Weak a.s. convergence VS a.s.weak convergence Let's consider a sequence $(\mu_n)_n$ of random probability measures on $\mathbb R$, and let $C_b$ be the Banach space of bounded continuous functions on $\mathbb R$. I am considering the following types of convergence, where $\mu$ is some non-random probability measure :
(A)
$$ \forall f\in C_b, \quad \mathbb P \left(\int f(x)d\mu_n(x)\rightarrow_{n\rightarrow\infty}\int f(x)d\mu(x)\right)=1$$
and 
(B)
$$ \mathbb P \left(\forall f\in C_b,\quad \int f(x)d\mu_n(x)\rightarrow_{n\rightarrow\infty}\int f(x)d\mu(x)\right)=1.$$
Whereas (B) clearly implies (A), what about the other direction ?
1) Do you have a counter-example where (A) does not implies (B) ?
2) Do we have a sufficient criterion so that (A) implies (B) ?
(I have in mind the use of the Borel-Cantelli lemma to improve the probability convergence to the almost sure one)
 A: I believe A and B are equivalent.
First, I believe that when the underlying state space (in this case $\mathbb{R}$) is locally compact, then probability measures $\nu_n$ converge weakly to $\nu$ iff $\int f\, d\nu_n \to \int f\,d\nu$ for all continuous functions $f$ which vanish at infinity, i.e. for all $f \in C_0(\mathbb{R})$.  I don't remember the proof off the top of my head, but you can find it in Billingsley's Convergence of Probability Measures.
Now $C_0(\mathbb{R})$ is separable, so one can choose a countable dense subset $\{f_1, f_2,\dots\}$.  Then using the triangle inequality and the fact that the measures $\nu_n$ are uniformly bounded in total variation (since they are all probability measures), you can show that $\nu_n \to \nu$ weakly iff $\int f_k \,d\nu_n \to \int f_k \,d\nu$ for all $k$.
Now if A holds, then by countable additivity
$$\mathbb{P}\left( \int f_k\,d\mu_n \to \int f_k\,d\mu \text{ for all } k\right) = 1.$$
But we have just argued that on this event, $\mu_n \to \mu$ weakly.
