Is $f(y)=\chi\!\left(\bigcup\limits_{k=1}^{\infty}\left[q_{k},q_{k}+\frac{1}{4^{k}}\right]\right)(y)$ Riemann-Integrable on $\mathbb{R}$?

Question

Is $$f(y)=\chi\!\left(\bigcup\limits_{k=1}^{\infty}\left[q_{k},q_{k}+\frac{1}{4^{k}}\right]\right)(y), \hspace{0.2cm} \{q_{k}\}_{k \in \mathbb{N}} \subseteq \mathbb{Q}$$

Riemann-Integrable ?

Studying Integration, I came across the following problem that was unable to approach by any angle. I think there isn't an explicit answer, it may depens on how the $\{q_{k}\}$ are distributed.

I'd like to know wether the Riemann-Theory can solve this problem,

If so,I obvioulsy don't think there's a general method but there could be observations that have to be made to approach this types of problems.

Any hint,tip or solution, even using Negligible sets considerations would be appreciated,

Thanks

Answer 1

Partial Answer

Claim: $f(y)$ is Lebesgue integrable.

Proof: Note that, for any $q_j$ and $q_k,$ we have that $m([q_j,q_j+4^{-j}]\cup[q_k,q_k+4^{-k}])\le 4^{-j}+4^{-k},$ since the intervals might overlap; if the intervals do not overlap, we get equality. Let $$U=\bigcup\limits_{k=1}^{\infty}\left[q_{k},q_{k}+\frac{1}{4^{k}}\right]. $$ So we have $$ \int\chi\!\left(\bigcup\limits_{k=1}^{\infty}\left[q_{k},q_{k}+\frac{1}{4^{k}}\right]\right)(y)\,dm=m(U) \le\sum_{k=1}^{\infty}\frac{1}{4^k} =\frac13 <\infty.\qquad \text{QED.} $$

So is it Riemann integrable? Well, the easiest theorem to use, when applicable, is the Riemann-Lebesgue Theorem. However, on an infinite domain, the Riemann-Lebesgue Theorem is not applicable. We do know that if the Riemann integral exists, then it is equal to the Lebesgue integral. And since the Lebesgue integral has been proven to exist by being finite, if the Riemann integral exists, it is finite. However, it is possible, as per zhw.'s answer, for $f$ to be discontinuous at every point in a closed and bounded interval.

Therefore, my current conclusion is that there is no conclusion. I think the answer is highly dependent on the location of the $q_k$'s.

Hopefully, the ideas in this answer will stimulate someone else to a full, correct answer.

Answer 2

As I asked in a comment, Riemann integrable (RI) where? Unless you specify a closed bounded interval, the question doesn't make much sense. (Now you say RI on $\mathbb R,$ but what does that mean?)

Edit The below is edited from the original, where I misunderstood the hypotheses to imply $\{q_k\}=\mathbb Q.$

There are certainly examples where the choice of $q_k$ lead to counterexamples: Assume $\{q_k\}$ is all of $\mathbb Q.$ Then the resulting $f$ is not RI on any interval $[a,b]$ with $b-a>1/3.$

Proof: Suppose $b-a>1/3.$ Let $E=(\bigcup [q_k,q_k+1/4^k])\cap [a,b].$ Then $m(E) \le \sum 4^{-k} = 1/3.$ Thus $m([a,b]\setminus E)=b-a - m(E) >0.$ At every $x\in [a,b]\setminus E,$ we have $f(x) = 0.$ But this $x$ is the limit of a sequence of rationals, along which $f=1.$ Thus $f$ is discontinuous at each such $x.$ By the Lebesgue criterion, $f$ is not RI on $[a,b].$