show that if $\lim (f(x_1)+\cdots +f(x_n))/n$ exists whenever $\lim (x_1+\cdots +x_n)/n$ exists, then $f$ is continuous 
Show that if $f:[0,1]\to \mathbb{R}$ and$$\lim \limits _{n\to \infty}\frac{f(x_1)+\cdots + f(x_n)}{n}$$exists whenever$$\lim \limits _{n\to \infty}\frac{x_1+\cdots + x_n}{n}$$exists, where $(x_n)$ is a sequence of real numbers, then $f$ is continuous.

It suffices to show that if $x_n\to x$, then $f(x_n)\to f(x)$. It might be useful to show this via a contradiction; suppose there is a sequence $x_n$ so that $x_n\to x\in \mathbb{R}$ but $f(x_n)\not \to f(x)$. We know that $\dfrac{f(x_1)+\cdots +f(x_n)}n$ exists and equals $L$, say. By definition, there exists $\varepsilon >0$ so that for all $N$, $\exists n\geq N$ such that $|f(x_n)-f(x)|\geq \varepsilon$. But I'm not sure how to get a contradiction from here.
 A: First show that if $x_n\rightarrow x$, then $\frac{\sum_{i=1}^n x_i}{n}\rightarrow x$ (and if $0\le x_n \le 1$, so is $0\le\frac{\sum_{i=1}^n x_i}{n}\le1$, we don't have to worry about the domain at all).
Now by the condition, $a:=\lim\frac{\sum_{i=1}^n f(x_i)}{n}$ exists. If $f(x)\not=a$, then the strategy is clear: Adding any number of $x$ to $x_n$ to create $\{y_i\}$ won't change the fact that the average $\frac{\sum_{i=1}^ny_i}{n}$ still converges, but it would make the average $\frac{\sum_{i=1}^n y_i}{n}$ oscillate around $a$ and $f(x)$.
To be more specific, first we pick $y_1=x_1, \cdots, y_n=x_n$ such that $\frac{\sum_{i=1}^n y_i}{n}$ is very close to $a$, now we set $y_{n+1}=\cdots = y_m = x$ for sufficiently large $m$ to bring the average close to $f(x)$, and then back to set $y_{m+1}=x_{n+1}, \cdots$ to bring the average once again close to $a$, and so on.
This shows $f(x) = \lim \frac{\sum_{i=1}^n f(x_i)}{n}$. Now if $f(x_n)$ doesn't converge to $f(x)$, we may use similar strategy to pick a subsequence such that the average oscillate.
A: Let $x_n \to x$
$$L := \liminf\limits_{n\to\infty} f\left(x_n\right)$$
There is an increasing sequence of integers $\left(\ell_p\right)_p$ such that $$\lim_{p\to\infty} f\left(x_{\ell_p}\right) = L$$
I will assume that $-\infty <L < \infty$, Using the fact that:

$$\forall m\in \Bbb N;\; \lim_{n\to\infty} \frac{1}{n+m}\left(\sum_{k=1}^n f\left(x_{\ell_k}\right)+ mf(x)\right) = L$$ and $$\forall n\in \Bbb N;\; \lim_{m\to\infty} \frac{1}{n+m}\left(\sum_{k=1}^n f\left(x_{\ell_k}\right)+ mf(x)\right) = f(x)$$

you can easily construct (by induction) two increasing sequences of integers $\left(m_p\right)_{p\in \Bbb N}$, $\left(n_p\right)_{p\in \Bbb N}$, such that $m_0= n_0 = 0$ and


*

*$\forall p\in\Bbb N_{\ge 1},$
$$\left|\frac1{n_{p}+m_p}\left(\sum_{k=1}^{n_{p}} f\left(x_{\ell_k}\right) + m_pf(x)\right) - f(x)\right| < \frac1{p};$$ and $$\left|\frac1{n_{p+1}+m_p}\left(\sum_{k=1}^{n_{p+1}} f\left(x_{\ell_k}\right) + m_pf(x)\right) - L\right| < \frac1{p}$$

Let $$z_k = \begin{cases}x_{\ell_{k-m_p}} & \text{if $n_p+m_p < k \le n_{p+1}+m_p$}\\
x & \text{if $n_{p+1} + m_p < k \le n_{p+1} + m_{p+1}$}\end{cases}$$
So $$\limsup_{k\to \infty}\left|z_k-x\right|\le \limsup_{k\to \infty} \left|x_{\ell_k}-x\right| = 0$$
This proves that: $\lim\limits_{k\to\infty} z_k = x$, So $$\lim_{n\to\infty} \frac1n \sum_{k=1}^n f\left(z_k\right)$$ exists. On the other hand $$\frac1{n_p + m_p}\sum_{k=1}^{n_p+m_p}f(z_k) = \frac1{n_{p}+m_p}\left(\sum_{k=1}^{n_{p}} f\left(x_{\ell_k}\right) + m_pf(x)\right) \underset {p\to \infty} \to f(x)$$
and $$\frac1{n_{p+1} + m_p}\sum_{k=1}^{n_{p+1}+m_p}f(z_k) = \frac1{n_{p+1}+m_p}\left(\sum_{k=1}^{n_{p+1}} f\left(x_{\ell_k}\right) + m_pf(x)\right) \underset {p\to \infty} \to L$$
By unicity of the limit $L=f(x)$. By doing the same thing for the $\limsup$, you have $$f(x) = \liminf_{n\to\infty} f\left(x_n\right) = \limsup_{n\to \infty} f\left(x_n\right).$$
