# Continuity of $\mu \mapsto \mu(E)$ for $\mu$ probability measure and $E$ Borel subset

Let $X$ be a topological space endowed with the Borel sigma-algebra, let $\mathcal{P}(X)$ be the set of all Borel probability measures on $X$, endowed with the weak* topology. Fix $E$ Borel subset of $X$. Is the function $f\colon \mathcal{P}(X) \to [0,1]$, defined by $f(\mu)=\mu(E)$, a continuous function? Or, at least, is it measurable (wrt the Borel sigma algebra of $\mathcal{P}(X)$)?

For Borel measurability, the answer should be yes, at least if $X$ is a metric space. Here's an argument using Dynkin's $\pi$-$\lambda$ lemma.

Let $\mathcal{B}$ be the Borel $\sigma$-algebra on $X$, and for each $E \in \mathcal{B}$ let $f_E(\mu) = \mu(E)$. Let $$\mathcal{L} = \{ E \in \mathcal{B} : f_E \text{ is Borel}\}.$$ We will use Dynkin's lemma to show $\mathcal{B} \subset \mathcal{L}$. It's easy to show that $\mathcal{L}$ is a Dynkin system:

• $X \in \mathcal{L}$ since $f_X$ is the constant function 1, which is certainly Borel.

• Suppose $E \in \mathcal{L}$, so that $f_E$ is Borel. But $f_{E^c} = 1 - f_E$ which is thus also Borel. So $E^c \in \mathcal{L}$.

• Suppose $E_1, E_2, \dots \in \mathcal{L}$ are pairwise disjoint, and let $E = \bigcup_n E_n$. Then for any $\mu \in \mathcal{P}(X)$, we have $f_E(\mu) = \mu(E) = \sum_n \mu(E_n) = \sum_n f_{E_n}(\mu)$ by countable additivity of $\mu$. So $f_E$ is a pointwise limit of Borel functions, hence is also Borel, and so $E \in \mathcal{L}$.

Now we need to show that $\mathcal{L}$ contains a $\pi$-system that generates $\mathcal{B}$. Let's show that $\mathcal{L}$ contains all the closed sets. So let $E \subset X$ be closed. Define $\varphi : X \to [0,1]$ by $\varphi(x) = \max(1 - d(x,E), 0)$, so that $\varphi$ is continuous, $\varphi(x) = 1$ for $x \in E$ and $\varphi(x) < 1$ for $x \notin E$. Then $\lim_{n \to \infty} \varphi(x)^n = 1_E(x)$. So by dominated convergence, for every $\mu \in \mathcal{P}(X)$ we have $$f_E(\mu) = \int 1_E\,d\mu = \lim_{n \to \infty}\int \varphi^n \,d\mu.$$ But since $\varphi^n$ is bounded and continuous for each $n$, by definition of the weak-* topology, the map $\mu \mapsto \int \varphi^n$ is continuous. So we have written $f_E$ as a pointwise limit of continuous functions, hence $f_E$ is Borel. Thus we have shown $E \in \mathcal{L}$, so $\mathcal{L}$ contains all the closed sets.

By Dynkin's lemma, $\mathcal{B} \subset \mathcal{L}$ and we are done.

More generally, the same proof works if $X$ is perfectly normal.

For the continuity, the answer in no in general. An easy counterexample is the following: $X=[0,1]$ with the euclidean topology, and $E=\{0\}$. The sequence of Dirac measures $\delta_{1/n}$ converges to $\delta_0$ w.r.t. weak* topology, but $\delta_{1/n}(E)=0$ for all $n$, while $\delta_0(E)=1$.

• +1. But please use \{ and \} to produce { and }. – Did Feb 2 '15 at 3:16
• Oh sorry, a forgetfulness. – Capublanca Feb 2 '15 at 3:18