I have a problem with a proof which seems to be easy, but I don't have any idea how to prove in the opposite side (from the end to the beginning).
We know that $\{x_n\}$ has uniform distibution on $[0,1]$ when $$ \lim_{n\rightarrow \infty}\dfrac1n \sum\limits_{j=1}^n f(x_j)=\int\limits_0^1 f(x) dx $$ is fulfilled for every function $f$ which is continuous on $[0,1]$. I have to prove that above condition is equivalent to the condition $$ \forall_{k\in \mathbb{N}} \lim_{n\rightarrow \infty} \dfrac1n \sum\limits_{j=1}^n x_j^k=\dfrac{1}{k+1}. $$
My attempt
Proof.
($\Rightarrow$) Taking $f(x_j)=x_j^k$ we obtain $$\lim_{n\rightarrow \infty} \dfrac1n \sum\limits_{j=1}^n x_j^k=\int\limits_0^1x^k dx=\dfrac{x^{k+1}}{k+1}|_0^1 =\dfrac{1}{k+1} \forall_{k\in \mathbb{N}}.$$
($\Leftarrow$) ???
Thanks for any ideas, because this side of the proof is kind of enigmatic for me.