The Unapologetic Mathematician

Mathematics for the interested outsider

Lebesgue Decomposition

As we’ve said before, singularity and absolute continuity are diametrically opposed. And so it’s not entirely surprising that if we have two totally \sigma-finite signed measures \mu and \nu, then we can break \nu into two uniquely-defined pieces, \nu_c and \nu_s, so that \nu_c\ll\mu, \nu_s\perp\mu, and \nu=\nu_c+\nu_s. We call such a pair the “Lebesgue decomposition” of \nu with respect to \mu.

Since a signed measure is absolutely continuous or singular with respect to \mu if and only if it’s absolutely continuous or singular with respect to \lvert\mu\rvert, we may as well assume that \mu is a measure. And since in both cases \nu is absolutely continuous or singular with respect to \mu if and only if \nu^+ and \nu^- both are, we may as well assume that \nu is also a measure. And, as usual, we can break our measurable space X down into the disjoint union of countably many subspaces on which both \mu and \nu are both totally finite. We can assemble the Lebesgue decompositions on a collection such subspaces into the Lebesgue decomposition on the whole, and so we can assume that \mu and \nu are totally finite.

Now that we can move on to the proof itself, the core is based on our very first result about absolute continuity: \nu\ll\mu+\nu. Thus the Radon-Nikodym theorem tells us that there exists a function f so that

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

for every measurable set E. Since 0\leq\nu(E)\leq\mu(E)+\nu(E), we must have 0\leq f\leq1 (\mu+\nu)-a.e., and thus 0\leq f\leq1 \nu-a.e. as well. Since

Let us define A=\{x\in X\vert f(x)=1\} and B=\{x\in X\vert0\leq f(x)<1\}. Then we calculate

\displaystyle\nu(A)=\int\limits_A\,d\mu+\int\limits_A\,d\nu=\mu(A)+\nu(A)

and thus (by the finiteness of \nu), \mu(A)=0. Defining \nu_s(E)=\nu(E\cap A) and \nu_c(E)=\nu(E\cap B), then it’s clear that \nu_s\perp\mu. We still must prove that \nu_c\ll\mu.

If \mu(E)=0, then we calculate

\displaystyle\int\limits_{E\cap B}\,d\nu=\nu(E\cap B)=\int\limits_{E\cap B}f\,d\mu+\int\limits_{E\cap B}f\,d\nu=\int\limits_{E\cap B}f\,d\nu

and, therefore

\displaystyle\int\limits_{E\cap B}(1-f)\,d\mu=0

But 1-f\geq0 \nu-a.e., which means that we must have \nu_c(E)=\nu(E\cap B)=0, and thus \nu_c\ll\mu.

Now, suppose \nu=\nu_s+\nu_c and \nu=\bar{\nu}_s+\bar{\nu}_c are two Lebesgue decompositions of \nu with respect to \mu. Then \nu_s-\bar{\nu}_s=\bar{\nu}_c-\nu_c. We know that both singularity and absolute continuity pass to sums, so \nu_s-\bar{\nu}_s is singular with respect to \mu, while \bar{\nu}_c-\nu_c is absolutely continuous with respect to \mu. But the only way for this to happen is for them both to be zero, and thus \nu_s=\bar{\nu}_s and \nu_c-\bar{\nu}_c.

July 14, 2010 Posted by | Analysis, Measure Theory | Leave a comment

   

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