Continuity from below of outer measure extending an algebrea
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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)leqs...