Show that $\mathbb Q\cap [0,1]$ is not Jordan measurable.

Question

I'm trying to show two things:
1. $J^*([0,1])=1$, where $J^*$ is the Jordan outer measure.
2. $\mathbb Q\cap [0,1]$ is not Jordan measurable - i.e. the Jordan outer and inner measures do not agree.
The Jordan measures here are defined on finite union of open intervals.

Any hints or proof help would be greatly appreciated.

Answer 1

Hints:

$1). \underline c(\mathbb Q\cap I)=0$ because $\mathbb Q\cap I$ has empty interior.

$2).$ To see that $\overline c(\mathbb Q\cap I)=1$, note that for any $\epsilon>,\ m([0,1+\epsilon)=1+\epsilon$, and that any finite sequence of intervals that covers $\mathbb Q\cap I$ must in fact cover all of $I$.

A useful fact which has been alluded to in the comments, is that $E$ is Jordan measurable if and only if the Riemann integral $\int \chi_E$ exists.