# Inner point of zero set with positive measure

Let $$f:B_r(0)\rightarrow\mathbb{R}$$ with $$B_r(0)\subset \mathbb{R}^2$$ the open Ball of radius $$r>0$$. Let us further denote the set of all zeros of $$f$$ by $$M:=\{x\in B_r(0):\ f(x)=0\}.$$ Let us also assume that the (outer) Lebesgue measure of that set is positiv, i.e. $$\mathcal{L}^2(M)>0.$$ Now my question is the following: If we impose some kind of regularity on $$f$$, for example continuity, differentiability or Hölder continuity, can we find an inner point of $$M$$?

Without any regularity this would probably be false, because we could construct something with the fat Cantor set (see e.g. https://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set), but I'm not sure how we could rule out such behaviour.

• If $E$ is any closed set of measure $0$ then $x \to d(x,E)$ is a Holder continuous (hence absolutely continuous) function whose zero set is exactly $E$. I believe we can even construct a $C^{\infty}$ function whose zero set is $E$ but I don't have a proof at this moment. – Kavi Rama Murthy Sep 25 '18 at 9:14
• @Kavi Rama Murthy : A very good example, didn't think of that one. By any chance, did you mean that the measure of $E$ is positive, because for the Hölder continuity we should only need that $E$ is closed? – humanStampedist Sep 25 '18 at 9:24
• Yes, I meant measure greater than $0$. Sorry for the error. – Kavi Rama Murthy Sep 25 '18 at 9:27

The answer is no. First note that for any closed set $$E \subset \mathbb{R}^n$$ there is a smooth function $$f \colon \mathbb{R}^n \rightarrow [0,\infty)$$ with zero set $$f^{-1}(0) = E$$, see e.g. here. Now take a 'fat' cantor set $$C$$ with positive measure and define $$E = C^n$$ - by rescaling we may also suppose that $$E \subset B_r(0)$$. By the previous remark, there exists a smooth function $$f$$ with zero set $$E$$. However, $$E$$ has no inner points.
Only in the case of analytic functions we can deduce that $$f \equiv 0$$, because the set of zeros is always countable and discret.