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 -finite signed measures and , then we can break into two uniquely-defined pieces, and , so that , , and . We call such a pair the “Lebesgue decomposition” of with respect to .
Since a signed measure is absolutely continuous or singular with respect to if and only if it’s absolutely continuous or singular with respect to , we may as well assume that is a measure. And since in both cases is absolutely continuous or singular with respect to if and only if and both are, we may as well assume that is also a measure. And, as usual, we can break our measurable space down into the disjoint union of countably many subspaces on which both and 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 and 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: . Thus the Radon-Nikodym theorem tells us that there exists a function so that
for every measurable set . Since , we must have -a.e., and thus -a.e. as well. Since
Let us define and . Then we calculate
and thus (by the finiteness of ), . Defining and , then it’s clear that . We still must prove that .
If , then we calculate
But -a.e., which means that we must have , and thus .
Now, suppose and are two Lebesgue decompositions of with respect to . Then . We know that both singularity and absolute continuity pass to sums, so is singular with respect to , while is absolutely continuous with respect to . But the only way for this to happen is for them both to be zero, and thus and .
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