I have two questions, the first is based on Q: Royden Chapter 2, Theorem 11.
In the book, it is said "Now consider the case that $m^*(E) = \infty$. Then, $E$ can be expressed as the disjoint union of a countable collection $\{E_k\}$ of measurable sets, each of which has finite measure." I know why already. However, why is the word "Then" there? It might suggest that the statement is true because of the fact that the outer measure of $E$ is infinity, but it is not true because we can do the same thing with finite case, right? Or am I missing something? If I am right, I mean the word "Then" should not be there (better change it with "Observe" or something haha). For now, the way I understand the proof is that we can actually assume any measurable $E$ (not necessarily of finite/infinite outer measure) and work the argument out to arrive at something that deals with finite outer measure. Hopefully anyone can correct me, if any.
I wonder about this equivalence in $\mathbb{R}^n$. What is the idea to show the direction that measurable $E$ can be approximated from the outside? I can only imagine the $R$ case as in no.1.
Thanks in advance.
If $E$ has infinite measure, it causes issues with the proof given for the finite case since it is not true that $$m^{\star}(O \setminus E) = m^{\star}(O) - m^{\star}(E) < \epsilon$$ because you can't subtract an infinite quantity from both sides of the inequality.
Therefore, we use the fact that the measure is $\sigma$-finite to break $E$ down into a countable sum of finite sets so we can use the finite argument.
(The word 'then' is used in the proof to mean "In that case, ...")
The argument in $R^{d}$ is exactly the same. You can find a sequence of open boxes* $(B_{i})_{i = 1}^{\infty}$ such that $\sum_{i = 1}^{\infty} m(B_{i}) \leq m^{\star}(E) + \epsilon$ and then take their union to obtain an open set $O$ satisfying $$m^{\star}(O) \leq m^{\star}(E) + \epsilon$$
*Here a box is an "d-dimensional interval" $B = \prod_{j = 1}^{d} I_{j}$ where each $I_{j}$ is an interval.