Proof that : if $B$ is Borel Set, then $B$ measurable set.
I know the definition of measureable sets, for all $A\subseteq \mathbb{R}$, $m^*(A)=m^*(A\cap E)+m^*(A\cap E^C)$, which $m*$ denote the outer measure. I cannot associating with the definition of Borel sets(\sigma algebra containing open sets)
I have read from this https://mathcs.org/analysis/reals/integ/proofs/borelmbl.html.
I can't understand it. Anyone can help me to explain how to proof it?
For a family $F$ of subsets of a set $X$ let $\Sigma_X(F)$ be the set of every $\sigma$-algebra $S$ on $X$ such that $F\subset S.$
First, $\Sigma_X(F)\ne \emptyset$ because $P(X)$ (the power-set of $X, $ i.e. the set of all subsets of $X$) belongs to $\Sigma_X(F).$
Second, if $\emptyset \ne A\subset \Sigma_X(F)$ then $\cap A\in \Sigma_X(F).$ In particular, with $A=\Sigma_X(F)$ we have $\cap \Sigma_X(F)\in \Sigma_X(F).$ So $\cap \Sigma_X(F)$ is the $\subset$-least member of $\Sigma_X(F),$ and it is called the $\sigma$-algebra on $X$ generated by $F.$
Let $X=\Bbb R$ and let $F$ be the set of all open subsets of $\Bbb R.$ Then the set $B$ of Borel sets on $\Bbb R $ is, by definition, $\cap \Sigma_{\Bbb R}(F).$
Now the set $L$ of all Lebesgue-measurable subsets of $\Bbb R$ is a $\sigma$-algebra on $\Bbb R$ and we have $ F\subset L.$ So $L\in \Sigma_{\Bbb R}(F).$ Therefore $$L\supset \cap \Sigma_{\Bbb R}(F)=B.$$