Cfrgtkky

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 $A_nuparrow A$ 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 $$R_n:=bigcup_{i=1}^{k_n} P_{n,i}inmathcal{A}$$ We then
have:
$$Asubseteqbigcup_{n=1}^{infty}(R_nsetminusbigcup_{j=1}^{n-1}R_j)cupbigcup_{n=1}^{infty}bigcup_{i=k_n+1}^{infty}P_{n,i}$$
And so by additivity of $mu$:
$$mu^*(A)leqsum_{n=1}^{infty}mu(R_nsetminusbigcup_{j=1}^{n-1}R_j)+sum_{n=1}^{infty}sum_{i=k_n+1}^{infty}mu(P_{n.i})leq
epsilon +
lim_msum_{n=1}^{m}mu(R_nsetminusbigcup_{j=1}^{n-1}R_j)=\
epsilon+lim_mmu(bigcup_{n=1}^{m}R_n)$$

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 $nlt m$, 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
$$ mu^*(E) le widetilde{mu}(liminf_{n rightarrow infty} B_n) le liminf_{n rightarrow infty} , widetilde{mu}(B_n) = lim_{n rightarrow infty} mu^*(A_n).$$