Asking for clarification on showing that the set of $\mu$ measurable sets is a $\sigma$-algebra Let $F$ be a collection of $\mu$-measurable subset of a set $A$, and suppose that it is already known that for any $S \subset X$ $\mu[S] \geq \sum_{i=1}^n\mu[S\cap B_i] + \mu[S\setminus B], k = 1, 2, 3,\dots$ for pairwise disjoint sets $B_n$ such that $B = \bigcup_{n=1}^\infty, B_n \in F, \forall n \in \mathbb{N}$. Then in order to show that $\mu[S] = \mu[S\cap B] + \mu[S\setminus B]$, by reading material argues as follows

$\mu[S] \geq \sum_{i=1}^n\mu[S\cap B_i] + \mu[S\setminus B]$ for all $n$ implies that
$\mu[S] \geq \lim_{n\to \infty} \sum_{i=1}^n\mu[S\cap B_i] + \mu[S\setminus B] \Longleftrightarrow  \mu[S] = \sum_{i=1}^\infty\mu[S\cap B_i] + \mu[S\setminus B] = \mu[S\cap B] + \mu[S\setminus B]$

and provides no further remarks/notes on the proof. I am wondering that is the justification for the $\Longleftrightarrow$ in the proof that since the inequality holds for all $n$, the RHS forms a monotone sequence which is bounded above by $\mu[S]$. Thus, if $\mu[S] < \infty$, then by completeness axiom it follows that this sequence has a limit, and this limit is $\mu[S]$. And if $\mu[S] = \infty$, then the sequence approaches $\infty$, i.e. $\mu[S]$?
 A: You have misunderstood some of what J. Kinnunen has presented.

*

*$\mu$ is outer measure on $X$, not measure. In fact, on page 11, the concept of measure has not yet been introduced. Kinnunen uses the notation $\mu^*$ for a general outer measure, but I will use $\mu$ to match your notation.


*$F$ is the collection of all $\mu$-measurable subsets of $X$, not a collection. The mission is to show that $F$ is a sigma-algebra of subsets of $X$.


*You are right that Kinnunen has shown on page 11 that

for any $S \subset X,\;\;$ $\mu[S] \geq \sum_{i=1}^n\mu[S\cap B_i] + \mu[S\setminus B],\;\; \forall n = 1, 2, 3,\dots,$ for pairwise disjoint sets $B_i$ such that $B = \bigcup_{i=1}^\infty B_i,$ with each $B_i \in F.\quad$ (This is the first bit of "Step 5")



*You are also right to see a monotone bounded sequence (indexed by $n$) at the RHS of the inequality. It is a monotone increasing sequence in $[0,\infty],$ and thus it has a limit, some element of $[0,\infty];$ in fact an element of $[0,\mu[S]].$ But based on what we've said so far, there is no reason to conclude that the limit is $\mu[S]$, as you claimed.


*Kinnunen has not argued as you wrote above. (In fact I think your placement of the $\Longleftrightarrow$ is misleading, and does not correspond to the text.) Rather the logic is as follows. We have now $$\mu[S] \geq \sum_{i=1}^\infty\mu[S\cap B_i] + \mu[S\setminus B].$$
We next apply the earlier result of Step 4, pages 10-11:
$$\mu[S] \geq \mu[S\cap \bigcup_{i=1}^\infty B_i] + \mu[S\setminus B]=\mu[S\cap  B] + \mu[S\setminus B].$$
So why, then, is it true that
$$\mu[S] \leq \mu[S\cap  B] + \mu[S\setminus B]\;\;?$$
This last relation is a consequence of the definition of outer measure. See page 7, Remarks 1.5.
