# Set with measure zero

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?

Hint: The measure of $$l(X)$$ of $$X$$ is smaller than or equal to $$1-\sum_{n\in\mathbb N}l(I_n)$$.
$$[0,1] \subset \cup_n I_n \cup ([0,1] \setminus \cup_n I_n)$$. 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.