$[0, 1]\setminus \mathbb Q$ can't be exhausted by Jordan sets

Question

We wish to show that $[0, 1]\setminus \mathbb Q$ can't be exhausted by Jordan sets. What I mean by this is that there's no nested sequence of Jordan sets $J_1 \subseteq J_2 \subseteq \dots$ such that $\bigcup_{k=1}^{\infty}J_k = [0,1]\setminus\mathbb Q$

Jordan in this sense means Jordan Measurable (boundary is negligible)

Sadly, the fact that the boundary of the union is a subset of the union of the boundaries is only true for finite union, else this would be piece of cake.

How would you tackle this? It may be related to improper integration

Answer 1

Better answer than the above one as I don't invoke measures.

Notice that $X = [0,1]\setminus \mathbb Q$ doesn't have any interior points, $\text{int}(X) = \emptyset$, so if $J \subset X$ then also $\text{int}(J) = \emptyset$.

This means that $J \subseteq \partial J$. Now, if $J$ is Jordan measurable, then by definition $\partial J$ is negligible, so $J$ is negligible.

Thus if $X = \bigcup_{k=1}^{\infty}J_k$ then it follows that $X$ is negligible, since it's a countable union of negligible sets, and this is a contradiction, since $X$ is not negligible.

Answer 2

Note that the inner Jordan measure of any subset of $A:=[0,1]\setminus\mathbb{Q}$ is $0$, since $A$ does not contain any positive length intervals. Hence, any Jordan subset of $A$ is of Jordan measure $0$, and hence of Lebesgue measure $0$. Hence, any countable union of such sets is still of Lebesgue measure $0$. Since $A$ has Lebesgue measure $1$, this union can not equal $A$.