Formal proof: induced outer measure defined via infimum of measure. Let $(\Omega, \mathscr{S}, \mu)$ be a measure space and $\nu$ the outer measure induced by $\mu$. Then for any $A \subseteq \Omega$ we have $$\nu(A)=\inf(\mu(B):B\in\mathscr{S}, A\subseteq B)$$
I just don't see how this statement is equivalent to the definition for the induced outer measure that we have:
$$\mu^*(A) = \inf\{\sum_n \mu(A_n): A_n\in \mathscr{S}, A\subseteq \bigcup_nA_n\}$$
Intuitively I understand that we are looking for the minimal cover of $A$ using sets from the $\sigma$-Algebra. But I don't see a path to formally prove that the statements are in fact equivalent. 
How can we formally show the equivalence of these statements?
Edit: Because $A \subseteq B$ it is clear that $\nu(A)\le\mu^*(A)$ as $\mu(B)$ is in the sum. Is it possible to show the reverse inequality, to complete the proof?
 A: Let $A \subset \Omega$. For any $B \in \mathscr{S}$ with $A \subset B$ we have $\mu^*(A) \leq \mu (B)$ since $B$ is a countable cover of $A$ of measurable sets. Taking the infimum over all such $B$ yields $\mu^*(A) \leq \nu(A)$. Conversely, for any countable collection $\{A_n\}$ of sets $A_n \in \mathscr{S}$ with $A \subset B:=\cup_{n} A_n$ (notice that $B \in \mathscr{S}$) we have, using subadditivity of $\mu$,
$$\nu(A) \leq \mu(B) \leq \sum_{n} \mu(A_n).$$
Taking the infimum over all such countable covers by measurable sets yields $\nu(A) \leq \mu^*(A)$. Since $A$ was arbitrary, this proves that $\nu = \mu^*$.
A: Let $\nu(A) = \inf\{\mu(B) \mid B \in \mathscr{S}, B \supset A\}$ for all $A \subset \Omega$ and $\mu^*$ be the outer measure induced by $\mu$. We claim, that $\mu^* = \nu$. First note, that $$\mu^*(A) = \inf\bigg\{\sum \mu(A_n) \mid A_n \in \mathscr{S}, A \subset \bigcup A_n\bigg\} =  \inf\bigg\{\sum \mu(A_n) \mid A_n \in \mathscr{S}, A \subset \bigsqcup A_n \bigg\}$$ where $\bigsqcup$ means that the union over the $A_n$ is pairwise disjoint. From this it is obvious, that $\nu \leq \mu^*$ (do you see why?) So now let $A \subset \Omega$. If $\nu(A) = \infty$, then by $\nu \leq \mu^*$ we must have $\nu(A) = \mu^*(A)$, so assume without loss of generality, that $\nu(A) < \infty$. Now let $\epsilon > 0$ be arbitrary, then by definition there exists $B \supset A$ such that $\nu(A) + \epsilon > \mu(B)$. But as $\{B\}$ is a $\mathscr{S}$-cover for $A$ this readily implies $\nu(A) + \epsilon > \mu^*(A)$ and since $\epsilon$ was arbitrary we have $\nu(A) \geq \mu^*(A)$, from which it follows that $\mu^* = \nu$.
