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.

  • 2
    $\begingroup$ Hint: there are no intervals that contain only rational numbers. $\endgroup$ Oct 4, 2018 at 0:17
  • $\begingroup$ I'm not sure if I follow. $\endgroup$
    – Mog
    Oct 4, 2018 at 0:26
  • $\begingroup$ Compute upper and lower sums for the function $=1$ on the rationals, $=0$ on the irrationals. $\endgroup$
    – GEdgar
    Oct 4, 2018 at 0:33
  • $\begingroup$ What is the measure of a singleton? Say $J ( \{ q\ } )$, where $q \in \mathbb{Q}$. $\endgroup$ Oct 4, 2018 at 0:33
  • $\begingroup$ @GEdgar What do you mean lower and upper sums? Measure of a singleton is $0$. $\endgroup$
    – Mog
    Oct 4, 2018 at 0:35

1 Answer 1



$1). \underline c(\mathbb Q\cap I)=0$ 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.

  • $\begingroup$ $I= [0,1]$ and yes it has outer measure $1$ $\endgroup$ Oct 4, 2018 at 2:23
  • $\begingroup$ Could you explain a bit more in detail please? I'm having a hard time following every step. :/ $\endgroup$
    – Mog
    Oct 4, 2018 at 2:23
  • $\begingroup$ $1).$ is just the definition of the inner measure. There are two parts to $2).$ The first will show that the outer measure is $\le 1$, and the second will show that the outer measure cannot be strictly less than $1$. $\endgroup$ Oct 4, 2018 at 2:34
  • $\begingroup$ 1). How do I show that $\mathbb Q\cap I$ has an empty interior? 2). Why should the finite sequence cover all of $I$? $\endgroup$
    – Mog
    Oct 4, 2018 at 2:41
  • $\begingroup$ It has empty interior because it contains no interval. To understand the second part, here is a basic explanation of Jordan measure than should help you with the definitions. It's in $\mathbb R^2$ but the definitions are mutatis mutandis, the same in $\mathbb R$. $\endgroup$ Oct 4, 2018 at 2:54

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