Lemma 4.3.7.label Let $X$ be a topological space and $Y\suf X$ be a subspace, then the Borel $\s$-algebra on $Y$ coincides with the $\s$-algebra on $Y$ induced by $\cali{B}_{X}$.
Proof. Since $\curl{A\in\cali{B}_X:A\cap Y\in\cali{B}_Y}$ is the $\s$-algebra that contains all open sets in $X$, $\cali{B}_{Y}$ contains the induced $\s$-algebra.
OTOH, the induced $\s$-algeba contains all open sets in $Y$, so it contains $\cali{B}_{Y}$.$\square$
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