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Assume $(f_n)$ is a monotically increasing sequence of functions on $\mathbb{R}$ with $0\leq f_n(x)\leq 1$ for all $n$ and all $x$. There are two parts to this. In the first, I show that $\exists f$ and a sequence $(n_k)$ such that $f(x)=\lim\limits_{k\to\infty}f_{n_k}(x)$ for all $x$. In my work so far, I think I have sufficiently shown this.

However, if I also assume that $f$ is continuous, I want to show that $f_{n_k}\to f$ uniformly on compact sets.

My thoughts: Take any compact set $K\subseteq\mathbb{R}$ and since $K$ is closed and bounded, there exists a closed and bounded interval $[a,b]$ such that $K\subseteq [a,b]$. If we can show uniform convergence on $[a,b]$, then it follows for $K$ and we are done. I then want to say something like:

"Let $\varepsilon >0$ and choose $\delta$ such that $\{x_1,\dots x_l\}$ is a $\delta$-dense set in $[a,b]$ where, by continuity and monotonicity, $x_m\leq x\leq x_{m+1}$ implies $f(x_m)-\varepsilon\leq f_{n_k}(x_m)\leq f_{n_k}(x)\leq f_{n_k}(x_{m+1})\leq f(x_{m+1})+\varepsilon$."

in order to begin showing this convergence within $\varepsilon.$ I am painfully aware of how incorrect the above is, but can't find a the right way to phrase my goal. Any help is greatly appreciated.

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  • $\begingroup$ Is $f_{n}$ continuous also (for the uniform convergence case)? $\endgroup$
    – user284331
    Nov 28, 2017 at 22:28

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For the uniformly convergence case, I assume further that those $f_{n}$ are continuous.

As what you have said, we need only to deal with $[a,b]$. For any $\epsilon>0$ and any $x\in[a,b]$, choose $n_{x}>0$ and $\delta_{x}>0$ such that $n\geq n_{x}$ implies $|f_{n}(x)-f(x)|<\epsilon$ and that $y\in B_{\delta_{x}}(x)$ implies $|f(y)-f(x)|<\epsilon$ and $|f_{n_{x}}(y)-f_{n_{x}}(x)|<\epsilon$. Compactness of $[a,b]$ implies some $\delta_{x_{1}},...,\delta_{x_{N}}$ such that $[a,b]\subseteq B_{\delta_{x_{1}}}(x_{1})\cup\cdots\cup B_{\delta_{x_{N}}}(x_{N})$. Now take $n_{0}=\max\{n_{x_{1}},...,n_{x_{N}}\}$ so for any $n\geq n_{0}$ and any $y\in[a,b]$ we have some $y\in B_{\delta_{x_{k}}}(x_{k})$ and hence
\begin{align*} |f_{n}(y)-f(y)|&=f(y)-f_{n}(y)\\ &=f(y)-f(x_{k})+f(x_{k})-f_{n_{x_{k}}}(x_{k})+f_{n_{x_{k}}}(x_{k})-f_{n_{x_{k}}}(x)+f_{n_{x_{k}}}(x)-f_{n}(y)\\ &\leq f(y)-f(x_{k})+f(x_{k})-f_{n_{x_{k}}}(x_{k})+f_{n_{x_{k}}}(x_{k})-f_{n_{x_{k}}}(y)\\ &<3\epsilon. \end{align*}

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