The Unapologetic Mathematician

Mathematics for the interested outsider

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 | 5 Comments

   

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