Let $G\subset\Bbb R$ be a Borel set, show that the Borel sets of $G$ (as a subspace) are the same as the Borel subsets of $\Bbb R$ included in $G$

Question

Let $B$ be the Borel sigma-algebra over $\Bbb R$ (real numbers). Let $G\subset R$ be a Borel-set. And $A_0$ the family of all subsets of $G$ which have the form $G\cap O$ for $O$ being an open subset of $R$.

Let $A_1$ be the sigma algebra over $G$ generated by $A_0$

and $A_2 = \{X\in B\mid X \subset G\}$

How to show that $A_1 = A_2$?
I would be especially interested in the direction $A_2 \subset A_1$

Answer 1

Consider $\{X \in B: X \cap G \in A_1\}$. Verify that this is sigma algebra. It contains all open sets in $\mathbb R$. Hence it contains all Borel sets. In particular, if $X$ is Borel and contained in $G$ then $X \in A_1$.

Answer 2

Let $\iota:G\to\mathbb R$ denote the inclusion and let $\tau$ denote the usual topology on $\mathbb R$.

We observe that:

• $\mathcal B=\sigma(\tau)$
• $\mathcal A_0=\iota^{-1}(\tau)$
• $\mathcal A_1=\sigma(\mathcal A_0)$
• $\mathcal A_2=\iota^{-1}(\mathcal B)$

The last bullet is not direct but also not difficult to prove: if $X\in\mathcal A_2$ then $X=\iota^{-1}(X)$ with $X\in\mathcal B$ and if conversely $X=\iota^{-1}(B)=G\cap B$ for some $B\in\mathcal B$ then also $X\in\mathcal B$.

So actually you are asked to prove that:$$\sigma(\iota^{-1}(\tau))=\iota^{-1}(\sigma(\tau))$$

This is a special case of a theorem that is nice to get acqainted with (and that is my aim with this answer).

See this answer for an outline of the proof of it.