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 $\mu\neq \nu$?
 A: An interesting example is here (but not exactly what is asked here):
R. O. Davies, Measures not approximable or not specifiable by means of balls.
Mathematica 18 (1971) 157--160.
A metric space X, two Borel probability measures defined on X, the two measures agree on all balls of the metric, but the two measures do not agree on all Borel sets.
A: 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.
