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Let \((E,\mathscr{E}, \mu)\) be a measure space. If \(A,B\in \mathscr{E}\) are such that \(\mu(A)+\mu(B)> \mu(E)\) then \(A\cap B \neq \emptyset\). This is essentially a measure-theoretic consequence of De-Morgan’s theorem (which for two sets \(A\) and \(B\) says \(|A\cap B| = |A| + |B| - |A\cup B|\).)

This has a neat application. Suppose \(A\subset [0,1]\) is measurable with \(\mu(A)= \frac{1+\alpha}{2}\). Then for any \(\alpha'<\alpha\), the set \(A' =(\alpha' + A)\cap [0,1]\) has measure \(\geq \mu(A)-\alpha'\), hence \(\mu(A)+\mu(A')>1\). It follows that there is an \(x\) with \(\{x, x-\alpha'\}\in A\). Conclude that \((-\alpha, \alpha)\subset A-A\).