Indeed, let be a Hahn decomposition for . We find that is a -finite measure on , while is a -finite measure on .
As it turns out that on , while on . For the first case, let be a set for which . Since , we must have , and so . Then by absolute continuity, we conclude that , and thus on . The proof that on is similar.
So now we can use the Radon-Nikodym theorem to show that there must be functions on and on so that
We define a function on all of by for and for . Then we can calculate
which in exactly the conclusion of the Radon-Nikodym theorem for the signed measure .