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I know few facts,

  • if $f : [0,1] \to \mathbb{R}$ is continuous, $Z(f) \triangleq f^{-1}(\{0\})$ is closed,
  • there are continuous functions whose zeros are nowhere dense,
  • there are nowhere dense sets of positive measure.

From these facts, I cannot conclude that if $f:[0,1] \to \mathbb{R}$ is continuous and its zeros $Z(f)$ form a nowhere dense set, this set $Z(f)$ is of null measure. Can we prove that it is indeed the case, or exhibit a counter-example?

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1 Answer 1

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Answer to the title question: Yes! There exists such a function. For example, consider the function $$f(x)=\frac{d(x,A)}{d(x,A)+d(x,B)}$$ where $A$ is the fat cantor set in $[0,1]$ and $B$ is any closed singleton set $\{b\}$ where $b \notin A$. Then $f$ is continuous whose zero set is $A$, which is nowhere dense in $[0,1]$ of positive measure!


Edit: The function $x \mapsto d(x,A)$ alone works.

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  • $\begingroup$ Thanks. Never heard about "fat" cantor set. $\endgroup$ Apr 19, 2019 at 4:00
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    $\begingroup$ See this paper .It explains what a fat cantor set is $\endgroup$ Apr 19, 2019 at 4:05
  • $\begingroup$ Thanks so much. $\endgroup$ Apr 19, 2019 at 4:08
  • $\begingroup$ Thank you, this is brilliant and simple! $\endgroup$
    – Cryme
    Apr 19, 2019 at 4:13
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    $\begingroup$ Both of you are very welcome! $\endgroup$ Apr 19, 2019 at 4:19

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