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Does the unit interval have a finite partition $P$ such that no element of $P$ contains an open interval? I would think that the answer is no, because each element of $P$ would have Lebesgue measure zero, but what if the elements of $P$ are not measurable?

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There are many sets of positive measure, indeed measure $1$, that contain no interval. –  André Nicolas Feb 4 '13 at 8:56
    
That's intriguing, do you have references to a few examples? –  Herng Yi Feb 4 '13 at 12:10
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Well, there is an example in the answer you accepted. If the complement of a counable dense set is too easy an example, one can produce fancier ones. –  André Nicolas Feb 4 '13 at 17:24
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up vote 4 down vote accepted

Indeed, you could use the Vitali set for such a partition but that would be overkill. Let $P_1=[0,1]\cap \mathbb Q$ and $P_2=[0,1] \setminus P_1$ then $P_1 \sqcup P_2=[0,1]$ but neither $P_1$ nor $P_2$ contain an open interval because any open interval contains both rational and irrational elements.

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Thanks for the answer! Haha I feel silly for overlooking this counterexample –  Herng Yi Feb 4 '13 at 12:11
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