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Suppose $X$ is a space with measure $\mu$, and $\{A_i\}$ are measurable sets such that $\mu(\bigcap_{i = 1}^n A_i) = 0$, $n>1$. Is it true that $ \mu(\bigcup_{i = 1}^n A_i) \geq \frac{\sum_{i=1}^n \mu (A_i)}{n-1}$? If not, can we replace $n-1$ with some $a_n < n$?

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It turns out this is easy. For future reference:

Since any $x$ (except for those in a set of measure zero) belongs to at most $n-1$ sets, we have:

$(n-1)\mathbb{1}_{\cup A_i}(x) \geq \sum \mathbb{1}_{A_i}(x)$

Integrating gives the desired result.

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