Let $X$ is a metric space and $\mathcal{A}$ the canonical Borel sigma algebra on $X$. Let $(\mu_n)_{n \in \mathbb{N}}$ be a sequence of probability measures on a measurable space $(X,\mathcal{A})$ and suppose $\mu_n \to \mu$ weakly for $n \to \infty$ i.e.
$$
\int_X f d\mu_n \underset{n \to \infty}{\longrightarrow} \int_X f d\mu
$$
for all $f \in C_b(X)$.
- What can we say about $\mu$? Is it a probability measure? A finite measure?
Feel free to link to theorems etc.
Best Answer
Provided $X$ is a Polish (i.e. complete and separable) metric space, $\mu$ is a probability measure. First, we test against the constant function $1$, explicitly
$$ 1 \equiv \int_X 1\,\mathrm{d}\mu_n \to \int_X 1\,\mathrm{d}\mu =\mu(X)$$
It remains to check that $\mu$ is a positive measure. By Hahn's decomposition theorem, write $X=X_+\cup X_-$ which is a disjoint union, with $\mu(E)\ge0$ for all $E\subseteq X_+$ and $\mu(E)\le0$ for all $E\subseteq X_-$. This means also $\mu=\mu_+-\mu_-$ where $\mu_+,\mu_-$ are the positive measures
$$\mu_\pm(E) := \mu(E\cap X_\pm) $$
Then, for any compact $K\subset X_-$ and $N\in\mathbb{N}_+$, consider the non-negative, bounded continuous function
$$ f_{K,N} := \max\{1-N\,\mathrm{dist}(x,K),0\} $$
By weak convergence,
$$ \int_Xf_{K,N}\,\mathrm{d}\mu = \lim_{n\to\infty}\int_Xf_{K,N}\,\mathrm{d}\mu_n\ge0$$
So
$$\int_Xf_{K,N}\,\mathrm{d}\mu_+ - \int_Xf_{K,N}\,\mathrm{d}\mu_-\ge0$$
Now take $N\to\infty$ and apply dominated convergence theorem,
$$-\mu_-(K) = \mu_+(K) - \mu_-(K) \ge0$$
Since $K$ is an arbitrary subset of $X_-$, by regularity of finite Borel measures on Polish spaces we see that $\mu_-$ is the zero measure. Hence $\mu=\mu_+$ is a positive measure.