Based on the following definition of outer measure of a set $E \subset \mathbb{R}^d$
$$m_*(E) = \inf \sum_{j=1}^{+\infty}|Q_j|, \quad \text{where} \ \ E \subset \bigcup_{j=1}^{+\infty}Q_j$$
and the $Q_j$ are closed cubes, the author of my text makes the following claim:
It is important to note that it would not suffice to allow finite sums in the definition of $m_*(E)$. The quantity that would be obtained if one considered only coverings of $E$ by finite unions of cubes is in general larger than $m_*(E)$. (See exercise 14.)
This seems intuitive enough (how else could you cover a disk?).
Exercise 14. The purpose of this exercise is to show that covering by a finite number of intervals will not suffice in the definition of the outer measure $m_*$.
The outer Jordan content $J_*(E)$ of a set $E \subset \mathbb{R}$ is defined by $$J_*(E) = \inf \sum_{j=1}^{N} |I_j|$$ where the infimum is taken over every finite covering $E \subset \bigcup_{j=1}^N I_j$ by intervals $I_j$.
(a) Prove that $J_*(E) = J_*(\bar{E})$
(b) Exhibit a countable subset $E\subset [0,1]$ such that $J_*(E) = 1$ while $m_*(E) = 0$.
(a) By noting that $|I_j| = (b_j - a_j) = |\bar{I_j}|$, it's clear that $J_*(E) = J_*(\bar{E})$.
(b) In order to do this, I need to create a subset of $[0,1]$ that has intervals that we can shrink to zero, but where the closure at any step has outer measure of one.
Consider for example the set of rationals $E$ in $[0,1]$. Covered by finite intervals (all contained in $[0,1]$) we have to have $J_*(E) = 1$, because if we didn't that would imply that there was some interval of positive length not covered by the $I_j$, which is impossible since there will be a rational number contained in that interval.
Now, considering coverings of $E$ by infinite intervals, we can choose a countably infinite set of overlapping closed intervals (closed cubes) such that each one is centered around a unique rational number $r \in [0,1]$. Now, the intersection of the sets in $I$ always covers the entire interval $[0,1]$, because in order to be an interval each $I_j$ must contain an infinite number of rationals. Notice that the size of the intervals was arbitrary. This implies that the infimum of each $|I_j| = 0$.
I was having a hard time explicitly constructing this last covering. I'm not sure if that's what the author wanted or if it's a simple formula. I know it's just intervals shrinking around their associated rational. We can make their length a function of $N$, then take $N \to \infty$ but I ran into some confusion writing the interval formulas down.
Edit: For clarity, I would like help with this proof. Thank you.