Consider the following problem:
Let $\mathcal{A}$ be an Algebra with an additive function $\mu$ and let $\mu^*$ be the outer measure corresponding to $\mu$. Show that if $A_n\uparrow A$ then $\mu^*(A_n)\rightarrow \mu^*(A)$.
After much effort I've crafted this attempt at a solution:
For every $n$ we have a cover by sets from $\mathcal{A}:\space\{P_{n,i}\}^{\infty}_{i=1}$ such that $\sum_{i}\mu(P_{n,i})<\mu^*(A_n)+\epsilon$. We then also have a $k_n$ such that: $$\sum_{i=k_n+1}^{\infty}\mu(P_{n,i})<\frac{\epsilon}{2^n}$$ We will also define $$R_n:=\bigcup_{i=1}^{k_n} P_{n,i}\in\mathcal{A}$$ We then have: $$A\subseteq\bigcup_{n=1}^{\infty}(R_n\setminus\bigcup_{j=1}^{n-1}R_j)\cup\bigcup_{n=1}^{\infty}\bigcup_{i=k_n+1}^{\infty}P_{n,i}$$ And so by additivity of $\mu$: $$\mu^*(A)\leq\sum_{n=1}^{\infty}\mu(R_n\setminus\bigcup_{j=1}^{n-1}R_j)+\sum_{n=1}^{\infty}\sum_{i=k_n+1}^{\infty}\mu(P_{n.i})\leq \epsilon + \lim_m\sum_{n=1}^{m}\mu(R_n\setminus\bigcup_{j=1}^{n-1}R_j)=\\ \epsilon+\lim_m\mu(\bigcup_{n=1}^{m}R_n)$$
This is where I'm stuck. Essentially I want to say that the union above isn't much larger than $R_m$ because $R_m$ covers most of $A_n$ for $n\lt m$, and $R_n$ isn't much larger than $A_n$, but I'm not sure this can actually be done.
Is this solution salvageable? Or is there another approach that might work better?
This is only a partial answer. I am not sure, if the statement is in general true!
If the additive measure $\mu$ on $\mathcal{A}$ has the property that it extends to a measure $\widetilde{\mu}$ on a $\sigma$-algebra $\mathcal{F} \supset \mathcal{A}$ with the property that for any $A \subset X$ there exists a $B \in \mathcal{F}$ with $A \subset B$ and $$\mu^*(A) = \widetilde{\mu}(B),$$ then the statement is true. For example, this is the case if $\mu$ is the Lebesgue-measure on $\mathbb{R}$ and $\mathcal{A}$ is the $\sigma$-algebra of all Borel-sets.
Proof: Take for $A_n$ the corresponding $B_n \in \mathcal{F}$ with $\mu^*(A_n) = \widetilde{\mu}(B_n)$. Then $x \in A$ implies that $x \in A_m$ for all $m \ge n$, starting at some $n \in \mathbb{N}$. Thus $x \in \liminf_{n \rightarrow \infty} B_n$, because $A_n \subset B_n$, and $$ \mu^*(E) \le \widetilde{\mu}(\liminf_{n \rightarrow \infty} B_n) \le \liminf_{n \rightarrow \infty} \, \widetilde{\mu}(B_n) = \lim_{n \rightarrow \infty} \mu^*(A_n).$$