Exercise 5, § 2

Let X be a compact space. An infinite sequence $(x_n)_{n \in \mathbf{N}}$ of points of X is said to have a limit distribution if the sequence of measures with finite support $\left( \sum_{i=0}^{n-1} \varepsilon_{x_i} \right)/n$ tends vaguely to a limit $\mu$ in $\mathscr{M}_+(X)$; one also says that the sequence $(x_n)$ is equidistributed for the measure $\mu$. If T is a total set in the Banach space $\mathscr{C}(X; \mathbf{C})$, show that for a sequence $(x_n)$ to have a limit distribution, it is necessary and sufficient that the sequence $\left( \left( \sum_{i=0}^{n-1} f(x_i) \right)/n \right)$ tend to a limit in $\mathbf{C}$ for every function $f \in T$. In particular, show that if X is a metrizable compact space, then for every sequence $(x_n)$ of points of X there exists a subsequence $(x_{n_k})$ that has a limit distribution.

If X is the interval $[0, 1]$ of $\mathbf{R}$ and if $\mu$ is Lebesgue measure on X, show that for a sequence $(x_n)$ of points of X to be equidistributed for $\mu$, it is necessary and sufficient that it be equidistributed mod 1 in the sense of GT, VII, §1, Exer. 14. For this, it is necessary and sufficient that for every integer $m \in \mathbf{Z}$, the sequence $\left( \left( \sum_{k=0}^{n-1} e^{2i\pi mx_k} \right)/n \right)$ tend to 0 for $m \neq 0$. From this deduce anew, in particular, that the sequence $x_n = n\theta - [n\theta]$ ($n \in \mathbf{N}$) is equidistributed mod 1 for every irrational number $\theta$ (GT, VII, §1, Exer. 14). What is its limit distribution when $\theta$ is rational? Cf. also Ch. IV, §5, No. 12.