Let $A,B \subset \mathbb{R}$ and $m^*(A)=m^*(B)=1$ and $m^*(A\cup B)=2$.
Prove that $m^*(A\cap B)=0$.
I tried every way I can think of but I do not know how to figure this out.
Only properties that I am aware of are monotonicity, countable subadditivity, the outer measure of empty set is zero, translation invariant.
Thank you in advance.
Best Answer
Lebesgue outer measure satisfies the inequality $m^* (A \cup B) + m^* (A \cap B) \le m^* (A) +m^* (B)$. Substitute to get the desired result.
The above inequality should not be difficult to prove no matter how you define Lebesgue outer measure. Basically it's an approximation argument. For instance, if you already know that $m^* (A)$ is the infimum of $m(G)$ over all open sets $G$ containing $A$, then you just pick an $\varepsilon >0$, take open sets $G_A$ and $G_B$ which $\varepsilon$-approximate $A$ and $B$ (in the above sense) and then $G_A \cup G_B$ and $G_A \cap G_B$ will approximate well $A \cup B$ and $A \cap B$ (respectively). Since Lebesgue measure is additive on open sets (and more generally, on measurable sets), the inequality will then follow in a straightforward manner.
Similarly, you can prove that Lebesgue inner measure satisfies the reverse inequality.