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So, let $f:\mathbb{R} \to \mathbb{R}$ be a continuous nonzero function(and not a zero on any interval(for God's sake I want something serious)) and let $Z_f = \{x\in \mathbb{R}| f(x) = 0\}$ be its set of zeros. I know that $Z_f$ has to be closed. But, what I need is: does it need to be of a measure zero? If it does, I need some kind of a proof, if not I need a counterexample.

The only functions that do come in my mind with a lot of zeros are sine, cosine and tangens functions, but all of those have countable zero set, which has measure zero(even union of countable family of zero sets is a zero set). So, help I would appriciate.

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    You might want to ask that the restriction to any interval of $f$ is not the zero function. Otherwise there are plenty of trivial counterexamples. – 2017-01-18
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    So, are you asking for a continuous function $f:\mathbb R\to\mathbb R$ for which $f^{-1}(0)$ contains no (nontrivial) interval but has positive measure? – 2017-01-18
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    It is not good to change the question after it has received answers... – 2017-01-18
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    If you want something analytic as you wrote in a comment to some answer then the zero set always has measure zero unless it is the zero function. This is immediate from the theory of holomorphic functions. – 2017-01-18
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    @LeBtz The theory of holomorphic functions is not applicable here since $f$ is assumed only continuous, and real with real variable... – 2017-01-18
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    @mathbeing analytic real functions can be extended to a holomorphic function on some open complex domain. We can then conclude for the extension that it must be zero. – 2017-01-18
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    @LeBtz Yes, but what the OP meant "with something analitically defined" in Umberto's answer was not that he wanted analityc functions, but that he wanted the definition of the functions considered to be given in an "analytic formula". I agree with you that this makes no much sense though. – 2017-01-18
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    Ah, thanks for the clarification. @mathbeing – 2017-01-18

2 Answers 2

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The distance function to a set $E$ is defined by $$ d(x,E) = \inf_{y \in E} |x-y|.$$

$d(x,E)$ is Lipschitz continuous, and $d(x,E) = 0$ if and only if $x \in \overline{E}$. In particular if $E$ is closed then $E = \{x : d(x,E) = 0\}$.

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    Umberto, that is really good and thank you, but I need something analytically defined. – 2017-01-18
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    What is the context requiring an analytic definition? – 2017-01-18
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A simple counter example would be $$ f(x)=\left\{\matrix{x\ \text{for }x>0\\0\ \text{for }x\leq 0} \right.$$

it is continuous and your set $Z_f=\mathbb{R}^{\leq 0}$.

Edit: The question was edited which makes my answer irrelevant. The new version of the question is asking for functions, that are not zero on any interval. In that case the set $Z_f$ is always a zero set, as if the the function is zero on a set that is dense in some interval and is continuous it has to be zero in the whole of that interval.

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    If I understand OP's question, he doesn't want $Z_f$ to contain any intervals – 2017-01-18