Can we always find $x_n\to\infty$ with $f_n(x_n)\to0$ if the $(f_n)$ are non-decreasing and pointwise convergent to zero? Suppose we have a sequence of functions $(f_n)$ with each $f_n:[0, \infty) \to \mathbb{R}$, each $f_n$ is monotone non-decreasing and $f_n\to 0$ pointwise. Is it always possible to choose a sequence $(x_n)$ with $x_n\to \infty$ slow enough such that we still have $f_n(x_n) \to 0$? If not, is this possible with additional hypotheses on the $(f_n)$ (e.g. continuity perhaps)?
 A: There is no need for the $f_n$ to be monotone non-decreasing.
Given $n\in \Bbb Z_+$, let $$m(n):=\min\{m\in\Bbb Z_+:|f_k(j)|<1/n \;(j=1,...,n)\;\text{for}\, k\geqslant m\}.$$ Now define $M(0):=0$ and  $M(n):=\max\{n,M(n-1)+1,\,m(1),...,m(n)\}\;(n\in\Bbb Z_+).$ Then $$|f_{M(n)}(n)|<\frac1n\quad(n\in\Bbb Z_+).$$Now $M(\centerdot)$ is strictly increasing and so has an inverse $M^{-1}:M(\Bbb N)\to\Bbb N$ which we can extend to a map $L:\Bbb N\to\Bbb N$ by defining $L(0):=0$ and $$L(k):=M^{-1}(l)\quad\text{for}\;k=M(l)+1,...,M(l+1)\quad(l\in\Bbb N).$$It follows that $$|f_k(L(k))|<\frac1{L(k)}\quad(k\in\Bbb Z_+).$$Note that $L$ is increasing and unbounded. Thus, by taking $x_n:=L(n)\;(n\in\Bbb Z_+)$, we obtain a sequence with the required property that $f_n(x_n)\to0$ as $n\to\infty$.
A: I believe the answer to be Yes.
Do you want $(x_n)_n$ to be strictly increasing or not necessarily?
If not, it suffices to only try to construct the sequence via the integers, i.e. $\forall n, \space x_n \in \mathbb{N} .$ It is therefore sufficient to look at $f_n|_{\mathbb N}=\hat f_n: \mathbb N \to \mathbb R$.
Let us look at the auxiliary functions $\hat f_n^k(x) = \chi_{x\le k} \hat f_n(x) +  \chi_{x> k}\hat f_n(k)$ where $\chi_{(.)}$ is the indicator function. One can see that for fixed $k$, $\hat f_n^k \to \hat f^k (\equiv 0)$ uniformly in $n$.
Namely given a pair of positive integers $(m,k)$, we can find a positive integer $n(m,k)$ such that for all $n\ge n(m,k)$ we have $\|\hat f_n^k\| \le \frac{1}{m} $. Now we can choose, in an increasing way, a sequence $(t_r)_{r \in \mathbb N} =(r+\max\{ n(1,1),\cdots, n(r,r)\})_r$ so that $t_r \to \infty$ and increases strictily. Finally, choose $x_n = r$ for $n \in [t_r,t_{r+1})$.
Remark: The construction may look difficult, but it's simply "waiting at the k-th step until you are sure that the (k+1)-st step becomes small enough, then stepping on the (k+1)-st one"

Now if you want $(x_n)$ to be strictly increasing, I feel like a key insight can be drawn from
Lusin-Egorov's theorem: If $f_n$ converges almost surely to $f$ (which is the case here with $f=0$) then, on every compact interval $I$ and for any $\varepsilon>0$ there exists a compact subset $I_\varepsilon \subset I$ with $\lambda(I_\varepsilon)\ge \lambda(I) - \varepsilon$ and $f_n|_{I_\varepsilon} \to f|_{I_\varepsilon}$ uniformly.
Choosing $\varepsilon<1$ and approaching the problem as above should yield a solution.
