Show that $\lim_{n \to \infty} (1/n)\sum_{i=1}^n f(x_i)$ exists Problem: Let $(x_i)$ be a sequence in (0,1) such that $\lim_{n \to \infty} (1/n)\sum_{i=1}^n x_i^k$ exists for every $k=0,1,2,...$ and $f\in C[0,1]$. Show that $\lim_{n \to \infty} (1/n)\sum_{i=1}^n f(x_i)$ exists.
I've used the fact that $B_n(f)$ converges uniformly to $f$ to get 
$\lim_{n \to \infty} (1/n)\sum_{i=1}^n f(x_i) = \lim_{n \to \infty} (1/n)\sum_{i=1}^n \sum_{k=0}^n f(\frac{k}{n}) {n \choose_k }x_i^k (1-x_i)^{n-k}$
And then you can arrange to get 
$\lim_{n \to \infty} (1/n) \sum_{k=0}^n \left[ f(\frac{k}{n}) {n \choose_k }\sum_{i=1}^nx_i^k (1-x_i)^{n-k}\right]$
I want to use the fact that $(1-x_i) < 1$ to consolidate some of the equation but I get nowhere. 
 A: To show convergence, we must show that
$$A_n = \frac1n\sum_{i=1}^n f(x_i)$$
is a Cauchy sequence.
Let $\varepsilon > 0$ be given. By the Weierstraß approximation theorem, there is a polynomial $p$ with $\lVert p-f\rVert_\infty < \varepsilon/4$. For all $n$, we have
$$\left\lvert\frac1n\sum_{i=1}^n f(x_i) - \frac1n\sum_{i=1}^n p(x_i) \right\rvert < \frac{\varepsilon}{4}.$$
Since by assumption $\frac1n\sum\limits_{i=1}^n x_i^k$ exists for all $k$, and $p$ is a polynomial,
$$P = \lim_{n\to\infty} \underbrace{\frac1n \sum_{i=1}^n p(x_i)}_{B_n}$$
exists. Choose an $N \in \mathbb{N}$ such that
$$\left\lvert\frac1n\sum_{i=1}^n p(x_i) - P \right\rvert < \frac{\varepsilon}{4}$$
for all $n \geqslant N$.
Then, for $n,m \geqslant N$ we have
$$\begin{align}
\lvert A_n - A_m\rvert &= \lvert A_n - B_n + B_n - P + P - B_m + B_m - A_m\rvert\\
&\leqslant \lvert A_n - B_n\rvert + \lvert B_n - P\rvert + \lvert P- B_m\rvert + \lvert B_m - A_m\rvert\\
&< 4\frac{\varepsilon}{4} = \varepsilon.
\end{align}$$
So $(A_n)$ is a Cauchy sequence.
On a more abstract level, consider the linear functionals $T_n \colon C([0,1]) \to \mathbb{R}$ (or $\mathbb{C}$) given by
$$T_n(f) = \frac1n \sum_{i=1}^n f(x_i).$$
We have $\lVert T_n\rVert = 1$ for all $n \geqslant 1$, so $\mathscr{T} = \{ T_n : n \in \mathbb{Z}^+\}$ is an equicontinuous family. By assumption, $\lim\limits_{n\to\infty} T_n(p_k)$ exists for all monomials $p_k \colon x \mapsto x^k$, hence $\lim\limits_{n\to\infty} T_n(p)$ exists for all polynomials $p$ by linearity. Since the subspace of polynomials is dense in $C([0,1])$ and $\mathscr{T}$ is equicontinuous, $\lim\limits_{n\to\infty} T_n(f)$ exists for all $f\in C([0,1])$, and $T = \lim\limits_{n\to\infty} T_n$ is a continuous linear operator (with $\lVert T\rVert \leqslant 1$). Note that it does not follow that the $T_n$ converge to $T$ in the norm topology of $C([0,1])^\ast$, only the convergence in the weak topology of $C([0,1])^\ast$ is guaranteed.
