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 Bendit May 2 at 13:54

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  • $\begingroup$ 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... $\endgroup$ – Xander Henderson May 2 at 3:57
  • 1
    $\begingroup$ Ok. Prove that (1) open intervals are measurable, and (2) the measurable sets form a sigma algebra. $\endgroup$ – 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 $$L\supset \cap \Sigma_{\Bbb R}(F)=B.$$

  • $\begingroup$ Thank you for your answer! I try to understand it. $\endgroup$ – Ongky Denny Wijaya May 3 at 6:03

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