Moving a result from the unconditional to the conditional I'm generally wary when lifting a result stated unconditionally to a situation where I'm conditioning on a random variable.  Consider the following classical result in weak convergence:
Theorem. Let $\phi,\phi_1,\phi_2,\dotsc$ be random measures on a Polish space $S$. Then $\phi_n \to \phi$ in distribution if and only if $\phi_n f \to \phi f$ in distribution for every bounded continuous $f$.
Now consider this application: Let $\eta$ be a random variable, and define $\zeta_n = \Pr[\phi_n \in \cdot \,| \eta]$ and $\zeta = \Pr[\phi \in \cdot \,| \eta]$. I would like to show $\zeta_n \to \zeta$ weakly almost surely.  
Naively, one might think to show that, for every bounded continuous $f$, we have 
$$\Pr [\phi_n f \in \cdot \,|\eta] \to \Pr [\phi f \in \cdot \, | \eta] \text { weakly a.s.}$$ But this fails because of the uncountable number of bounded continuous functions.  However, if I knew that there was some countable class $\mathcal C$ of bounded continuous functions such that convergence on this class implied convergence for all bounded continuous functions, then I could establish the a.s. weak convergence on this countable class and then conclude that convergence holds simultaneously, and then I can apply the Theorem above.
Is there anything that I have missed?
Are there example where one might expect to transfer an unconditional result to a conditional setting, but trouble lurks?
 A: There's a fairly well known result due to Varadarajan (I think in this paper, though I can't get through the paywall at the moment), that provides your countable class $\mathcal{C}$ you desire: If $S$ is a Polish space, then there exists a countable family $\mathcal{C}$ of continuous bounded functions of $S$ such that $\mu_n \rightarrow \mu$ in $\mathcal{P}(S)$ if and only if $\mu_n f\rightarrow\mu f$ for every $f \in \mathcal{C}$.
A: To fill in some details in Dan's answer:
If $S$ is Polish then there does exist a countable $\mathcal{C} \subset C_b(S)$ which determines weak convergence in $\mathcal{P}(S)$, as you desire.
It's a standard fact that if $S$ is Polish then the weak topology on $\mathcal{P}(S)$ is separable and metrizable; see the paper of Varadarajan cited by Dan, or Billingsley's Convergence of Probability Measures which anybody working in this field surely must have on their shelf ;-).  As such it is second countable.
Now by definition of the weak topology, open sets of the following form are a basis for $\mathcal{P}(S)$:
$$U_{\mu, f_1, \dots, f_n, \epsilon} := \{\nu : |\mu f_i - \nu f_i| < \epsilon,\; i = 1,\dots,n\}$$
where $\mu \in \mathcal{P}(S)$, $f_1, \dots, f_n \in C_b(S)$ (the bounded continuous functions), and $\epsilon > 0$.
By second countability, there is a basis $\mathcal{U}$ consisting of countably many such sets.  Let $\mathcal{C}$ consist of all the functions $f_i$, used in defining all the $U \in \mathcal{U}$; then $\mathcal{C}$ is a countable union of finite sets and hence countable. 
Now suppose $f \nu_j \to f\nu$ for all $f \in \mathcal{C}$.  Suppose $U \in \mathcal{U}$ is a neighborhood of $\nu$.  We can write $U = U_{\mu, f_1, \dots, f_n, \epsilon}$ for some  $\mu, f_1, \dots, f_n, \epsilon$. Since $\nu \in U$, we have $|f_i \nu - f_i \mu| < \epsilon$ for all $i$. Let $\delta = \max_i |f_i \nu - f_i \mu| < \epsilon$.
Now $f_1, \dots, f_n \in \mathcal{C}$, so by assumption we have $f_i \nu_j \to f_i \nu$ for each $i$.  There is thus $J$ so large that for all $j > J$, we have $|f_i \nu_j - f_i \nu| < \epsilon - \delta$ for each $i$.  Then for such $j$ we have
$$|f_i \nu_j - f_i \mu| \le |f_i \nu_j - f_i \nu| + |f_i \nu - f_i \mu| < (\epsilon - \delta) + \delta = \epsilon.$$
Hence $\nu_j \in U$.  Since $U$ ranges over a basis, it follows that $\nu_j \to \nu$ weakly.
