Integrability of a function defined using an enumeration of the rationals in $[0,1)$

Question

We have $(r_k),k\geq1$ as an enumeration for the rationals in $[0,1)$ and $f$ is a function defined as $$\large f=\sum_{k\geq 1}\frac{1}{k^2}\mathbf{1}_{[0,r_k)}$$ where $\mathbf{1}$ is the indicator function. Show that $f$ is Riemann integrable

I would like to do this using the condition where we find a sequence of step-functions $(\phi_n)$ such that $$\sup\lvert f-\phi_n\rvert\to0,\;n\to\infty.$$ I have though of defining $\phi_n$ on a partition $P_n=\{\frac{m}{n}:m=0,1,...,n\}$ enumerated ascendingly as $x_m$ defining $$\phi_n(x)=f(x_m),\forall x\in(x_{m-1},x_m).$$ This gives
$\sup\lvert f-\phi_n\rvert=\sup_{0\lt m\leq n}\lvert f(x_{m-1})-f(x_m)\rvert$ as $f$ is increasing but I do not know where to go from here.

Any hints would be great!

Answer 1

1. Correct argument. You showed that for each $n$, $r_n$ is a jump discontinuity.
2. As for Riemann integrability, the function is bounded and nondecreasing. This implies Riemann integrability from Vitali-Lebesgue Theorem (a function is Riemann integrable on iff it is bounded and continuous a.e.), but we can also do it directly (note that with the appropriate variations the argument below works for any locally bounded non-decreasing function).

Step I. Fix $\epsilon$ and let $N$ be large enough so that a. $1/k^2 > \epsilon$ only if $k\le N$; b. $\sum_{k>N} \frac{1}{k^2} < \epsilon$.

Step II. Now partition $[a,b]$ to $a=p_1<p_2 <\dots <p_L=b$ in such a way such that each of the elements $r_1,\dots,r_N$ is in the interior of exactly one partition interval $[p_t,p_{t+1}]$ (special attention if $r_j$ is one of the endpoints $a$,$b$, details of which I'll leave), and that the total length of all of the corresponding $N$ partition intervals is $<\epsilon$. Label these $N$ partition intervals by $I_1,\dots, I_N$. The last condition can be written as $\sum_{j} |I_j| < \epsilon$. Label all remaining closed intervals $[p_t,p_{t+1}]$ from the partition by $J_1,\dots, J_M$.

Step III. Note that $\sup_{x} f(x)<2$. Now for each of the intervals $I_1,\dots,I_N$ we have $\sup_{x \in I_j} f(x) - \inf_{x\in I_j} f(x) \le \sup_{x} f(x) <2$, and for each of the remaining intervals we have $\sup_{x\in J_j} f(x) - \inf_{x\in J_j} f(x)\le \epsilon$.

Step IV. Use the bounds above to conclude that the difference between the upper Riemann sum and the lower Riemann sum is bounded above by $2 \sum_{j} |I_j| + \epsilon \sum_{j} J_j \le 2 \epsilon + \epsilon (b-a)$. As $\epsilon$ is arbitrary, we conclude that the function is Riemann integrable.

Hope this helps.