TITLE: Weak-* Convergence of functionals defined by probability measures $\{\delta_n\}$
QUESTION [5 upvotes]: Consider the sequence of Dirac delta probability measures $\{\delta_n\}_{n=1}^\infty$ on $\mathbb R$, i.e. the measures which assign probability 1 to the point $n\in\mathbb R$.
These give the associated linear functionals $\Lambda_n$ defined by $\Lambda_n(f) = \int f\,d\delta_n = f(n)$ for all bounded continuous functions on $\mathbb R$, i.e. all $f\in C_b(\mathbb R)$. All these $\Lambda_n$ therefore have norm $\leq 1$.
So while this sequence is not tight and does not have a subsequence converging to a probability measure, it is contained in the compact unit ball of the weak-* topology by the Banach-Alaoglu theorem. So it has a subsequence which converges in weak-* to some linear functional $\Lambda$.
I am having a hard time seeing how this can be so. What can this limit be? How can we have that $\lim_{k\to\infty} f(n_k) = \Lambda(f)$ even exists for every bounded function?

REPLY [6 votes]: I will summarize the given comments to answer this question.
The unit ball in every dual space is weak-* compact. However, only for separable spaces, the unit ball in the dual space is sequentially weak-* compact.
The space $C_b(\mathbb R)$ is not separable.
Thus, your sequence $\{\Lambda_n\}$ may not have a weak-* convergent subsequence (and indeed, it does not have any), but it does have a weak-* convergent subnet.
If I am not mistaken (please correct me if I am wrong!), the existence of this weak-* subnet requires the axiom of choice, hence you cannot write down an explicit subnet which converges weak-*.