Showing that the two definitions of a regular outer measure in Wikipedia actually are equivalent

Question

Consider the two allegedly equivalent definitions of a regular outer measure listed in Wikipedia:

For any subset $A$ of $X$ and any positive number $\epsilon$, there exists a $\mu$-measurable subset $B$ of $X$ which contains $A$ and with $\mu(B) < \mu(A) + \epsilon$

For any subset $A$ of $X$, there exists a $\mu$-measurable subset $B$ of $X$ which contains $A$ such that $\mu(A) = \mu(B)$

I agree that the second definition implies the first one, but as for the other way around, I wanted to take a more detailed look into the argument by proving it myself! My question is: Is the following argument a.) correct, b.) rigorous enough?

Namely, let $X$ be a set and $\mu$ be an outer measure for $X$ and suppose the first condition holds. Let $\epsilon > 0$ be given. Monotonicity of $\mu$ implies that $\mu(A) \leq \mu(B)$ for all $B$ s.t. $A \subset B$ and if $\mu(A) = \infty$ then surely $\mu(B) \leq \infty = \mu(A)$. Therefore we may assume that $\mu(A) < \infty$. Let $B_i$ be the set such that $A \subset B_i$ and $\mu(B_i) - \frac{\epsilon}{2^i} < \mu(A)$. Then by assumption $\mu(B_i) < \infty$ for all positive integers $i$. Construct a sequence $S_k$ for positive integers $k$ by $S_k = \bigcap_{i=1}^kB_i$. Then $S_k$ satisfies $A \subset S_k$ and $k \leq l$ if and only if $S_l \subset S_k$. Moreover by construction $\mu(S_k) \leq \mu(B_k)$. By taking the limit $\lim_{k\to\infty}$ we see that $\lim_{k\to\infty}\mu(S_k) - \frac{\epsilon}{2^k} = \mu(S) - 0 \leq \lim_{k\to\infty}\mu(B_k) < \mu(A)$ with $S := S_\infty = \bigcap_{i=1}^\infty B_i$. Because $\mu$-measurable sets form a $\sigma$-algebra it follows that the countable intersection of $\mu$-measurable sets is still $\mu$-measurable. Hence $S$ is $\mu$-measurable and $\mu(S) = \mu(A)$.

Answer 1

For any subset $A$ of $X$ and any positive number $\epsilon$ , there exists a $\mu$-measurable subset $B$ of $X$ which contains $A$ and with $\mu(B) <\mu(A) +\epsilon$

$\forall n\in \Bbb{N} $ , let $\epsilon=\frac{1}{n}$ then there exists a sequence $(B_n) $ of $\mu$-measurable subsets such that $A\subset B_n$ and $\mu(B_n) <\mu(A) +\frac{1}{n}$ for all $n\in\Bbb{N}$

Let $B=\bigcap_{n\in\Bbb{N}} B_n$.

Claim: $\mu(B) =\mu(A) $

Step $1$: $A\subset B_n$ for all $n\in\Bbb{N}$ implies $$A\subset \bigcap_{n\in\Bbb{N}} B_n =B$$Then by monotonicity of measure $\mu$ , we have $\mu(A) \le \mu(B) $

Step $2$ : $B\subset B_n$ for all $n\in\Bbb{N}$

Then $\mu(B) \le \mu(B_n) \le \mu(A) +\frac{1}{n}$ for all $n\in \Bbb{N}$

Hence $\mu(B) \le \mu(A) $

Thus from step $1$ and step $2$ , we have $\mu(B) =\mu(A) $