Lemma 9:30 Introduction to Mathematical Analysis by Browder

Question

In the Lemma proof,They have written "Let $I$ be bounded open interval containing $\overline{B}$;there exixts an open $V$ with $I\supset V\supset I\setminus B$ such that $m(V)<m(I\setminus B)+\epsilon/2$ "My question is why $V$ is contained in I?

I mean I know, By definition of $m(I\setminus V)$,we can find a open interval $V$ such that $I\setminus V \subset V$ and $m(V)<m(I\setminus B)+\epsilon/2$" for given $\epsilon$.But Why this $V$ we will be cointained in $I$?

Answer 1

If you have $I\setminus B \subset V_0$ with $m(V_0)<m(I\setminus B)+\epsilon /2$ the $V=I \cap V_0$ is such that $V \subset I$ and $I\setminus B \subset V$ with $m(V)<m(I\setminus B)+\epsilon /2$.