Help with proof: given outer measure $m^*$, if $A = \bigcup_{n =1}^\infty A_n$, then $m^*A \leq \sum\limits_{n =1}^\infty m^*(A_n)$ 
Prove given outer measure $m^*$, if $A = \bigcup_{n  =1}^\infty A_n$,
  then $m^*A \leq \sum\limits_{n =1}^\infty m^*(A_n)$

Definition of outer measure:
$$m^*(A) = \inf\{\sum\limits_k|I_k|: \{I_k\} \text{ is a covering of A by open intervals}\}$$
Given $I = (a,b), |I| = b-a$
Proof:
Given $\epsilon >0$, exists for each $n$ a covering $\{I_{k,n}: k \in \mathbb{N}\}$ of $A_n$ such that 
$\sum\limits_{k = 1}^\infty |I_k,n| < m^*(A) + \epsilon/2^n$ 
The collection $\{I_{k,n}: k \in \mathbb{N}\}$ covers $A$ and
$\sum\limits_{k,n} |I_{k,n}| = \sum\limits_{n=1}^\infty\sum\limits_{k=1}^\infty |I_{k,n}| \leq \sum\limits_{n=1}^\infty (m^*(A_n) + \epsilon/2^n)  = \sum\limits_{n=1}^\infty m^*(A_n) + \epsilon$
Thus the infimum of the total lengths of coverings of $A$ by open intervals is $\leq \sum\limits_n m^*(A_n) + \epsilon$ and since $\epsilon > 0$ is arbitrary, the infimum is $\leq \sum\limits_n m^*(A_n) $
This completes the proof. 
I have some questions about the proof:


*

*How can you possibly know that $\sum\limits_{k = 1}^\infty |I_k,n| < m^*(A) + \epsilon/2^n$? Because we know $\sum\limits_{k = 1}^\infty |I_k,n| > \inf\{\sum\limits_{k = 1}^\infty |I_k,n|\} = m^*(A)$ directly following infimum, how can you add a $\epsilon/2^n$ and reverse the sign? Seems impossible to me.

*Suppose above is true, then $\sum\limits_{k,n} |I_{k,n}| = \sum\limits_{n=1}^\infty\sum\limits_{k=1}^\infty |I_{k,n}| \leq \sum\limits_{n=1}^\infty (m^*(A_n) + \epsilon/2^n)  = \sum\limits_{n=1}^\infty m^*(A_n) + \epsilon$. But each inequality is strict:  ($\sum\limits_{k = 1}^\infty |I_k,n| < m^*(A) + \epsilon/2^n$), how could you sum them up and get unstrict inequality $(\leq)$?

*When $\sum\limits_{n=1}^\infty (m^*(A_n) + \epsilon/2^n)$, this entire thing should equal to $\sum\limits_{n=1}^\infty m^*(A_n) + 2\epsilon$ by geometric series, instead of what is in the proof?

(3) The sum starts at $n=1$ and $\frac{1}{2} + \frac{1}{2^2} + \cdots = 1$. Even if the sum really were $2\epsilon$, which would be an actual error in the proof, it would be trivial to fix (just carry forward the $2\epsilon$, having a $2\epsilon$ i nthe final inequality is just as good.
(2) If you know that $a<b$ then also $a\le b$. Which kind of inequality you have doesn't really matter in this case.
(1) It's by the definition of infimum! If $\inf\; T = y$, then for any number larger than $y$, say $y + \epsilon/2^n$ you have that there is some element $t \in T$ with $t < y + \epsilon/2^n$. If not, then $y + \epsilon/2^n$ would be a lower bound for $T$ that is larger than $y$.
