Prove that $ {\mathscr{B}}([0,1]) = {\mathscr{B}}(\mathbb{R})|_{[0,1]}$. Prove that $ {\mathscr{B}}([0,1]) = {\mathscr{B}}(\mathbb{R})|_{[0,1]}$. That is, the Borel set generated by [0,1] is that same as the Borel set generated by $\mathbb{R}$ then restricted on [0,1]. 
I have shown that $ {\mathscr{B}}([0,1]) \subseteq {\mathscr{B}}(\mathbb{R})|_{[0,1]}$, but have no idea the other direction. 
 A: You want to show that $A \cap [0,1]$ is Borel-measurable in $[0,1]$ for every Borel set $A$ in $\mathbb R$. This is equivalent to the inclusion $[0,1] \hookrightarrow \mathbb R$ being measurable with respect to the Borel $\sigma$-algebras and follows by combining the following facts:


*

*The inclusion $[0,1] \hookrightarrow \mathbb R$ is continuous.

*If $f: X \to Y$ is a continuous map between topological spaces, then $f$ is measurable with respect to the Borel $\sigma$-algebras.


If you want to prove the second fact, you consider the set $S$ of subsets $A \subseteq Y$ for which $f^{-1}(A)$ is a Borel set in $X$. You verify that $S$ is a $\sigma$-algebra and contains all open sets, hence contains the Borel $\sigma$-algebra.
A: Considering the subspace topology on $[0,1]$, open sets of $[0,1]$ are obtained by $U\cap [0,1]$ where $U$ is an open set in the original topology. Borels set of $[0,1]$, i.e. $\sigma-$algebra of its open sets are open, closed and $G_{\delta}$ and $F_{\sigma}$ sets of $[0,1]$, which is exactly Borel sets of $\mathbb R$ limited to $[0,1]$.
