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I'm reading some extreme value theory and in particular regular variation in Resnick's 1987 book Extreme Values, Regular Variation, and Point Processes, and several times he has claimed uniform convergence of a sequence of functions because "monotone functions are converging pointwise to a continuous limit". I am finding this reasoning a little dubious.

EDIT: Thanks to some comments below I realized I was confused, I really want each $f_n$ to be a monotone function on $[a,b]$, so it isn't quite like Dini's theorem. Formally I am wondering:

Let $f_n:[a,b]\rightarrow\mathbb{R}$ be a sequence of non-decreasing functions converging pointwise to the continuous function $f$. Is the convergence uniform?

My thoughts:

I have been thinking that since the set of discontinuities of each function $f_n$ is at most countable, the set of discontinuities of the entire sequence is at most countable...something something? Any thoughts would be greatly appreciated!

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    $\begingroup$ @DavidMitra Isn't the limit then $f(0) = 1$, $f(x)=0$ otherwise and therefore not continuous? $\endgroup$ – mlk Jun 14 '14 at 17:41
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    $\begingroup$ @DavidMitra But then $f_n$ is not monotone, which was one of the requirements. $\endgroup$ – mlk Jun 14 '14 at 17:43
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    $\begingroup$ @mlk $f_n$ is monotone, fix $x$ and increase $n$. $\endgroup$ – BlueBuck Jun 14 '14 at 17:44
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    $\begingroup$ Ahh, sorry, the "sequence of non-decreasing functions" was a bit misleading, I thought you wanted a theorem for functions monotonic in $x$, not in $n$. (Which I think, I have seen somewhere, but cannot remember the name of) $\endgroup$ – mlk Jun 14 '14 at 17:48
  • $\begingroup$ @DavidMitra Can anything be done if we add the hypothesis that each $f_n$ is a monotone function? $\endgroup$ – BlueBuck Jun 14 '14 at 17:48
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Let $f:=\lim f_n$, and let $\varepsilon >0$ be given. We have to show that if $n$ is large enough, then $$\vert f_n(t)-f(t)\vert\leq\varepsilon\quad \hbox{for all $t\in[a,b]$}\, .$$

Since $f$ is assumed to be continuous, it is uniformly continuous on the compact interval $[a,b]$. So we may find a subdivision $a=t_0<t_1\dots <t_K=b$ of $[a,b]$ such that the oscillation of $f$ on each interval $[t_i,t_{i+1}]$ is less than $\varepsilon/2$.

Since $f_n(t_i)\to f(t_i)$ as $n\to\infty$ for $i=0,\dots ,K$, one can find $N$ such that if $n\geq N$, then $$\vert f_n(t_i)-f(t_i)\vert\leq\varepsilon/2\quad\hbox{for $i=0,\dots K$}\, .$$

Let us check that $\vert f_n(t)-f(t)\vert\leq \varepsilon$ for every $n\geq N$ and all $t\in [a,b]$.

Fix $n\geq N$, and take any $t\in [a,b]$. One can choose $i$ such that $t\in [t_i,t_{i+1}]$. Since the functions $f_n$ are non-decreasing we have $$f_n(t_i)\leq f_n(t)\leq f_n(t_{i+1})\, . $$ Since $\vert f_n(t_i)-f(t_i)\vert$ and $\vert f_n(t_{i+1})-f(t_{i+1})\vert$ are not greater than $\varepsilon/2$, it follows that $$f(t_i)-\varepsilon/2\leq f_n(t)\leq f(t_{i+1})+\varepsilon/2\, . $$ Moreover, since the oscillation of $f$ on $[t_i,t_{i+1}]$ is less than $\varepsilon/2$ and since $t\in [t_i,t_{i+1}]$, we also have $f(t_i)\geq f(t)-\varepsilon/2$ and $f(t_{i+1})\leq f(t)+\varepsilon/2$. Altogether, this gives $$f(t)-2\varepsilon/2\leq f_n(t)\leq f(t)+2\varepsilon/2\, , $$ which concludes the proof.

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  • $\begingroup$ f(ti)−ε/2≤fn(t)≤fn(ti+1)+ε/2. this line is wrong, as the right side of the inequality should be symmetric and just be the limit function. That gives the desired end form $\endgroup$ – Charlie Tian Nov 29 '17 at 5:04
  • $\begingroup$ @NikPronko: thanks for pointing out the typo. It should be OK now. $\endgroup$ – Etienne Dec 9 '17 at 10:59
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    $\begingroup$ @CharlieTian Thanks, for the same reason. $\endgroup$ – Etienne Dec 9 '17 at 11:00

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