I don't understand the following as I read along a proof in a paper (Page 66, "Asymptotic Behaviour of some interacting systems", by Sylvie Meleard):

We denote by $\mathcal{P}({M})$ the space of probability measures on a metric space $M$, equipped with the weak topology.

Let $E$ be a metric space. Let $\{ \mu_n \}$ be a sequence of random measures on $E$, i.e. for each $n$, $\mu_n$ is a $\mathcal{P} (E)$-valued random variable. Also, let $\mathbb{Q}$ be a deterministic probability measure on the same probability space. We can then treat $\delta_{\mathbb{Q}}$ be a constant $\mathcal{P}(E)$-valued random variable.

Since both $\text{Law} ( \mu_n) $ and $\text{Law} ( \delta_{\mathbb{Q}})$ are measures on $\mathcal{P}(E)$, the paper defines that $\mu_n$ converges in law to $\mathbb{Q}$ if $$\text{Law} ( \mu_n) \implies \text{Law} ( \delta_{\mathbb{Q}}). \quad \quad \quad \quad \, \, \, \, (*)$$ However, in the paper, the fact that $\mu_n$ converges in law to $\mathbb{Q}$ is concluded by establishing that $$\mathbb{E} \bigg[ \bigg| \int_E f \,d \mu_n - \int_E f \,d \mathbb{Q} \bigg| \bigg] \rightarrow 0, \quad \quad \quad (**)$$ for all continuous bounded functions $f$ on $E$.

By definition of $(*)$, this is equivalent to saying that $$ \int_{\mathcal{P}(E)} f \, d\text{Law} ( \mu_n) \rightarrow \int_{\mathcal{P}(E)} f \, d \text{Law} ( \delta_{\mathbb{Q}}),$$ for all continuous bounded functions $f$ on $\mathcal{P}(E)$. How does this follow from $(**)$? Any ideas?


Let $F : \mathcal{P}(E) \to \mathbb{R}$ be bounded and continuous, and let $M := \sup |F|$. In adjusted notation, we wish to show $\mathbb{E} F(\mu_n) \to \mathbb{E} F(\delta_{\mathbb{Q}}) = F(\mathbb{Q})$. To save me some typing, let's suppose without loss of generality that $F(\mathbb{Q}) = 0$.

Fix $\epsilon > 0$. By continuity of $F$, there is a weakly open set $U \subset \mathcal{P}(E)$, containing $\mathbb{Q}$, such that if $\nu \in U$ then $|F(\nu)| < \epsilon$. By definition of the weak topology, there are finitely many bounded continuous $f_1, \dots, f_k : E \to \mathbb{R}$ and $\delta > 0$ such that if $\left| \int f_i\,d\nu - \int f_i \,d\mathbb{Q}\right| < \delta$ for $i = 1,\dots,k$ then $\nu \in U$.

Now for each $i$ we have from (**) that $\mathbb{E} \left| \int f_i\,d\mu_n - \int f_i\,d\mathbb{Q}\right| \to 0$. By Chebyshev's inequality we therefore have $\mathbb{P}\left(\left| \int f_i\,d\mu_n - \int f_i\,d\mathbb{Q}\right| \ge \delta\right) \to 0$. So we may find a large enough $N$ such that for all $n \ge N$ and all $i = 1,\dots,k$ we have $\mathbb{P}\left(\left| \int f_i\,d\mu_n - \int f_i\,d\mathbb{Q}\right| \ge \delta\right) < \epsilon/k$. For such $n$, a union bound gives $\mathbb{P}\left(\exists i : \left| \int f_i\,d\mu_n - \int f_i\,d\mathbb{Q}\right| \ge \delta\right) < \epsilon$. In particular, for $n \ge N$ we have $\mathbb{P}(\mu_n \notin U) < \epsilon$.

Now we can write $$|\mathbb{E} F(\mu_n)| \le |\mathbb{E}[F(\mu_n) ; \mu_n \in U]| + |\mathbb{E} [F(\mu_n); \mu_n \notin U]|.$$ The first term is bounded by $\epsilon$, since we have $|F| < \epsilon$ on $U$. And for $n \ge N$ the second term is bounded by $M \mathbb{P}(\mu_n \notin U) < M \epsilon$. Thus we conclude $$\limsup_{n \to \infty} |\mathbb{E} F(\mu_n)| \le \epsilon + \epsilon M.$$ Letting $\epsilon \to 0$ we have the desired conclusion.

Note this used the fact that the limiting distribution $\delta_\mathbb{Q}$ was degenerate. I'm not sure how to do it without that. We might need more assumptions on $E$ in that case.


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