Weak convergence of $\mu_n = \frac{1}{2n +1} \sum_{k=-n}^{n} \delta_{\frac{k}{n}}$

Question

Show that sequence of probabilities $$ \mu_n = \frac{1}{2n +1} \sum_{k=-n}^{n} \delta_{\frac{k}{n}} $$ converges weakly. To which probability?

Now, I've written down the first elements: $$ \mu_1 = \frac{1}{3}(\delta_{-1} + \delta_{0} + \delta_{1}) $$ $$ \mu_2 = \frac{1}{5}(\delta_{-1} + \delta_{\frac{-1}{2}} + \delta_{0} +\delta_{\frac{1}{2}} + \delta_{1}) $$ and so on. It is a convex combination of Dirac's deltas which "denses" around $[-1, 1]$. It can't be $U(-1,1)$ because no irrational numbers will ever fall into $\mu_n$.

What probability is that? How can I show the weak convergence?

Answer 1

Hint: let $f\colon \mathbb R\to \mathbb R$ be a continuous and bounded function. Compute the limit $\lim_{n\to +\infty}\int_{\mathbb R}f(x)d\mu_n(x)$ by writing the integral as a Riemann sum: $$ \int_{\mathbb R}f(x)d\mu_n(x)=\frac 1{2n+1}\sum_{k=-n}^nf\left(\frac kn\right). $$