I was reading the proof of countable sub-additivity of exterior measure and ran into interesting moment which seems to me very confusing and I cannot understand this. 
Let me clarify my question: Since $E\subset \bigcup\limits_{j,k=1}^{\infty}Q_{k,j}$ then my monotonicity of $m_*$ we have the following: $$m_*(E)\leq m_*\left(\bigcup\limits_{j,k=1}^{\infty}Q_{k,j}\right)$$ and why the last measure is equal to this sum? I was thinking on this for a couple days but no results.
P.S. This is excerpt from Shakarchi-Stein's book.
Would be very grateful for detailed answer!
I'm in my first course in measure theory and using the same book. I think I know the answer, but I apologize if I'm wrong.
By definition $$m_{\ast}(E) = \inf \sum|Q_{i}|$$ where the infimum is taken over all countable covering $E \subset \bigcup Q_{j}$ by closed cubes. The covering $\bigcup Q_{k,j}$ is countable by choice. So, $$\inf \sum |Q_{i}| \leq \sum |Q_{k,j}|$$