Why does the proof of countable subadditivity of the outermeasure require translating to intervals? When trying to show:
$m^*(\cup A_n) \leq \sum m^*(A_n)$
where $n$ ranges over some countable indexing set. The standard proof goes like this:
1.) if the RHS is infinite then the inequality is true
2.) prove that the LHS is $\leq$ RHS + $\epsilon$ for arbitrary $\epsilon > 0$, to do this we consider a sequence of coverings of each set $A_n$, each covering being made up of a sequence of $I_m^n$ intervals which has measure $\leq m^*(A_n) + \frac{\epsilon}{2^n}$.
We then line up all the intervals, sum them up and conclude that since $\cup A_n \subseteq \cup I_m^n$ that $m^*(\cup A_n) \leq \sum m^*(I_m^n) \leq \sum m^*(A_n)+ \epsilon$ then let $\epsilon$ go to $0$ and we're done.
My question is why do we need to go through the medium of using intervals? It feels like there's some subtlety that I'm missing. If we can just assert that:
$\cup A_n \subseteq \cup I_m^n \implies m^*(\cup A_n) \leq \sum m^*(I_m^n)$
then why can't we skip the whole interval kerfuffle and just write:
$\cup A_n \subseteq \cup A_n \implies m^*(\cup A_n) \leq \sum m^*(A_n)$
What is the proof of the first assertion? Why can't we just use the same logic for the second?
 A: Which is the definition of outer measure which you are using, in what space is it defined and which is the function that induces the outer measure? 
The most common example which I think you are using is the outer measure in $\mathbb{R}$, which is defined by
\begin{equation}\tag{1}
   m^*(A)=\text{inf }\bigg\{\sum_{j=1}^{\infty} \lambda(I_j):  A\subseteq \bigcup\limits_{j=1}^\infty I_j, I_j \text{intervals}  \bigg\}
   \end{equation}
where $A\subseteq \mathbb{R}$ and $\lambda$ is the function which assigns to each interval it's  length.
So you are basically asking: Why can't one replace the intervals
   with the sets $A_n$ in the proof? Why can't someone take covers of $A_n$ instead of intervals and follow all the above steps?
Well, the quickest and most correct answer is that the definition doesn't allow you to do so. The definition needs intervals to have a meaning, and you have to use them to prove a theorem which is based on this definition. 
But why is that? Following your way of thinking, one can wonder:
   Why does someone even need intervals in the definition $(1)$? Why can't someone use covers of arbitrary subsets of $\mathbb{R}$ in the right-hand side of relation $(1)$ to define the outer measure? 
Answer: Even if the definition allowed you to use any subsets of $\mathbb{R}$ as covers and follow the steps that you described above, you would have to encounter a very serious problem: using the right-hand side of the relation $(1)$ to reach your proof, eventually you would have to talk about $\lambda(A_n)$, or, in other words, you would have to talk about the "lengths" of $A_n\subseteq\mathbb{R}$. However, this is a thing you want to avoid when  $A_n\subseteq\mathbb{R}$ are arbitrary subsets of $\mathbb{R}$, because there is a chance that they are a little "weird", like the Vitali sets (in fact there are uncountable many such sets in $\mathbb{R}$). 
You always have to remember that there are sets in $\mathbb{R}$ that you can't talk about their "length" or "mass". Vitali sets are examples of sets that are not measurable with respect to the Lebesgue measure.
A: The reason why you can do it with intervals is because the definition of this outer measure is: (assuming that you already know how it is defined for intervals)
$m^*(A)=\inf\{\sum_{n=1}^\infty m^*(I_n): A\subseteq \cup_{n=1}^\infty I_n, \ I_n$ are intervals$\}$ 
So obviously if $\cup A_n\subseteq \cup I_m^n$ when $(I_m^n)_{n=1}^\infty$ is a sequence of intervals then $m^*(\cup A_n)\leq \sum m^*(I_m^n)$, simply because the infimum of a set can't be greater than an element of that set. But if $(I_m^n)$ is a sequence of sets which are not intervals then it is not so trivial that the same inequality holds. This is actually what you have to prove. 
