show the $\LaTeX$ source.