# Uniqueness of extension of zero measure

Let $(\Omega,\mathscr F)$ be a measurable space with two probability measures $\mu, \nu: \mathscr F\to[0,1]$ defined over it. Suppose that $\mathscr C\subset\mathscr F$ is some class of sets and $$(\mu-\nu)|_\mathscr C = 0.$$ Which necessary and sufficient conditions are known on $\mathscr C$ in order to assure that $\mu = \nu$?

For example, it is sufficient for $\mathscr C$ to be a ring of sets generating $\mathscr F$, as it follows from Caratheodory's extension theorem. However, is that condition also necessary?

In particular, if $\sigma(\mathscr C) = \mathscr F$ can it be the case that $\mu\neq \nu$?

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A sufficient condition is that $\mathcal{C}$ is a set of generators of $\mathcal{F}$ closed under pairwise intersections (a $\pi$-system). The proof goes via Dynkins lemma and can be found in "Probability with Martingales" by Williams.

Edit: The first version was not quite right. This works: Let $\Omega=\{1,2,3,4\}$ and $\mathcal{F}=2^\Omega$. Let $$\mathcal{C}=\big\{\{1,2\},\{3,4\},\{1,3\},\{2,4\}\big\}.$$ Clearly $\mathcal{F}=\sigma(\mathcal{C})$. Define $\mu$ by $\mu\{1\}=\mu\{4\}=1/6$ and $\mu\{2\}=\mu\{3\}=1/3$. Define $\nu$ by $\nu\{1\}=\nu\{4\}=1/3$ and $\nu\{2\}=\nu\{3\}=1/6$. Then $\nu$ and $\mu$ agree on $\mathcal{C}$, but are different probability measures. The counterexample is from "Counterexamples in Probability" by Stoyanov.

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Just to update: this is Lemma 1.6 in the book mentioned. Strangely, I didn't find it in the book by Bogachev. –  Ilya Aug 23 '12 at 8:26