Does the set of zeros of an absolutely continuous function contain an open interval?

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?

• What is your approach to this problem? Is $\mu$ the Lebesgue measure? – Jonas Nov 8 '18 at 19:56
• yes, the Lebesgue measure – M.A Nov 8 '18 at 20:24

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.
No, any closed subset of $$[0,1]$$ can be the zero set of a $$C^\infty$$ function, and such subsets can have positive measure but no interior (e.g. a fat Cantor set).