 Solution

The Caratheodory measurability criterion yields a complete sigma-algebra ${\mathfrak{M}}_{\mu }^{\ast }$, 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 $E$:Subadditivity applied to the first two pieces impliesThe reverse inequality is outer subadditivity. Hence $A\cup B$ is measurable, proving closure under finite unions and therefore the algebra of sets property. The argument never subtracts infinite values.