= Solution
The <Caratheodory measurability criterion> yields a complete sigma-algebra $\mathfrak M_\mu^*$, and $\mu$ restricted to it is a countably additive <measure>. <Completeness> means that every subset of a measurable <null set> is measurable.
For the requested algebra proof, $X$ and $\varnothing$ plainly satisfy the criterion. Replacing $A$ by $X\setminus A$ merely exchanges the two terms, proving closure under complements. If $A,B$ are measurable, apply their splitting identities successively to any $E$:
$$
\mu(E)=\mu(E\cap A)+\mu((E\setminus A)\cap B)+\mu(E\setminus(A\cup B)).
$$
Subadditivity applied to the first two pieces implies
$$
\mu(E)\geq\mu(E\cap(A\cup B))+\mu(E\setminus(A\cup B)).
$$
The reverse inequality is outer subadditivity. Hence $A\cup B$ is measurable, proving closure under finite unions and therefore the \b[<algebra of sets> property]. The argument never subtracts infinite values.
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