Prove that a subset is measurable is and only if the measurable of the set equal to the sum of that subset and its complement

Question

Let $X$ be a set and $\mathscr{A}$ a collection of subsets of $X$ that form an algebra of sets. Suppose $l$ is a measure on $\mathscr{A}$ such that $l(X) < \infty$. Define $\mu^{*} $ as $$ \mu^{*}(E) = \inf \; \{\Sigma \, l(C_{i}) \;\; | \;\; C_{i} \in A \; , E \subset \cup_{i=1}^{\infty}C_{i} \}. $$ Prove that a set $A$ is $\mu^{*}$ measurable if and only if $$ \mu^\star(A) = l(X) - \mu^\star(A^{c}) $$

Can I use the fact that by definition, if $A$ is $\mu^{*}$-measurable, then for all $B\subset X$,

$$ \mu*(B)=\mu*(B\cap A)+\mu*(B\cap A^c) $$

What is the relationship between $\mu^{*}(X)$ and $l(X)$, I'm very unsure about this chapter (constructing measure).

Thanks for any input

Answer 1

Don't overthink it. Yes, you can use the Caratheodory criterion $\mu^*(B) = \mu^*(B\cap A) + \mu^*(B\cap A^c)$.

Try writing $\mu^*(A) + \mu^*(A^c) = l(X)$ instead. Does it give you any ideas?