If $\{f_n(x)\}_{n=1}^{\infty}$ is formed by non-decreasing functions. And for any $x\in [a,b] $, it converges to $f(x)$ If $\{f_n(x)\}_{n=1}^{\infty}$ is formed by non-decreasing functions. And for any $x\in [a,b] $, it converges to $f(x)$  which $f(x)$ is continuous. Prove $\{f_n(x)\}_{n=1}^{\infty}$ converges uniformly to $f(x)$.  
I know the condition we got is $\forall \epsilon \gt0$, and $\forall x_0\in[a,b]$, there exists a $N(x_0,\epsilon)$ such that $|f_n(x)-f(x)|\lt \epsilon$ for $n\gt N(x_0,\epsilon)$. And we need to proof  $\forall \epsilon \gt0$, and $\forall x_0\in[a,b]$, there exists a $N(\epsilon)$ such that $|f_n(x)-f(x)|\lt \epsilon$ for $n\gt N(\epsilon)$. But I don't know how to use the condition $f_n$ is non-decreasing.
 A: This fails. Counterexample : $f_n(x) = x^n$ on $[0,1]$, and $f_n\to f$, where $f[0,1) =0$, $f(1) =1$, but $f_n\not \rightrightarrows f$. 
UPDATE 
If you are dealing with Dini theorem, then the statement should be 

Suppose $(f_n)_1^\infty$ is a sequence of continuous functions on $[a,b] \subset \Bbb R$, and $f_n \to f$ where $f$ is also continuous on $[a,b]$. If for each $x\in[a,b], (f_n(x))_1^\infty$ is a non-decreasing numerical sequence, then $f_n \rightrightarrows f [n \to \infty, x \in [a,b]]$. 

To prove this, for fixed $\varepsilon > 0$ you could find for each $x \in [a,b]$ an $N_x$ s.t. $0 \leqslant (f - f_{N_x})(x) < \varepsilon$. By continuity, you could make the inequality be valid in an open interval containing $x$. Now you get an open covering of $[a,b]$. Use the Heine-Borel finite open covering theorem, you would find a common $N$ for all $x$. Now use the monotonicity. 
A: The idea is that monotonicity together with the (uniform) continuity of the pointwise limit allows a uniform control over convergence speed by means of squeezing.
Fix arbitrary $\epsilon > 0$. Then we prepare $\delta > 0$ and $K, N \geq 1$ as follows:


*

*Pick $\delta > 0$ such that $|f(x) - f(y)| < \epsilon/2$ whenever $x, y \in [a, b]$ satisfy $|x -y| < \delta$. This is possible since $f : [a, b] \to \mathbb{R}$ is uniformly continuous.

*Pick a partition $\{a = x_0 < x_1 < \cdots < x_K = b\}$ so that $|x_k - x_{k-1}| < \delta$ for each $k = 1, \cdots, K$.

*Pick $N$ such that $|f_n(x_k) - f(x_k)| < \epsilon/2$ whenever $n \geq N$ and $k = 0, \cdots, K$.
Now we claim that $\sup_{x\in[a,b]} |f_n(x) - f(x)| < \epsilon$ whenever $n \geq N$, which is enough to establish the uniform convergence. Indeed, for each $n \geq N$ and $x \in [a, b]$, pick $k$ such that $x \in [x_{k-1}, x_k]$. Since both $f_n$ and $f$ are non-decreasing,
\begin{align*}
f_n(x) - f(x)
&\leq f_n(x_k) - f(x_{k-1}) \\
&= \left( f_n(x_k) - f(x_k) \right) + \left( f(x_k) - f(x_{k-1}) \right) \\
&\leq \frac{\epsilon}{2} + \frac{\epsilon}{2}
= \epsilon.
\end{align*}
Similar argument shows that
\begin{align*}
f(x) - f_n(x)
&\leq f(x_k) - f_n(x_{k-1}) \\
&= \left( f(x_k) - f(x_{k-1}) \right) + \left( f(x_{k-1}) - f_n(x_{k-1}) \right) \\
&\leq \frac{\epsilon}{2} + \frac{\epsilon}{2}
= \epsilon.
\end{align*}
Consequently we obtain $|f_n(x) - f(x)| < \epsilon$. Since this is true for all $n\geq N$ and $x \in [a, b]$, the desired conclusion follows.
