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Let $S$ be a subset of the set of real numbers $\mathbb{R}$, let $\mathcal{B}$ be the Borel $\sigma$-algebra generated by all open subsets of $\mathbb{R}$. Consider $S$ as a topological space endowed with the subspace topology (i.e., the topology inherited from $\mathbb{R}$), and let ${\mathcal{B}}_S$ denote the Borel $\sigma$-algebra generated by all open subsets of S. Is it true that ${\mathcal{B}}_S\subset \mathcal{B}$ ? (Here, the set $S$ is not necessarily a Borel set in $\mathbb{R}$.) I think, in general, this is not true. However, I could not find a counter-example.

serenus
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  • https://math.stackexchange.com/a/2693750/659976 this might help – JaberEdgar May 26 '20 at 13:51
  • The answers in this reference show that if $S$ is not a Borel set in $\mathbb{R}$ then the inclusin $\mathcal{B}_S\subset \mathcal{B}$, in general, is NOT true. Am I right? – serenus May 26 '20 at 14:29

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You are right in that it is not true in general. Let $S$ be a subset of $\mathbb{R}$ which is not Borel measurable. But since $S$ is open with the subspace topology, $S \in \mathcal{B}_S$ but not in $\mathcal{B}$.

However, if $S$ is open, then this is true. Let $S$ be open in $\mathbb{R}$ and let $\mathcal{B}_S$ be Borel sigma algebra on the subspace topology. Then since all subsets of $S$ which are open w.r.t. $S$ are also open w.r.t. $\mathbb{R}$, each open subset of $S$ also lies in $\mathcal{B}$ which is thus a sigma algebra containing all open subsets of $S$ and thus $\mathcal{B}_S \subseteq \mathcal{B}$.

If $S$ is closed, the proof is similar.

G. Chiusole
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  • I think that, more generally, when $S$ is a Borel set in $\mathbb{R}$, then we have $\mathcal{B}_S\subset \mathcal{B}$. – serenus May 26 '20 at 16:36
  • Yes, that's true. Let $S$ be Borel, then every open subset of $S$ is of the form $S \cap U$ for some $U \subseteq \mathbb{R}$ open. Hence it is also Borel, thus all sets which are open in $S$ lie in $\mathcal{B}$ and thus $\mathcal{B}_S$ does as well. – G. Chiusole May 26 '20 at 16:54