show that $\lim_{n\to+\infty}\frac{1}{n}\sum_{k=0}^{n}f(x_{k})$ is exists let $f:[0,1]\to R^{+}$ real-valued continuous functions,and such
$$\int_{0}^{1}f(x)dx=2019,~~\int_{0}^{1}f^2(x)dx=20181027$$
(1):show that:there exists unique sequence $x_{0},x_{1},\cdots,x_{n}\in [0,1]$,such $x_{0}<x_{1}<\cdots<x_{n}$ and for any postive integer $k=1,2,\cdots,n$.such
$$\int_{x_{k-1}}^{x_{k}}f(t)dt=\dfrac{1}{n}\int_{0}^{1}f(t)dt$$
(2):and show that $$\lim_{n\to+\infty}\dfrac{1}{n}\sum_{k=0}^{n}f(x_{k})$$ is exists.and find this value.
I can do it $(1)$ exists.I think it is First mean value theorem for integration。but How to this sequence is unique. How to solve this (2),if this limt has exists.it seem use Stolz-Cesaro's Lemma:$$\lim_{n\to+\infty}\dfrac{1}{n}\sum_{k=0}^{n}f(x_{k})=\lim_{n\to\infty}f(x_{n})$$
this problem is my teacher gave me the exercise, these two problems I can not solve all , so I want to ask the teacher here,Thanks
 A: *

*First note that if the stated conditions hold, then
\begin{align}
     \int_{x_0}^{x_k}f(x)dx&=\sum_{i=1}^k\int_{x_{i-1}}^{x_i}f(x)dx\\
    &=\frac{k}{n}\int_0^1f(x)dx\ ,
   \end{align}
and, in particular,
$$
     \int_{x_0}^{x_n}f(x)dx=\int_0^1f(x)dx\ .
   $$

*Subtracting the left side of the final equation above from both sides gives
\begin{align}
    0&=\int_0^1f(x)dx-\int_{x_0}^{x_n}f(x)dx\\
  &=\int_0^{x_0}f(x)dx+\int_{x_n}^1f(x)dx\ ,
\end{align}
and since $\ 0\le x_0\le x_n\le1\ $, and $\ f\ $ is positive and continuous, it follows from this that $\ x_0=0\ $ and $\ x_n=1\ $ if the stated conditions hold

*Now, following the suggestion of Kelenner in the first of the comments, put
$$
    F(x)=\int_0^xf(t)dt\ .
 $$
Then $\ F\ $ is strictly increasing and continuous on $\ [0,1]\ $, and if the stated conditions hold, then the first equation above tells us that $\ x_k\ $ must satisfy the equation
$$
 \ F\big(x_k\big)=\frac{k}{n}\int_0^1f(x)dx\ .
$$
But since
$$
 F(0)=0<\frac{k}{n}\int_0^1f(x)dx<\int_0^1f(x)dx=F(1)\ ,
   $$
for $\ k=1,2,\dots,n-1\ $ , then by the intermediate value theorem, therefore, there must exist $\ x_{nk}\in(0,1)\ $ with
$$ F\big(x_{nk}\big)=\frac{k}{n}\int_\limits{0}^1f(x)dx\ ,$$
and since $\ F\ $ is strictly increasing, $\ x_{nk}\ $ is uniquely determined.  Moreover,
\begin{align}
  F\big(x_{nk}\big)-F\big(x_{n\,(k-1)}\big)&=\int_{x_{n\,(k-1)}}^{x_{nk}}f(x)dx\\
&=\frac{1}{n}\int_0^1f(x)dx\\
&\ge m\big(x_{nk}-x_{n\,(k-1)}\big)\ ,
   \end{align}
where $\ m=\min_\limits{x\in[0,1]}f(x)>0\ $.  Therefore, $\ 0<x_{nk}-x_{n\,(k-1)}\le$$\frac{1}{mn}\int_\limits{0}^1f(x)dx\ $, and so $\ x_{nk}-x_{n\,(k-1)}\rightarrow0\ $ as $\ n\rightarrow\infty\ $, as Paramanand Singh has observed in the comments.  For notational convenience, put $\ x_{n0}=x_0=0\ $ and $\ x_{nn}=x_n=1\ $.

*We are now in a position to prove that
$$ \lim_{n\rightarrow\infty}\sum_{k=0}^nf\big(x_{nk}\big)=\frac{\int_\limits{0}^1f(x)^2dx}{\int_\limits{0}^1f(x)dx}=\frac{20181027}{2019}\ .
 $$
Let $\ M=\max_{x\in[0,1]}f(x)\ $ and $\ \epsilon\ $ be any positive real number.  Since $\ f\ $ must be uniformly continuous on $\ [0,1]\ $, there exists a positive real number $\ \delta\ $ such that $\ \left|f(x)-f(y)\right|\le\frac{\epsilon}{M}\ $ for all $\ x,y\in[0,1]\ $ with $\ |x-y|\le\delta\ $.  Therefore, if $\ n>\frac{1}{m\delta}\int_\limits{x=0}^1f(x)dx\ $, then $\ \left|x-x_{nk}\right|\le\delta\ $, and hence $\ \left|f(x)-f\big(x_{nk}\big)\right|\le\frac{\epsilon}{M}\ $, for all $\ x\in\left[x_{nk}, x_{n\,(k+1)}\right]\ $. Hence, for all such $\ n\ $ we have
\begin{align}
\left|\,\int_\limits{0}^1f(x)^2dx-\right.&\left.\frac{1}{n}\int_\limits{0}^1f(x)dx \sum_{k=0}^{n-1}f\big(x_{nk}\big)\right|\\
&=\left|\,\int_\limits{0}^1f(x)^2dx- \sum_{k=0}^{n-1}\int_{x_{nk}}^{x_{n\,(k+1)}}f\big(x_{nk}\big)f(x)dx\right|\\
&=\left|\,\sum_{k=0}^{n-1}\int_{x_{nk}}^{x_{n\,(k+1)}}\left(f(x)-f\big(x_{nk}\big)\right)f(x)dx\right|\\
&\le\sum_{k=0}^{n-1}\epsilon\big(x_{n\,(k+1)}-x_{nk}\big)\\
  &=\epsilon\ .
  \end{align}
Thus,
$$
\lim_{n\rightarrow\infty}\frac{1}{n}\int_\limits{0}^1f(x)dx \sum_{k=0}^{n-1}f\big(x_{nk}\big)=\int_\limits{0}^1f(x)^2dx\ ,
$$
from which, since $\ \int_\limits{0}^1f(x)dx\ne0\ $, and $\ \frac{f\left(x_{nn}\right)}{n}=\frac{f(1)}{n}\rightarrow0\ $ as $\ n\rightarrow\infty\ $, it follows that
$$
\lim_{n\rightarrow\infty}\frac{1}{n} \sum_{k=0}^nf\big(x_{nk}\big)=\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=0}^{n-1}f\big(x_{nk}\big)=\frac{\int_\limits{0}^1f(x)^2dx}{\int_\limits{0}^1f(x)dx}\ .
$$
