How can I show that the convergence of a sequence of functions is strictly increasing?

Question

I am taking an advanced course in calculus and I am having hard times with this question. I have that $\{ r_1, r_2,...\}$ is an enumeration of $Q$ and $f_n(x)= \frac{1}{n^2} $ if $x>r_n$ and $f_n(x)= 0 $ if $x\le r_n$ and f(x) = $\sum_{n=1}^{\infty}f_n(x)$ and the question is to show that $f$ is strictly increasing and continuous at every $x \in I$, irrationals. I really dont know where to start. I have looked at my literature (advanced calculus) and im not getting any smarter. Any help is welcome.

Answer 1

Hints:

$1).\ $ for the first part, fix $x<y$ and an $f_n$. Consider three cases: $r_n\le x;\ x<r_n\le y;\ y\le r_n.$

$2).\ $For the second part, suppose $x$ is irrational and pick rationals $\ r_n<x<r_m.$ Since $f$ is increasing, $|f(x)-f(r_n)|<|f(r_m)-f(r_n)|=\left|\sum_{\{j|r_j>r_m\}}\frac{1}{j^2}-\sum_{\{j|r_j>r_n\}}\frac{1}{j^2}\right|=\left|\sum_{\{j|r_n<r_j\le r_m\}}\frac{1}{j^2}\right|.$ To finish note that there are arbitrary many rationals arbitrarly close to $x$, to the left and to the right of $x$.

Notice if $p=r_m$ is rational, then we can take a subsequence $r_{n_k}\to p^{-}$ and then

$|f(p)-f(r_{n_k})|=\left|\sum_{\{j|r_j>p\}}\frac{1}{j^2}-\sum_{\{j|r_j>r_{n_k}\}}\frac{1}{j^2}\right|=\left|\sum_{\{j|r_{n_k}<r_j\le p\}}\frac{1}{j^2}\right|>1/m^2$ so $f$ is not continuous at $p.$