$\pi$-system that generates the $\sigma$-algebra of Borel sets

Question

I have the following problem.

Give an example of a $\pi$-system of subsets of $\mathbb{R}$ that generates the $\sigma$-algebra of Borel sets $\mathscr{B}(\mathbb{R})$.

This seems trivial to me. Let C be the collection of all intervals of the form $(-\infty,m]$, where $m \in \mathbb{R}$. This is a $\pi$-system because if $a\in C$ and $b\in C$, then $a\cap b$ will be an interval of the form $(-\infty,m]$, which is in C. Furthermore, $\sigma(C)=\mathscr{B}(\mathbb{R})$, because the $\pi$-system C, itself, can be used to generated the Borel $\sigma$-algebra $\mathscr{B}(\mathbb{R})$.

I am not sure whether I am missing something or the question is just very straighforward.

I'd appreciate any help.

Thanks!

Answer 1

Indeed the family sets of the form $(-\infty,a]$ is a $\pi-$system. that generates the Borel sigma algebra.

Off course in this family you have to include the empty set.

Another $\pi-$system which generates Borel is $C=\{(a,b]:a<b\} \cup \{\emptyset\}$