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Let $(A_i)_{i=1}^k$ be a collection of subsets of $\mathbb{R}$ with the property that $\mathbb{R}=\bigcup_{i=1}^k A_i$. Is it the case that for some $j$, there exists a subset $S\subseteq A_j$ which is almost an interval (in the sense that $S\cup N$ is an interval for some null set $N$)?

This is clearly false without the "almost". For example, if we take $k=2$, $A_1=\mathbb{Q}$, and $A_2=\mathbb{R}\setminus{Q}$, there are no intervals. However, the weaker statement I propose above certainly holds. Is it true in general?

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    Assuming measurability of the partition, it's still false for $k=2$ since there are sets such that they and their complement both meet every open interval in a set of positive measure. See, e.g., this question and answer: http://math.stackexchange.com/questions/573172012-03-27
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    @ccc: How do you mean "assuming measurability"? The question is whether the statement is true in general -- if it's false assuming measurability, then it's false in general, no?2012-03-27
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    @joriki: Oh certainly, I just meant to imply that it's trivially false when nonmeasurable partitions are allowed. So I assumed that the question was asking about measurable partitions, in which case there's a little more work to do to build a counterexample. I should have phrased this less ambiguously.2012-03-27

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