# Proof borel sets are measurable sets [closed]

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?

## closed as off-topic by Xander Henderson, RRL, YuiTo Cheng, Cesareo, Theo BenditMay 2 at 13:54

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Xander Henderson, RRL, YuiTo Cheng, Cesareo, Theo Bendit
If this question can be reworded to fit the rules in the help center, please edit the question.

• Define what it means for a set to be "measurable". In most of the expositions of which I am aware, one starts with the open sets, defines a premeasure there, then generates a $\sigma$-algebra in a way which makes Borel sets measurable by construction. Thus I don't understand the question... – Xander Henderson May 2 at 3:57
• Ok. Prove that (1) open intervals are measurable, and (2) the measurable sets form a sigma algebra. – Andrés E. Caicedo May 2 at 4:06

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 $$L\supset \cap \Sigma_{\Bbb R}(F)=B.$$