Let $g: [0, \infty) \rightarrow [0,1]$ be a continuous, monotone increasing function where $g(0)=0$ and $g(x)\rightarrow 1$ as $x \rightarrow \infty$. Also let $f_n:[0, \infty) \rightarrow [0,1]$ be a monotone increasing function for each $n \geq 1$ (not necessarily continuous).
I want to prove that if $f_n \rightarrow g$ pointwise, then $f_n \rightarrow g$ uniformly on $[0, \infty)$.
I thought separating into two cases might be useful. Since $g: [0, \infty) \rightarrow [0,1]$, for any given $\epsilon >0$, there exists $M$ such that $\vert g(x) - 1 \vert < \epsilon$ for all $x > M$. Then $\sup \vert f_n (x) - g(x) \vert = \vert f_n(M) -g(M) \vert \rightarrow 0$ as $n \rightarrow \infty$. Does this prove the uniform convergence on $[M, \infty)$?
For the convergence on $[0,M]$, I cannot apply Dini's Theorem since $f_n$ are not assumed to be continuous. But $g$ is uniform continuous on $[0,M]$. Maybe this helps but I cannot see it.