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Given a bounded and measurable set $K\subset \mathbb R^N$ . Does there exist a smooth cutoff function $\eta\in C^\infty (\mathbb R^N)$ such that $\eta=1$ on $K$ and zero outside a neighbourhood of $K$.

I am trying to understand if smooth cutoffs exists for every bbounded measurable set in $\mathbb R^N$ .

Harish
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  • Do you mean you want something like $\forall \epsilon>0$ there exists an open set $U\supset K$ such that $m(U-K) <\epsilon$ and $\eta|{K} = 1$ and $\eta|{U^c} = 0$? Otherwise it is easy to find such a function. – Dayton Oct 19 '19 at 02:21
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    currently I am not worried about the measure of the set minus the neighbourhood. although it is an interesting question. please quote some references where it mentions the existence, since I am aware of existence of continuous cutoff by Urysohn lemma, but smoothness is not clear to me. – Harish Oct 19 '19 at 02:26
  • Then take a look here https://en.wikipedia.org/wiki/Mollifier the end of the article shows how to construct such a function – Dayton Oct 19 '19 at 02:39
  • outside a neighborhood ... do you pick the neighborhood in advance, so you are asking if for every neighborhood one could find a smooth cutoff function? (If you don't pick the neighborhood in advance, then take the constant $1$ function, it is $0$ outside of the neighborhood $\Bbb R^N$ of $K$.) – Mirko Oct 19 '19 at 02:48
  • How does Urysohn lemma provide a continuous cutoff? I just posted an answer showing that a continuous cutoff need not exist. In order to use Urysohn lemma your set $K$ must be closed, and certainly not every measurable set is closed. Did you mean to ask your question for closed (which along with bounded would of course imply compact) measurable sets $K$? But if your set is compact, then you need not say measurable, and the answer is given in the following post https://math.stackexchange.com/questions/198748/c-infty-version-of-urysohn-lemma-in-bbb-rn – Mirko Oct 19 '19 at 03:23

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I am a bit puzzled of the statement, but I would say the answer is no.

No neighborhood is specified in advance, in the statement of the question. So if we want to find a smooth cutoff function AND a neighborhood that works with this function, then the problem is unreasonably easy. Take $\eta=1$ on $\Bbb R^n$, the constant $1$ function. Then $\Bbb R^n$ is a neighborhood of $K$ and $\eta$ is $0$ outside of $\Bbb R^n$ (vacuously, and not very interestingly, so obviously this could not be what was meant to be asked).

So then perhaps the statement was meant to be:
Given a bounded and measurable set $K\subset \mathbb R^N$ and any neighborhood $U$ of $K$, does there exist a smooth cutoff function $\eta\in C^\infty (\mathbb R^N)$ such that $\eta=1$ on $K$ and $0$ outside $U$.

But this one is also rather easy to answer, with a counterexample.

Let $K$ be the set of all rational numbers in $[0,1]\subset\Bbb R$. Let $U$ be a neighborhood of $K$ such that $\emptyset\neq[0,1]\setminus U$ (that is $[0,1]\nsubseteq U$, for example $U=(-1,2)\setminus \{\frac1\pi\}$, or $U$ is any open set containing $K$ such that the Lebesgue measure of $U$ is some $\varepsilon<1$, such an $U$ exists for every $0<\varepsilon<1$).

Then there is no continuous cutoff function, much less a smooth one. Indeed since $K$ is dense in $[0,1]$ any continuous function that is $1$ on $K$ has to be $1$ on $[0,1]$ and hence cannot be $0$ outside $U$.

The answer would be positive if we in addition assume that $K$ is closed (and hence measurable, and being bounded would also be compact). There is a $C^\infty$ version of Urysohn Lemma in $\Bbb R^n$ (please see my comment to the OP above for a link).

Mirko
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