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Let $\{f_n\}$ be a sequence of monotonically increasing functions on $\mathbb{R}$.

Let $\{f_n\}$ be uniformly bounded on $\mathbb{R}$.

Then, there exists a subsequence $\{f_{n_k}\}$ pointwise convergent to some $f$.

Now, assume $f$ is continuous on $\mathbb{R}$.

Here, I want to prove that $f_{n_k}\rightarrow f$ uniformly on $\mathbb{R}$.

How do i prove this ?

I have proven that " $\forall \epsilon>0,\exists K\in\mathbb{N}$ such that $k≧K \Rightarrow \forall x\in\mathbb{R}, |f(x)-f_{n_k}(x)||<\epsilon \bigvee f_{n_k}(x) < \inf f + \epsilon \bigvee \sup f - \epsilon < f_{n_k}(x)$ ".

The argument is in the link below.

I don't understand why above statement implies "$f_{n_k}\rightarrow f$ uniformly on $\mathbb{R}$". Please explain me how..

Reference ; http://www.math.umn.edu/~jodeit/course/SP6S06.pdf

Thank you in advance!

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    @Learer That's not the original question. It says "Prove $f_{n_k}\rightarrow f$ uniformly on $\mathbb{R}$" on p.167 Rudin PMA – 2012-12-26
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    @Learner Yes, so i'm asking that how does "$f_{n_k}$ converges uniformly over compacts on $\mathbb{R}$l" imply that " $f_{n_k}$ converges uniformly on $\mathbb{R}$". – 2012-12-26
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    It's not true. For example, consider $f_n(x)=I(x-n)$ where $I(\le0)=0$ and $I(>0)=1$. [Reference](http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_analyst_2004;task=show_msg;msg=0657.0001) – 2013-02-16

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