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

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.

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

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.

• $I= [0,1]$ and yes it has outer measure $1$ Oct 4, 2018 at 2:23
• Could you explain a bit more in detail please? I'm having a hard time following every step. :/
– Mog
Oct 4, 2018 at 2:23
• $1).$ 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 $1$. Oct 4, 2018 at 2:34
• 1). How do I show that $\mathbb Q\cap I$ has an empty interior? 2). Why should the finite sequence cover all of $I$?
– Mog
Oct 4, 2018 at 2:41
• 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 $\mathbb R$. Oct 4, 2018 at 2:54