Let be $\mathcal{P}(M)$ the power set of $M$, $\mathcal{R}$ a ring of sets based on subsets of $M$ and $\mu:\mathcal{R}\to[0,\infty]$ a content. For an arbitrary but fixed $A\in\mathcal{P}(M)$ we define the following family of sets $$ \mathcal{U}(A):=\left\{\bigcup\limits_{n=1}^{\infty}A_n\mid A_n\in\mathcal{R}, A\subseteq \bigcup\limits_{n=1}^{\infty}A_n\mid A_n\right\} $$ and $\mu^{\star}:\mathcal{P}(M)\to[0,\infty]$ by $$ \mu^{\star}(A):=\inf\left\{\sum\limits_{n=1}^{\infty}\mu(A_n)\mid \bigcup\limits_{n=1}^{\infty}A_n\in\mathcal{U}(A)\right\}. $$ We assume that $\mu^{\star}$ satifies the properties of an outer measure. Show that $\mu^{\star}(A)$ restricted to $\mathcal{R}$ is finitely additive.
My approach:
Let be $\bigcup\limits_{n=1}^NA_n\in \mathcal{R}$ a disjoint union. Due to $\sigma$-subadditivity of outer measure $\mu^{\star}$ we see that $$ \mu^{\star}\left(\bigcup\limits_{n=1}^NA_n\cup \emptyset\cup \emptyset\dots\right)\leq \sum\limits_{n=1}^N\mu^{\star}(A_n)+\sum\limits_{n=N+1}^{\infty}\mu^{\star}(\emptyset)=\sum\limits_{n=1}^N\mu^{\star}(A_n). $$ Next, we choose a union/countable cover of $\bigcup\limits_{n=1}^NA_n$, i.e. $\bigcup\limits_{k=1}^{\infty}B_k\in \mathcal{U}\left(\bigcup\limits_{n=1}^NA_n\right)$. It is clear that for each of the disjoint $A_n$ it holds $\bigcup\limits_{k=1}^{\infty}(B_k\cap A_n)\in \mathcal{U}\left(A_n\right)$; note that indeed $(B_k\cap A_n)\in\mathcal{R}$. Moreover, we immediately see that $\bigcup\limits_{n=1}^N(B_k\cap A_n)\subseteq B_k$ for all $B_k$. The finite additivity and the monotonicity of $\mu$ allow us to conclude $$ \sum\limits_{k=1}^{\infty}\mu(B_k)\geq \sum\limits_{k=1}^{\infty}\sum\limits_{n=1}^N\mu(B_k\cap A_n)=\sum\limits_{n=1}^N\sum\limits_{k=1}^{\infty}\mu(B_k\cap A_n). $$ As each $\mu^{\star}(A_n)$ is a lower bound of the respective series of the $N$-many series, we know that $$ \dots=\sum\limits_{n=1}^N\sum\limits_{k=1}^{\infty}\mu(B_k\cap A_n)\geq \sum\limits_{n=1}^{N}\mu^{\star}(A_n). $$ But we also know that $\mu^{\star}\left(\bigcup\limits_{n=1}^NA_n\right)$ is the greatest lower bound which means that $\mu^{\star}\left(\bigcup\limits_{n=1}^NA_n\right)\geq \sum\limits_{n=1}^{N}\mu^{\star}(A_n)$. Consequently, $$ \mu^{\star}\left(\bigcup\limits_{n=1}^NA_n\right)= \sum\limits_{n=1}^{N}\mu^{\star}(A_n). $$ As the disjoint union $\bigcup\limits_{n=1}^NA_n$ was chosen arbitrarily $\mu^{\star}$ is finitely additive.
Is this correct?