"The exterior measure of a closed cube is equal to his volume." is the original question and we can also find the proof in the book real analysis (by Stein 2nd edition), p.11.
My question is, what is the role of "$\epsilon$" in the proof?
I can prove this in the following: (the proof for $|Q| \leq m_*(Q)$)
- Let $Q$ be arbitrarily covered by $Q\subset \bigcup_{j=1}^{\infty}Q_j$.
- By monotonicity, $|Q|\leq |\bigcup_{j=1}^{\infty}Q_j|$.
- By subadditivity, $|\bigcup_{j=1}^{\infty}Q_j|\leq \sum_1^{\infty}|Q_j|$
- SInce the covering is arbitrary, so $|Q| \leq \text{inf}\sum_1^{\infty}|Q_j|=m_*(Q)$
Can I prove this in that way, without using $\epsilon$ ?
I think the spirit of the proof is the $\epsilon$; however, it seems reasonable in the proof I provided.
No. First, the notation $|\bigcup_jQ_j|$ doesn't even make sense; $|E|$ is defined if $E$ is a cube, not for arbirary $E$. So steps 2 and 3 don't make much sense. The monotonicity and subadditivity you refer to are properties of $m_*$, not $|\cdot|$.
The crux of the matter is showing that $|Q|\le\sum_j|Q_j|$. But how do you prove that?
It seems obvious. But it also seems obvious for intervals of rationals, and in that context it's not true - it's easy to see that there exist intervals $I_j$ such that $[0,1]\cap\Bbb Q\subset\bigcup_j I_j$ but $\sum_j|I_j|<1$.
The point to the actual proof is to replace the $Q_j$ by slightly larger open cubes; now compactness shows that $Q$ is covered by finitely many of those open cubes. But the larger open cubes are larger; the point to that $\epsilon$ is to make sure that they're not too much larger...