Had a question about the proof laid out in Stein/Shakarchi's text that closed sets are measurable. Im hoping someone can explain the part in bold below, which is the only thing that's bugging me. They first assume F to be compact and show this is equivalent to it being closed, so take F to be compact:
Suppose $F$ is compact, which implies we can find a $O$ such that $F_*\subset O$ and $m_*(O)\leq m_*(F)+\epsilon$.
Since $F$ is closed, then $O\backslash F$ is open and we write it as a countable union of almost disjoint cubes $\implies O\backslash F = \bigcup_i Q_i$
Fix an $N$. The finite union $K=\bigcup_i^{N}Qi$ is compact, which means that $d(K,F)>0$
This implies that $m_*(O)\geq m_*(F)+m_*(K)= m_*(F)+\sum_i^{N}m_*(Q_i)$
Which works out to be equivalent to $\sum_i^{N}m_*(Q_i)\leq m_*(O)-m_*(F)\leq \epsilon$
Here's where i get lost: The authors then claim that "this holds as N goes to infinity"...Why?
Which gives us $m_*(O\backslash F)\leq \sum_i^{\infty}m_*(Q_i)\leq m_*(O)-m_*(F)\leq \epsilon$
The inequality is independent of $N$, that is, it holds for all $N$. Since $m_*$ is non negative, this implies that $$ \sum_{i=1}^\infty m_*(Q_i)=\sup_N\sum_{i=1}^Nm_*(Q_i)\le\epsilon. $$