For example let $X:= [0,+\infty]$ with the extended standard topology. Let $A_i, A_j$ are measurable subsets such that $A_i\cap A_j=\emptyset$.
Then I can get that
(1)Equation $\mu(S)=\mu(S\cap A_i\cup A_j)-\mu(S\backslash(A_i\cup A_j)$ for any $S$ And
(2)$A_i\cup A_j$ is measurable set.
First, $\bigcup\limits_{i=1}^{N} A_{i} = \bigcup\limits_{i=1}^{N-1} A_{i}\cup A_N$ implise induction.
So, I conclude that for all of $N\in \mathbb{N}$, Equation (1) holds. i.e.
$\mu(S)=$ $\mu(S\cap$$\bigcup\limits_{i=1}^{N} A_{i})$$+\mu(S\backslash\bigcup\limits_{i=1}^{N}A_{i}).$ Here, I happend to have a question.
When I wanted to claim the following $\mu(S)=$ $\mu(S\cap$$\bigcup\limits_{i=1}^{\infty} A_{i})$$+\mu(S\backslash\bigcup\limits_{i=1}^{\infty}A_{i})$,
[1]. Because for all of $N\in \mathbb{N}$, Equation (1) holds, can I put $\infty$ on $N$ ?or
[2]. Because I apply $\lim_{n\to\infty}$ both sides to $\mu(S)=$ $\mu(S\cap$$\bigcup\limits_{i=1}^{N} A_{i})$$+\mu(S\backslash\bigcup\limits_{i=1}^{N}A_{i}), $can I put $\infty$ on $N$?
In the latter case, To put "$lim$" in $\mu$, i think that there are more something condition. (for example continuous.)
Which is right?
And second question, For example In Euclidean space. $A, B$ are open sets. Then $A \cap B$ is also open. I think that from $\bigcap\limits_{i=1}^{N}A_{i} = \bigcap\limits_{i=1}^{N-1}A_{i}\cap A_N$ and induction then
for all $N\in\mathbb{N}$, $\bigcap\limits_{i=1}^{N}A_{i}$ is open.
If above [1] logic was right, i could put $\infty$ in $N$ and $\bigcap\limits_{i=1}^{\infty}A_{i}$ is open. but I already knew, it`s not correct. because for $A_i = (0-\frac{1}{i+1},1+\frac{1}{i+1})$,
$\bigcap\limits_{i=1}^{\infty}A_{i}$ is closed.
Where did i misunderstanding?
$\infty$ is not a positive integer. If a property holds for each $n$ you cannot put $n=\infty$ in it. In both cases you have to take limits.