This stackexchange answer stated, that
A bounded subset of $\mathbb{R}^n$ is Jordan measurable iff its boundary has Lebesgue measure $0$.
I can't figure out why this is the case, specifically for the following case:
Let $S = (0, 1) \cap \mathbb{Q}$. Then, $S$ is bounded. It also has a Lebesgue measure $0$ (as a subset of $\mathbb{Q}$).
But I don't quite understand why is $S$ Jordan measurable. By definition (that we were given in our university), for a set to be Jordan measurable, for every $\epsilon > 0$, it must be coverable by a finite system of intervals $I_1, ..., I_k$, so that $\big|\sum_{i=1}^{k}{I_i}\big| < \epsilon$. How exactly would I construct such system of intervals, which will cover all of $S$ and have a total volume $< \epsilon$?
My gut feeling tells me that this is not possilbe, as the amount of elements in $S$ is not finite. But I can't either prove or disprove this.
Edit: so, my mistake was confusing boundary of $S$ and $S$ itself. Boundary of $S$ does not equal to $S$ in this case.
It seems that your set $S$ has closure $[0,1]$ and empty interior, so I believe according to the standard definition $S$ has boundary of measure $1$.
Indeed, if you cover $S$ by a finite number of intervals of total size bounded by $\epsilon$, a subset of measure $1-\epsilon$ of $[0,1]$ is not covered, and there will be a rational number in this set. If $\bigcup_{i=1}^kI_i$ is an interval, this is easy; otherwise, there is some $i\in[k]$ and $a:=\sup I_i$ such that $$b:=\inf\{x\in\bigcup_{i=1}^kI_i:x>a\}>a.$$ Now choose some rational $q\in(a,b)$.
We can, however, cover $S$ by an infinite number of intervals of total measure bounded by $\epsilon$.