Caratheodory: Measurability Let $\mathcal{A}$ be an algebra over $X$ and $\mu:\mathcal{A}\to[0,\infty)$ a finite, positive and countably additive set function.
Consider the induced outer measure:
$$\mu^*(A):=\inf_{A\subseteq\bigcup_k E_k}\sum_k\mu(E_k)$$
Then the following are equivalent:
$$\forall B\subseteq X:\mu^*(A\cap B)+\mu^*(B\setminus A)=\mu^*(B)$$
$$\mu^*(A)+\mu^*(X\setminus A)=\mu^*(X)$$
How do I prove this and what happens if the assumption of an algebra is dropped?

Besides, the statement becomes wrong for noninduced outer measures:
$$X=\{1,2,3\}:\quad\mu^*(\varnothing)=0,\mu^*(X)=2,\mu^*(A)=1\text{ for }A\neq\varnothing,X$$
 A: To answer this question, it turns out to be easiest to forget about the definition of measurability and instead work with properties of the class of measurable sets.
Let us write $\mathcal{M}$ for the class of $\mu^\ast$-measurable sets and recall that this is a $\sigma$-algebra containing $\mathcal{A}$ and that $\mu^\ast |_{\mathcal{E}}$ is a complete measure.
The last ingredient that we need is that for each $B \subset X$ there is a set $M_B \in \mathcal{M}$ with $B \subset M_B$ and such that $\mu^\ast(B) = \mu^\ast(M_B)$ holds. To see this, note that the definition of $\mu^\ast$ yields for each $n \in \mathbb{N}$ a family $E_k^{(n)}$ of sets in $\mathcal{A} \subset \mathcal{M}$ with $B \subset \bigcup_k E_k^{(n)}$ and with
$$
\mu^\ast(B) \leq \mu^\ast\left(\bigcup_k E_k^{(n)}\right) \leq \sum_k \mu^\ast(E_k^{(n)}) \leq \mu^\ast(B) + \frac{1}{n}.
$$
Now take $M_B := \bigcap_n \bigcup_k E_k^{(n)}$.
Now assume that $B \subset X$ with
$$
\mu^\ast(X) = \mu^\ast(B) + \mu^\ast(B^c) = \mu^\ast(M_B) + \mu^\ast(M_{B^c}).
$$
Then $M_{B^c} \supset B^c$ and hence $M_{B^c}^c \subset B \subset M_B$ with
$$
\mu^\ast(M_{B^c}^c) = \mu^\ast(X) - \mu^\ast(M_{B^c}) = \mu^\ast(M_B).
$$
But this implies $B = M_{B^c}^c \cup N$ for a $\mu^\ast$-Null-set $N = B \setminus M_{B^c}^c \subset M_B \setminus M_{B^c}^c$.
Hence, $B \in \mathcal{E}$ is measurable.
The only properties really used in this proof is that each of the sets in $\mathcal{A}$ is measurable with $\mu^\ast(A) = \mu(A)$. This remains true when the algebra $\mathcal{A}$ is replaced by a semiring (as long as $\mu$ is still a "measure" on $\mathcal{A}$).
