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?

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


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