A theorem of Doob. Let \(\left(\mu_{t}ight)_{t \geqslant 0}\) and \(\left(u_{t}ight)_{t \geqslant 0}\) be two families of measures on the \(\sigma\)-finite measure space \((X, \mathscr{A})\) such that \(u_{t} \ll \mu_{t}\) for all \(t \geqslant 0\) and \(t \mapsto \mu_{t}(A), u_{t}(A)\) are measurable for all \(A \in \mathscr{A}\). Then there exists a measurable function \((t, x) \mapsto p(t, x),(t, x) \in[0, \infty) \times X\), such that \(u_{t}=p(t, \cdot) \mu_{t}\) for all \(t \geqslant 0\).

[ argue as in the proof of Theorem 25.2 :

\[p_{\alpha}(t, x):=\sum_{A \in \alpha} \frac{u_{t}(A)}{\mu_{t}(A)} \mathbb{1}_{A}(x)\]

and check that this function is jointly measurable in \(t\) and \(x\).]

Data from theorem 25.2

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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 (1,