Let $\phi:[0,1] \to \mathbb R$ be a bounded absolutely continuous function.
Assume we know the following:
$ \mu\{t \in [0,1]|\varphi(t)=0 \} >0 $ (i.e, it has non-zero measure)
Then does it follow that $\{ t | \varphi(t)=0 \}$ contains an open interval?
Let $A\subset[0,1]$ be a closed set of positive measure that doesn't contain any interval (this, for instance). Then the function $\phi(x)=d(x,A)$ is Lipshchitz, hence absolutely continuous, and $\{\phi=0\}=A$ doesn't contain any interval.