Understanding answer to question showing equality of two measures on the Borel $\sigma$ algebra on $\mathbb{R}$ I understand that this question has been asked before, however my question is in reference to a specific answer given here. 
Proving two measures of Borel sigma-algebra are equal
In particular it is the last answer (which the OP provided). Conveniently their answer has numbers so I when I am referring to numbers I am referring to the numbers they used.  
In part 5 they said that 
$$m(\cup_{i=1}^{\infty}(A_{i}\cap I))=\sum_{i=1}^{\infty}m(A_{i}\cap I)$$
This is only true if the $A_{i}\cap I$ are disjoint? Which they are not necessarily so. Also at the start they wrote $A_{i} \in A$, shouldn't it be in $S$. (It is my understanding that they are trying to show that $S$ is a $\sigma$-algebra). 
Also on a more fundamental level, what good does showing $S$ is a $\sigma$-algebra do? In part (7) they use the fact that $m(A\cap I)=n(A\cap I)$ but isn't this only true for elements in $S$? To me it seems that this shows $m(A)=n(A)$ for all elements in $S$ which is not necessarily all elements in $\mathcal{B}$. 
 A: Answer: I think his proof is wrong.

First, he says:

We need to prove that S is in the Borel σ-algebra

But this makes no sense, because $S \subset \mathcal{B}$ by definition. I think he wants to show that $S$ is a $\sigma$-algebra. But then, its unclear why he proves (2),(3),(4),(5),(6) because (2),(4) and (5) would be sufficent.
Further, you found a mistake in (5), it only works when the sets are disjoint.
One way how one could repair his proof would be to show that $S$ is a Dynkin system, i.e.
$$ X \in S $$
$$ A \in S \Rightarrow A^C \in S$$
$$ \text{For any sequence of disjoint sets } (A_i)_{i \in\mathbb{N}} \Rightarrow \cup A_i \in S$$
This has been proven in (2),(4) and (5). It remains now to show that $$A,B\in S  \Rightarrow A\cap B \in S$$
and then it follows that $S$ is a $\sigma$-algebra.
To part (7): I think he wanted to show that $S$ is a $\sigma$-algebra, because hr thought it would imply the following


*

*$S$ contains all open intervals (thats wrong)

*$S \subset \mathcal{B}$ by definition (thats correct)

*since $\mathcal{B}$ is the smalles $\sigma$-algebra which contains all open intervalls, one has $S = \mathcal{B}$ (wrong since 1 is wrong)


Then you would know that $$n(A \cap (-k,k)) = m(A \cap (-k,k))$$ for each Borel-Set $A\in\mathcal{B}$. He is using the monotonic convergence theorem to deduce that
$$n(A) = m(A).$$
He probably came up with that solution, because the OP ask in the comments

Thank you! Do you mind proving it using the Monotone Class Theorem
  instead? The text has the theorem in a chapter prior to this problem.
  Thank you again

However, by his definition of $S$ the fact $$1. \,S \text{contains all open intervals} $$ is certainly not true. So again, you are right, he cannot conclude that $S =\mathcal{B}$ and thus this argument doesn't work.
The only way I can think of to repair this, is by redefining  $S$ to
$$\tilde{S} = \{ A \in \mathcal{B}, \, I \text{open intervall} |n(A\cap I ) = m(A\cap I) \} $$
but that is the same as
$$\tilde{S}  = \{ A \in \mathcal{B} \, | \, n(A ) = m(A) \} .$$
This is exactly what nullUser proposes in his answer https://math.stackexchange.com/a/1135888/82101 (which also has most upvotes). He proposes to show that $\tilde{S}$ is a $\sigma$-algebra by showing first that its a Dynkin system. Then it follows that $\tilde{S}  = \mathcal{B}$ and you are done.
