I know that the $\sigma$-field generated by the class of all open subsets of $\mathbf{R}$.
I want to show that a field generated by $\pi (\mathbf{R} )$ is the Borel $\sigma$-field ( $\mathbf{B} (\mathbf{R} )$) on $\mathbf{R}$.
i.e. $(\pi (\mathbf{R} ))\ =\ \ \sigma (\pi (\mathbf{R} )$
considering $\pi (\mathbf{R} ))\ =\ \{ (-\infty ,a]\ :\ a\ \varepsilon \ \mathbf{R} )$
I read that I need to show $\mathbf{B} (\mathbf{R} )\ \subseteq \ \sigma (\pi (\mathbf{R} ))$
But I need more insight.