I believe the entire second-to-last paragraph can be replaced by:
If there is no such varphiinA_{mathbb{P}_1}, then mathbb{P}_2notinmathcal{A}.
Thus we can take U_2 to be the complement of mathcal{A}
(which is open since mathcal{A} is closed)
and let U_1 be the entire space.
I’m thinking that the proof becomes conceptually slightly clearer if you show that the graph is closed by showing that it contains the limit of any convergent sequence, though:
Suppose
) for all ninmathbb{N}, and \to(\mathbb{P}’,\mathbb{P})). Since mathcal{A} is closed, we have mathbb{P}inmathcal{A}. We must show that %20=%201). Thus, suppose that %3Cb). Since \to\mathbb{P}(\varphi)), for all sufficiently large n we have %3Cb) and therefore %20%3C%20b)%20=%201). Since mathbb{P}_ntomathbb{P}, it follows that %20%3C%20b)%20=%201). In other words, mathbb{P} assigns probability 1 to every element of A_{mathbb{P}′}, and hence to all of A_{mathbb{P}′} (since this set is countable).
Comments on the proof of Theorem 2:
I believe the entire second-to-last paragraph can be replaced by:
I’m thinking that the proof becomes conceptually slightly clearer if you show that the graph is closed by showing that it contains the limit of any convergent sequence, though:
I think this is a cleaner way of saying it; I’m going to take this approach. Thanks!