Let $\mu^{\ast} : \mathcal{H} \to [0,\infty]$ an outer measure, where $\mathcal{H}$ is an hereditary $\sigma$-ring. Suppose that for each $A \in \mathcal{H}$ there is an measurable set $E$ such that $A \subset E$ and $\mu^{\ast}(A) = \mu^{\ast}(E).$ We say in this case that $\mu^{\ast}$ has the enveloping property.
Prove that if a sequence $\{A_n\} \subset \mathcal{H}$ is such that $A_n \nearrow A$, then $$\mu^{\ast}(A) = \lim_{k \to \infty}\mu^{\ast}(A_k). $$
I tried several naive things like:
For each $k$ there is $\tilde A_k$ measurable such that $A_k \subset \tilde A_k$ and $\mu^{\ast}(A_k) = \mu^{\ast}(\tilde A_k).$ Then, I tried to show that $\tilde A_k \subset \tilde A_{k+1}$, but it seems not to be true.
Then, I considered $\tilde A_N := \cup_{k=1}^N\tilde A_k$. It is true that $\tilde A_N \supset \bigcup_{k=1}^N A_k$, but is not true necessarily that $$\mu^{\ast}(\tilde A_N) = \mu^{\ast}(\bigcup_{k=1}^NA_k),$$ so it led to nowhere.
Any hints?
Thanks in advance.
First, notice that $\lim_{n\rightarrow \infty}\mu^*(A_k)=L$ exists since $\mu(A_k)$ is a increasing real sequence. Since $\mu^*$ is monotonic, $L\leq \mu^*(A)$.
For each $k \in \mathbb N$, let $\tilde A_k\supseteq A_k$ be a measurable set such that $\mu(\tilde A_k)=\mu(A_k)$.
For each $k\in \mathbb N$, let $B_k=\bigcap_{l\geq k}\tilde A_l$. Each $B_k$ is measurable and $A_k\subseteq B_k\subseteq \tilde A_k$ (notice that if $l\geq k$, then $A_k\subseteq A_l\subseteq \tilde A_l$, so $A_k\subset B_k)$. Since $\mu^*$ is monotonic, $\mu(A_k)=\mu(B_k)$, and $B_k$ is increasing. Then:
$$\mu^*(A)=\mu^*(\bigcup_{k \in \mathbb N}A_n)\leq \mu^*(\bigcup_{k \in \mathbb N}B_n)=\lim_{n \rightarrow\infty}\mu^*(B_n)=\lim_{n \rightarrow\infty}\mu^*(A_n)=L.$$