Show that Theorem 24.6 is enough to prove the Radon-Nikodm theorem (Theorem 25.2 ) for a countably

Show that Theorem 24.6 is enough to prove the Radon-Nikodým theorem (Theorem 25.2 ) for a countably generated \(\mathscr{A}\), i.e. \(\mathscr{A}=\sigma\left(\left\{A_{n}ight\}_{n \in \mathbb{N}}ight)\).

[ set \(\mathscr{A}_{n}:=\sigma\left(A_{1}, A_{2}, \ldots, A_{n}ight)\) and observe that the atoms of \(\mathscr{A}_{n}\) are of the form \(C_{1} \cap\) \(\cdots \cap C_{n}\), where \(C_{i} \in\left\{A_{i}, A_{i}^{c}ight\}, 1 \leqslant i \leqslant n\).]

Data from theorem 25.2

Data from theorem 24.6

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Theorem 25.2 (Radon-Nikodm) Let u, v be two measures on the measurable space (X, A). If is o-finite, then the following assertions are equivalent: (1) v(4) = [ f(x) (dx) for some a.e. unique M(A), >0; (ii) . The unique function f is called the Radon-Nikodm derivative and it is tradition- ally denoted by f= dv/du. Above we have just verified that (i) (ii). The converse direction is less obvi- ous and we want to use a martingale argument for its proof. For this we need a few more preparations which extend the notion of a martingale to directed index sets. Let (I,