Outer Measure Equality and Limits: Why equality remains under limits for for limit of set?

Question

I am reading measure theory in Rudin Book and in one of the its proofs, it is derived from an equality which exists between four sequences that this equality should also exist on four Limits. I think it is resulted from this fact that sum (and so difference) of two convergent sequences converges to sum of limits But I feel somehow unsure in this case. Can some one describe equality on limit more for me? Thanks in advance This is the sequences and the limits:

Answer 1

For the benefit of people without access to the book:

1. Rudin builds up the Lebesgue measure by first considering an arbitrary additive, regular, nonnegative set function $\mu$ which is finite on $E$ (the set of elementary subsets of $\mathbb{R}^p$), then defining an outer measure $\mu^*$ on $\mathbb{R}^p$ in the usual way (i.e. taking infinums of the $\mu$-measures of coverings by open intervals).
2. (34) states that, given a sequence of elementary sets $A_n$ (i.e. each $A_n$ is the union of finitely many intervals) converging to $A$, then $\mu^*(A_n)$ converges to $\mu^*(A)$ as $n$ tends to infinity
3. Theorem 11.8(a) states that every elementary set $A$ (elementary defined as in 2.) satisfies $\mu^*(A) = \mu(A)$.

So the logic is as follows: $$\mu(A_n) + \mu(B_n) = \mu(A_n \cup B_n) + \mu(A_n \cap B_n)$$ Take limits of both sides: $$\lim_{n \to \infty} \left( \mu(A_n) + \mu(B_n) \right) = \lim_{n \to \infty} \left( \mu(A_n \cup B_n) + \mu(A_n \cap B_n) \right) $$ Distribute the limits (fine since these are strictly positive quantities, and we're happy to have $\infty + a = \infty$: $$\lim_{n \to \infty} \left( \mu(A_n) \right) + \lim_{n \to \infty} \left( \mu(B_n) \right) = \lim_{n \to \infty} \left( \mu(A_n \cup B_n) \right) + \lim_{n \to \infty} \left( \mu(A_n \cap B_n) \right) $$ Apply theorem 11.8(a), transforming $\mu$ to $\mu^*$ since the $A_n$ are elementary: $$\lim_{n \to \infty} \left( \mu^*(A_n) \right) + \lim_{n \to \infty} \left( \mu^*(B_n) \right) = \lim_{n \to \infty} \left( \mu^*(A_n \cup B_n) \right) + \lim_{n \to \infty} \left( \mu^*(A_n \cap B_n) \right) $$ Apply (34) to take simplify, noting that $A_n \cup B_n \to A \cup B$ and $A_n \cap B_n \to A \cap B$: $$\mu^*(A) + \mu^*(B) = \mu^*(A \cup B) + \mu^*(A \cap B)$$