# 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?

• Have I got it right that $\zeta_n$ are random measures on the space $\mathcal{P}(S)$ of measures? That is, they are $\mathcal{P}(\mathcal{P}(S))$-valued random variables? – Nate Eldredge May 4 '16 at 0:27
• Yes, that is correct. – D.R. May 5 '16 at 3:37
• If $S$ is compact, then $C(S)$ is separable, so you can pick a countable dense subset and use that as $\mathcal{C}$. I think you can do something similar if $S$ is locally compact (let $\mathcal{C}$ be dense in $C_c(S)$). – Nate Eldredge May 5 '16 at 3:56
• Interesting idea. And of course Polish does not imply locally compact, e.g., Baire space ${\mathbb N}^{\mathbb N}$. Perhaps there is a counterexample lurking in there. – D.R. May 5 '16 at 4:07

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 $f \in \mathcal{C}$.

• Oh right. This is really just the fact that the weak topology on $\mathcal{P}(S)$ is second countable (since it's separable and metrizable). – Nate Eldredge May 8 '16 at 18:08

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.