Outer Jordan content of closure is equal to Lebesgue measure---is this false?

Question

This is another problem from Pugh's Real Mathematical Analysis. This one is 6.16.

Prove that $J^*A = J^*\overline{A} = mA$ where $\overline{A}$ is the closure of $A$.

Here $J^*$ denotes the outer Jordan content and $m$ denotes Lebesgue measure. There are no restrictions on $A$, except that it's some subset of $\mathbb{R}$.

I know how to prove the first equality but the second equality appears to be false. For example, take $A = [0, 1] \cap \mathbb{Q}$; then the outer Jordan content of both $A$ and $\overline{A}$ is 1, but the Lebesgue measure is 0. I thought perhaps this was a typo, and that it should say $m\overline{A}$ instead of $mA$. But then the claim is still false unless $A$ is bounded; say we take $A = \overline{A} = \mathbb{N}$; the Lebesgue measure is again zero, and the outer Jordan content is infinite.

Is there any reasonable interpretation of this problem?

Answer 1

In the version of the text that I see on Google books, in exercise 16 they define Jordan content only for bounded sets. In problem 17, they ask you to show $J^*A=J^*\bar A=m\bar A$. It also says that it is a corrected printing 2003. So I think you are right, it was just a typo.

Answer 2

You have to assume $A\subseteq\Bbb R$ is bounded. Then use that $\bar A$ is bounded and closed, hence compact. Then open covers are really manageable. It should be clear that $J(A)\leqslant J(\overline A)$, so it remains you fiddle with $\varepsilon$s to get $J(\overline A)\leqslant J(A)+\varepsilon$ for any $\varepsilon$. I am thinking you can do some $\sum 2^{-i}$ trickery by taking balls around points of $\bar A$ that intersect $A$.