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This is a problem in Frank Jones's real variable's book.

Prove that $\lambda^* (A)+\lambda^*(B)\geq \lambda^*(A\cup B)+\lambda^*(A\cap B)$ where $\lambda^*$ is the outer measure.

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up vote 5 down vote accepted

$\lambda^\ast(A)$ is basically the infimum of $\lambda(U)$ over measurable sets $U \supset A$, and whether this statement can be taken for granted depends on the precise definition of outer measure that you are using. Actually, the minimum is attained on a set called a measurable hull of $A$, and it is unique up to a set of measure zero.

Now take $U$ and $V$ to be measurable hulls of $A$ and $B$. Clearly, $U \cup V$ and $U \cap V$ are measurable sets that contain $A \cup B$ and $A \cap B$, respectively. Hence $\lambda^\ast(A \cup B) + \lambda^\ast(A \cap B) \le \lambda(U \cup V) + \lambda(U \cap V) = \lambda(U) + \lambda(V) = \lambda^\ast(A) + \lambda^\ast(B)$.

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