Prove that a set $A$ is $\mu^\star$ measurable is and only if $\mu^\star (A) = l(X) - \mu^\star(A^{c})$ I am preparing for a quiz over Lebesgue Measure and the problem set I have in class is really tricky for me. 
The problem I have is the following. 
Let $X$ be a set and $A$ a collection of subsets of $X$ that form an algebra of sets. Suppose $l$ is a measure on $A$ such that $l(X) < \infty$. Define $\mu^\star $ as 
$$
\mu^\star(E) = inf \; \{\Sigma C_{i} \;\; | \;\; C_{i} \in A \; , E \subset \cup_{i=1}^{\infty}C_{i} \}.
$$  Prove that a set $B$ is $\mu^\star$ measurable if and only if 
$$
\mu^\star(B) = l(X) - \mu^\star(B^{c})
$$
What I know so far is that $l$ is a premeasure, so that $\mu(B) = l(B)$ $\;\;$  if $B \in A$. And $\Rightarrow$ direction is quite easy because it just follows by the definition of $\mu^\star$ measurable. 
Is there anybody who can help me to show the converse way? 
Thanks in advance for your suggestion(s). 
 A: I think the question is this: if $(X,\mathcal M,l)$ is a finite measure space and an outer measure $\mu^*$ is defined by $$\mu^*(E) = \inf \left\{ \sum_k \ell(C_k) : E \subset \bigcup_k C_k,\ C_k \in \mathcal M \right\},$$ then a set $E$ is $\mu^*$-measurable if and only if $$\mu^*(E) + \mu^*(X \setminus E) = \mu^*(X).$$
Here is a sketch of the proof. 


*

*Although $\mu^*(E)$ is defined for all $E \subset X$, it satisfies $\mu^*(E) = \ell(E)$ whenever $E \in \mathcal M$.

*If $E \subset X$ then there exists $C \in \mathcal M$ satisfying $E \subset C$ and $\mu^*(E) = \mu^*(C)$. 

*If $E$ satisfies the stated assumption, then there exists $D \in \mathcal M$ satisfying $D \subset E$ and $\mu^*(D) = \mu^*(E)$. This is because, for $C \in \mathcal M$ with $X \setminus E \subset C$ and $\mu^*(X \setminus E) = \mu^*(C)$ we have $$\mu^*(E) + \mu^*(X\setminus E) = \mu^*(X) = \mu^*(C) + \mu^*(X \setminus C)$$ so that $X \setminus C \subset E$ and $\mu^*(X\setminus C) = \mu^*(E)$.

*With $E$ and $D$ as above, we have $\mu^*(E) = \mu^*(E \cap D) + \mu^*(E \setminus D) = \mu^*(D) + \mu^*(E \setminus D)$ because $D \in \mathcal M$. Thus $\mu^*(E \setminus D) = 0$.

*Again with $E$ and $D$ as above we have $\mu^*(A \cap D) = \mu^*(A \cap E)$ for every set $A \subset X$ because $$\mu^*(A \cap E) \le \mu^*((A \cap E) \cap D) + \mu^*((A \cap E) \setminus D) \le \mu^*(A \cap D) + \mu^*(E \setminus D) = \mu^*(A \cap D).$$

*Finally, if $A \subset X$ then $$\mu^*(A \cap E) + \mu^*(A \setminus E) \le \mu^*(A \cap D) + \mu^*(A \setminus D) \le \mu^*(A)$$ because $D \in \mathcal M$. 


It follows that $E$ is $\mu^*$-measurable.
