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 $\mathcal{P}(X)$)?
 A: 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 $\delta_0(E)=1$. 
A: 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.
