Prove that the limit $\lim_{n \to \infty} \frac{1}{n} \sum_{k = 1}^{n} f(x_k)$ exists 
Prove that the limit $\lim_{n \to \infty} \frac{1}{n} \sum_{k = 1}^{n} f(x_k)$ exists given that $\lim_{n \to \infty} \frac{1}{n} \sum_{k = 1}^{n} x_k^i$ exists for every $i \in \{0,1,\dots \}$ and $x_k \in (0,1).$ and $f$ is a continuous function on $[0,1]$.

At first, the sums look like Riemann sums to he me, but I don't think it will help me prove the existence of the limit. I feel like this is a nested limit argument. The presence of the index $i$  also bothers me. Here are my thoughts,
\begin{align}
\lim_{n \to \infty} \frac{1}{n} \sum_{k = 1}^{n} f(x_k) &= f(\lim_{n \to \infty} \frac{1}{n} \sum_{k = 1}^{n} x_k) \\
&=f(x)
\end{align}
where $x = \lim_{n \to \infty} \frac{1}{n} \sum_{k = 1}^{n} x_k$ and the exchange of composition is justified by continuity. But this problem occurs in a chapter from the Weistrass Approximation thoerm, so I suspect something about him is needed? If I could get a hint it would be great.
Also it looks like $lim_{n \to \infty} \frac{1}{n} \sum_{k = 1}^{n} x_k^i$ is the polynomial I need. 
 A: Your reasoning is wrong because it assumes that $f$ is linear (or almost): you assumed 
$$
f\left(\frac1n\sum_{k=1}^nx_k\right)=\frac1n\sum_{k=1}^nf(x_k);
$$
This fails for most continuous functions and most choices of $x_1,\ldots,x_n$. 
If $p$ is a polynomial,  $p(t)=\sum_{j=0}^ma_jx^j$. Then, writing $y_j=\lim_{n\to\infty}\frac1n\sum_{k=1}^nx_k^j$,
$$
\lim_{n\to\infty}\frac1n\sum_{k=1}^np(x_k)=\lim_{n\to\infty}\frac1n\sum_{k=1}^n\sum_{j=0}^ma_jx_k^j
=\sum_{j=0}^ma_j\lim_{n\to\infty}\frac1n\sum_{k=1}^nx_k^j=\sum_{j=1}^na_jy_j
$$
exists. 
Now, by Weierstrass, for each $m$ there exists a polynomial $p_m$ with $|f(t)-p_m(t)|<1/m$ for all $t\in(0,1)$. Then
$$
\left|\frac1n\sum_{k=1}^np_{m_1}(x_k)-\frac1n\sum_{k=1}^np_{m_2}(x_k)\right|\leq\frac1n\sum_{k=1}^n|p_{m_1}(x_k)-p_{m_2}(x_k)|\\ \leq\frac1n\sum_{k=1}^n|p_{m_1}(x_k)-f (x_k)|+|f (x_k)-p_{m_2}(x_k)|\leq\frac1n\,\sum_{k=1}^n\frac1 {m_1}+\frac1 {m_2}=\frac1 {m_1}+\frac1 {m_2}.
$$
Thus the numbers $\frac1n\sum_{k=1}^np_{1/m}(x_k)$ converge, and an estimation similar to the above one shows that $\frac1n\sum_{k=1}^nf(x_k)$ converges to that number. 
