Suppose that $\{f_n\}$ is a sequence of nondecreasing functions which map the unit interval into itself. Suppose that $\lim_{n\rightarrow \infty} f_n(x)=f(x)$ pointwise and that $f$ is a continuous function. Prove that $f_n(x) \rightarrow f(x)$ uniformly as $n \rightarrow \infty$, $0\leq x\leq1$. Note that the functions $f_n$ are not necessarily continuous.
This is one of the preliminary exam from UC Berkeley, the solution goes like this:
Because $f$ is continuous on $[0,1]$, which is compact, it is then uniformly continuous. Hence there exists $\delta >0$ such that if $|x-y|<\delta$ then $|f(x)-f(y)|<\epsilon$.
We then partition the interval with $x_0=0, \cdots ,x_m=1$ such that the distance $x_{i}-x_{i-1}$ is less than $\delta$.
Note that since there are only finite number of $x_m$, there is $N\in \mathbb{N}$ such that if $n\geq N$ then $|f_n(x_i)-f(x_i)|<\epsilon$ where $i=0,\cdots, m$
Now if $x\in[0,1]$, then $x\in[x_{i-1},x_i]$ for some $i\in\{1, \cdots m\}$.
My question is how to use the nondecreasingness to arrived at this inequality, for $n\geq N$
$f(x_{i-1})-\epsilon
Can someone please help, I have been staring at the inequality for about a day now. Thanks.