Limsup of sequence of functions I have difficulty proving the following. Let $f_n:[0,1]\to [0,\infty)$ be a sequence of nondecreasing functions. First, how to show that there exists a sequence $x_n\downarrow 0$ such that $$\limsup_{n\to\infty}f_n(x_n)=\lim_{x\to 0}\limsup_{n\to\infty}f_n(x).$$
How to show then that $$\lim_{x\to 0}\limsup_{n\to\infty}f_n(x)=0$$ if and only if
$$f_n(x_n)\to 0$$ for every sequence $x_n\to 0$?
 A: 1) For $x \in [0,1]$ define 
$$H(x) = \limsup_{n\rightarrow\infty} f_n(x)$$ 
It is not difficult to show that $H(x)$ is a nondecreasing (and possibly infinite valued) function over $x\in [0,1]$. It follows that $\lim_{x\rightarrow 0^+} H(x)$ exists (possibly being $\infty$). 
2) Fix $x>0$. Let $\{x_n\}_{n=1}^{\infty}$ be any sequence of real numbers in $[0,1]$ that satisfies $\lim_{n\rightarrow\infty} x_n = 0$.  Then there is an integer $m$ such that
$$0\leq x_n \leq x \quad,  \forall n \geq m$$
From this (and the fact that each $f_n$ function is nondecreasing) it is easy to prove 
$$ \limsup_{n\rightarrow\infty} f_n(x_n) \leq \lim_{x\rightarrow 0^+} H(x) $$
3) We prove there exists a particular sequence $\{x_n\}_{n=1}^{\infty}$ for which the following reverse inequality holds: 
$$ \limsup_{n\rightarrow\infty} f_n(x_n) \geq \lim_{x\rightarrow 0^+} H(x) $$
For simplicity assume that $H(1)<\infty$, and so $H(1/k)<\infty$ for all $k\in\{1, 2, 3, ...\}$ (the general result is similar).


*

*Define integer $n[1]>0$ such that $f_{n[1]}(1) \geq H(1) - 1$. 

*For $k \in \{2, 3, 4, ...\}$, define integer $n[k]>n[k-1]$ such that $f_{n[k]}(1/k) \geq H(1/k) - 1/k$. 
Now consider a subsequence $\{x_n\}_{n=1}^{\infty}$ such that $x_{n[k]} = 1/k$ for each $k \in \{1, 2, 3, ...\}$, and the in-between values are held constant. 
