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$ \newcommand{\bR}{\mathbb{R}} \newcommand{\eps}{\epsilon} \newcommand{\bN}{\mathbb{N}} $ For a subset $O$ of $\bR$ and $\eps >0$, we define the $\eps$-neighborhood of $O$ as $$ \{x \in \bR : \exists y \in O \text{ such that } |x-y| < \eps \} . $$

Then I come up with the following result:

Lemma Let $(O_n)$ be a sequence of open subsets of $\bR$ such that $O_{n+1} \subset O_n$ for all $n \in \bN$. We also assume that $1_{O_n} \to 0$ pointwise everywhere as $n \to \infty$. Let $B_n$ be the $\frac{1}{n}$-neighborhood of $O_n$. Then $1_{B_n} \to 0$ pointwise everywhere as $n \to \infty$.

Proof From this thread, any open subset of $\bR$ is a countable union of disjoint open intervals. Thus we can safely assume that $O_n$ is an open interval. The claim then follows easily.

I would like to ask if above result holds for $\bR^d$. Thank you for your elaboration.

Akira
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    What are the reason for the close votes? – Akira Apr 23 '25 at 11:04
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    Does "everywhere" mean "pointwise"? And it might help if you spell out the "easy" argument. (I did not vote to close but these are some small issues with your post.) – Moishe Kohan Apr 23 '25 at 14:50
  • @Akira I am voting to reopen, but note that it is not clear whether you have tried to prove the result or not, and if so what progress you have made. – Dean Miller Apr 24 '25 at 07:52
  • I have to admit that I was sloppy in my proof which is indeed not correct (as seen in below answer). – Akira Apr 24 '25 at 07:54

1 Answers1

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Notice that if your open sets are dense somewhere, the limit of the $1_{B_n}$ will not converge to $0$. So you need a condition on the boundary. Indeed, the sequence of indicator functions of $B_n$ converge to $0$ iff the sequence of indicator functions of $\overline O_n$ goes to $0$. I will show $1_{B_n}\rightarrow 0 \implies 1_{ \overline O_n} \rightarrow 0$ (The converse implication also holds but is not needed for my argument): $$x\in \overline{O_n}\iff d(x,O_n)=0$$ If $x\not\in \bigcap_n B_n$ then for some $n$, $d(O_n,x)>0$. The rest of the argument should be straight forward. However, the sequence $O_n:=B_{1/n}(0)\setminus\{0\}$ is a counter example. Indeed, $\bigcap_n O_n=\emptyset$ but $\bigcap_n \overline{O_n}=\{0\}$. This also applies to $\mathbb{R}$ so it seems to contradict your original claim. A simpler version of the counter example for $\mathbb{R}$ is $O_n:=(0,1/n)$ where $\bigcap_n B_n=\{0\}$ and thus $1_{B_n}(0)=1$ for all $n$.

Final edit: The statement I made in the beginning holds more generally. In a metric space, given a sequence of decreasing sets $O_n$(need not be open), $\lim_n(1_{B_n})=\lim_n(1_{\overline O_n})$. Just something you could look into if you want to.

M.Hoss
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    As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. – Community Apr 23 '25 at 11:28
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    I'm afraid I'm unable to understand what your first two sentences mean. Your counterexample is a good one: it might be clearer to start with the counterexample $O_n = (0,1/n)$. – Mentastin Apr 23 '25 at 12:05