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Let $I_n$ a sequence of open intervals in $[0,1]$ such that $\sum \vert I_n \vert <1$. Prove that the set $X=[0,1]-(\bigcup_{n \in \mathbb{N}} I_n)$ it has no measure zero.

I thought about using the fact that $X$ is compact, to prove that $X$ has not zero content and that would imply what is being asked. However, I can not find a way to start, any suggestions?

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Hint: The measure of $l(X)$ of $X$ is smaller than or equal to $1-\sum_{n\in\mathbb N}l(I_n)$.

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$[0,1] \subset \cup_n I_n \cup ([0,1] \setminus \cup_n I_n) $. If $\cup ([0,1] \setminus \cup_n I_n)$ has measure $0$ this gives $1=l([0,1]) \leq \sum l(I_n)+0 <1$, a contradiction.

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