Prove that the probability triple $(\Omega, \mathcal{M}, \mathbb{P^*}) $ constructed from the Extension Theorem is complete.
Specifically, this problem asks us to show that, given any $A \subseteq \Omega$, such that $A \in \mathcal{M}$ and $ \mathbb{P^*} (A)=0$, and any $B \subseteq A$, then $$ \mathbb{P^*}(B \cap E) + \mathbb{P^*}(B^c \cap E) = \mathbb{P^*}(E).$$
The proof of this theorem posted below seemed fairly straightforward, but there are two issues (one of which has been previously asked by another user, without answer), which make me doubt it.
The text I am reading writes that, from this result, it follows $B$ has outer measure $0$, i.e.: $$B \in \mathcal{M} \implies \mathbb{P^*}(B)=0,$$ but doesn’t that result follow regardless of whether $B$ is in $\mathcal{M}$ or not?
The other issue is that the following proof does not seem to make use of the fact that $$A \in \mathcal{M}.$$
In fact, it seems we could rewrite the definition of a complete probability triple as follows:
Definition: If $A \subseteq \Omega$, such that $$\mathbb{P^*}(A)=0,$$ then $$S(A) \in \mathcal{M},$$ where $S(A)$ is the set of all subsets of $A$.
Proof:
By the monitonicity of outer measure, for any $E \subseteq \Omega$, $$\mathbb{P^*}(B \cap E) \leq \mathbb{P^*}(A \cap E).$$
Further, since $(A \cap E) \subseteq A$, $$\mathbb{P^*}(B \cap E)=0. \qquad (1)$$
Clearly, $(B^c \cap E) \subseteq E$, which implies $$\mathbb{P^*}(B^c \cap E) \leq \mathbb{P^*}(E).\qquad (2)$$
Putting $(1)$ and $(2)$ together $$\mathbb{P^*}(B \cap E) + \mathbb{P^*}(B^c \cap E) \leq \mathbb{P^*}(E).$$
Since outer measure is also subadditive: $$ \mathbb{P^*}(E) \leq \mathbb{P^*}(B \cap E) + \mathbb{P^*}(B^c \cap E).$$
Thus $\mathbb{P^*}$ is additive on the union $B \cap E$ and $B^c \cap E$, for all subsets $E$ of $\Omega$, from which we conclude $B \in \mathcal{M}$.
$\square$
Any set $A$ with $P^{*}(A)=0$ is measurable. The condition $A \in \mathcal M$ is not necessary.