Remark 2

Let $X$ be a topological space and let $Y$ be a Borel subset of $X$. Then the Borel sets of the space $Y$ are exactly the Borel sets of $X$ which are contained in $Y$. For (i) the Borel sets in $X$ which are contained in $Y$ form a $\sigma$-algebra on $Y$ which contains the closed sets in $Y$ and hence contains all Borel sets of $Y$; (ii) the subsets $A$ of $X$ such that $A \cap Y$ is a Borel set of $Y$ form a $\sigma$-algebra on $X$ which contains all the closed sets of $X$ and therefore contains all Borel sets of $X$.