Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
What have you tried? –  Alex Becker Oct 3 '12 at 0:28

1 Answer 1

up vote 4 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 $\lambda^\ast(A \cup B) + \lambda^\ast(A \cap B) \le \lambda(U \cup V) + \lambda(U \cap V) = \lambda(U) + \lambda(V) = \lambda^\ast(A) + \lambda^\ast(B)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.