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.


For any fixed $m \in \mathbb{N}$ you can find points $0=x_0< x_1 < \ldots < x_{m-1}$ with $g(x_k) = k/m$. By pointwise convergence there is $N$ such that $|g(x_k) - f_n(x_k)|<1/m$ for $n\ge N$ and for all $k$. Now $g$ maps the interval $[x_{k-1}, x_k]$ to $[(k-1)/m, k/m]$, and $f_n$ maps the same interval into $((k-2)/m, (k+1)/m)$, which implies $|f_n - g| < 2/m$ on $[x_{k-1}, x_k]$, for all $n \ge N$ and $k=1,\ldots,m-1$. A slight modification shows the same estimate on the interval $[x_{m-1}, \infty)$, so you get uniform convergence.

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    $\begingroup$ For the case $[x_{m-1}, \infty)$, does the uniform convergence follow from my argument in the post? $\endgroup$ – Karatug Ozan Bircan Nov 25 '14 at 1:11
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    $\begingroup$ Yes, it does, that is true. $\endgroup$ – Lukas Geyer Nov 25 '14 at 1:23

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