Infimum as an outer measure Let $X$ be a set, $A$ an algebra over $X$ and $\mu$ a finite measure on $A$. For $C \in P(X)$, let $\mu^*(C) = $inf$\{\sum \mu(C_k) | C \subset \cup C_k, C_k \in A\}$.
Why is $\mu^*(C) $ an outer measure? 
($\mu(\emptyset) = 0$ is clear.)
 A: In this answer $\mathcal A$ denotes the algebra.

Let $A=\bigcup_{n=1}^{\infty}A_{n}$. 
It is our aim to prove that
$\mu^{*}\left(A\right)\leq\sum_{n=1}^{\infty}\mu^{*}\left(A_{n}\right)$.
If $\sum_{n=1}^{\infty}\mu^{*}\left(A_{n}\right)=\infty$ then we
are are ready, so let us assume now that $\sum_{n=1}^{\infty}\mu^{*}\left(A_{n}\right)<\infty$.
Let $\epsilon>0$ and for every $n$ find a countable collection $\mathcal{C}_{n}\subseteq\mathcal{A}$
such that $A_{n}\subseteq\bigcup_{C\in\mathcal{C}_{n}}C$ and $\sum_{C\in\mathcal{C}_{n}}\mu C\leq\mu^{*}\left(A_{n}\right)+\epsilon2^{-n}$.
Now let $\mathcal{C}$ denote the union of the collections $\mathcal{C}_{n}$. 
Then $\mathcal{C}$ is countable with $\mathcal{C}\subseteq\mathcal{A}$
and $A\subseteq\bigcup_{C\in\mathcal{C}}C$, so that: $$\mu^{*}A\leq\sum_{C\in\mathcal{C}_{n}}\mu C=\sum_{n=1}^{\infty}\sum_{C\in\mathcal{C}_{n}}\mu C\leq\sum_{n=1}^{\infty}\left(\mu^{*}\left(A_{n}\right)+\epsilon2^{-n}\right)=\epsilon+\sum_{n=1}^{\infty}\mu^{*}\left(A_{n}\right)$$
This can be done for every $\epsilon>0$ so finally we conclude that:
$$\mu^{*}A\leq\sum_{n=1}^{\infty}\mu^{*}\left(A_{n}\right)$$
