# Intervals contained in finite cover for $\mathbb{R}$

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: math.stackexchange.com/questions/57317 – user83827 Mar 27 '12 at 15:15
@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? – joriki Mar 27 '12 at 16:45
@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. – user83827 Mar 27 '12 at 16:51

Lets $A_1:=\{ x=\lfloor x \rfloor. x_1x_2....\, |0.x_1x_3x_5.. \mbox{is periodic} \}$.
Let $A_2 = R \backslash A_1$. Then, I think, $R= A_1 \cup A_2$ doesn't have your property...
Consider $A_{1}=\bigcup_{p\in \mathbb{Z}} \left[p+\frac{1}{|p|+1},p+2-\frac{1}{|p|+1}\right]$ and consider $A_2=\mathbb{R}\setminus A_1$. These set cannot be written as a union of any interval!
You may have misunderstood the question. In this case $A_1$ is a union of intervals, so with $j=1$ there is clearly a subset $S\subseteq A_j$ which is an interval, for instance $S=[1+1/2,3-1/2]$. – joriki Mar 27 '12 at 17:26