I am following a measure theory course, and I am currently looking at approximation theorems for the first time. The lecturer starts by saying that given a set $A$ such that
$$ \pi^{*}(A) < \infty \quad A \in \mathcal{M} $$
there exists a set $F$ such that
$$ F \in \mathcal{F} \quad A \subseteq F \quad \pi^{*}(A) = \pi^{*}(F) $$
That is, for any measurable set, there is a set containing $A$ in the $\sigma$-algebra which has the same outer-measure as $A$. I cannot work out why this claim holds and don't really know where I should start with a proof.
I will try to define all the notation below:
$\mathcal{F}$ is the $\sigma$-algebra generated by an algebra $\mathcal{A} \subseteq \mathcal{P}(\Omega)$.
$\pi^*(A)$ is defined as follows:
$$ \pi^{*}(A) = \inf \sum_{i \geq 1} \nu(E_{i}) $$
where the infinum is taken over all subsets $\{E_{i}\} \subseteq \mathcal{A}$ which form a covering of $A$, and $\nu$ is a measure defined on the algebra $\mathcal{A}$.
Lastly $\mathcal{M}$, the set of measurable sets, is defined as follows:
$$ A \in \mathcal{M} \iff \forall E \subseteq \Omega, \quad \pi^{*}(E) = \pi^{*}(E \cap A) + \pi^{*}(E \cap A^{c}) $$
By infimum properties, for any $n\in\mathbb N$, there exists a collection of sets $A^*_n=\{A_{nk}\}_{k\in\mathbb N}$ such that $A\subset \bigcup A^*_n$ for all $n$, and $\pi^*(A)+\frac1n>\sum\limits_{k=1}^\infty \nu(A_{nk})$ for all $n$.
Let $B=\bigcap\limits_{n=1}^\infty\bigcup A_n^*$. Clearly, $A\subset B$, and $B$ is in the $\sigma-$algebra. Moreover, if $\pi^*(B)>\pi^*(A)$, then there exists some $m\in\mathbb N$ such that $\frac1m<\pi^*(B)-\pi^*(A)$. But that implies that $$\pi^*(B)\leq \pi^*\left(\bigcup A_m^*\right)<\pi^*(A)+\frac1m<\pi^*(B)$$which is a contradiction. Thus, $\pi^*(B)\leq \pi^*(A)$, and we are done.