Is every Lebesgue null set the intersection of a sequence of unions of intervals? If a set $U\subset\mathbb{R}$ has Lebesgue measure $0$, that means for every $\varepsilon>0$ there exists a set $\mathcal{I}$ of intervals such that $U\subset\cup\mathcal{I}$ and $\sum_{I\in\mathcal{I}}\ell(I)<\varepsilon$. My question is if this implies there exists a sequence $\{\mathcal{I}_i\}$ of sets of open intervals such that $U=\cap_i\cup\mathcal{I}_i$.
It seems to me this should be true. The Cantor set is already by definition the intersection of a sequence of unions of intervals. However whether the statement holds for the set of rationals is already way less trivial.
 A: No, except in the trivial sense that any set is the union of "intervals" that are single points $[x,x]$.  If your intervals are open intervals, any union of intervals is an open set and the intersection of a sequence of these is a $G_\delta$ set.  If your intervals are closed intervals (but not singletons), the union of their interiors is an open set, which is the union of countably many open intervals, and any other point in the union of the closed intervals is an endpoint of one of these open intervals, so you have the union of 
an open set and a countable set, which is an $F_\sigma$ set; the intersection of a sequence of these is an $F_{\sigma\delta}$ set.  There are sets that are Borel, hence Lebesgue measurable, but not $G_\delta$ or $F_{\sigma\delta}$.  There are also Lebesgue measurable sets that are not Borel.
A: Not true. It is well known that the set of all rational numbers cannot be written in this form. 
See Show that $\mathbb Q$ is not a $G_{\delta}$-set.
and the comments below the question. 
