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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.

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Have you been introduced to measurable sets? – Jonas Meyer Aug 12 '11 at 20:19
The outer measure satisfies the inequality $m^* (A \cup B) + m^* (A \cap B) \le m^* (A) +m^* (B)$. Substitute. – Mark Aug 12 '11 at 20:25
@Mark: I know that the OP's question is referring to Lebesgue measure, but is that inequality also only true for Lebesgue measure? – user1736 Aug 13 '11 at 1:23
@user1736: I think the inequality does not hold for general outer measures. For example consider $X=\{0,1,2\}$ and the function $\mu:\mathcal{P}(X)\to[0,\infty]$ with $\mu(\emptyset)=0$, $\mu(A)=1$, if $A\in\mathcal{P}(X)\setminus\{\emptyset,X\}$ and $\mu(X)=2$. If I'm not mistaken, $\mu$ is an outer measure on $X$ but, choosing $A=\{0,1\}$ and $B=\{1,2\}$, we see that the inequality does not hold. – LostInMath Aug 13 '11 at 6:41
@LostInMath: I haven't checked your example but you're right, it doesn't hold for general outer measures. It holds for regular outer measures and the property is called strong subadditivity or submodularity. For Lebesgue measure it holds and is not particularly hard to show. – t.b. Aug 13 '11 at 6:43
up vote 5 down vote accepted

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.

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Let $A$ and $B$ be two sets such that $m^{\ast}(A\cup B)=2,m^{\ast}(A)=m^{\ast}(B)=1$. Those assumptions imply for any given $\varepsilon>0$ by the definition of the outer measure the existence of intervals $(I_n),(J_n)$ and $(K_n)$ such that $A\cup B\subset \cup_n I_n,A\subset \cup_n J_n$ and $B\subset \cup_n K_n$ and \begin{equation*} \sum_nl(I_n)+2\varepsilon>2,\ \ \sum_nl(J_n)+\varepsilon/2>1,\ \ \sum_nl(K_n)=\varepsilon/2>1. \end{equation*} Now introduce $\tilde{I}_n= I_n\backslash (J_n\cup K_n)$, so $A\cap B\subset \tilde{I}_n$ and moreover $l(\tilde{I}_n)\geq l(I_n)-l(J_n)-l(K_n)$. Now we sum over $n$ and use the previous three inequalities involving $\varepsilon$ and get $\sum l(\tilde{I}_n)+\varepsilon>0$, for any $\varepsilon$.

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