If $B \subset A \subset \mathbb{R}$ such that $m^*(B)=0 $. Show that $m^*(A \setminus B)=m^*(A)$.

Question

I was stuck in a problem that required to show the above result where $m^*$ is Lebesgue measure.I proceeded in the following way. Since $A\setminus B \subset A$ therefore $m^*(A\setminus B) \le m^*(A)$ but I am not able to get other way inequality . I don't know I can say the measure of union of disjoint sets is sum of measure of individual sets. I think this result holds for only measurable set.That's why I am not able to prove the other way inequality.Can someone enlighten me on this?

Answer 1

By subadditivity of Lebesgue outer measure, $$m^*(A) \le m^*(A \setminus B) + m^*(A \cap B) \le m^*(A \setminus B) + m^*(B) = m^*(A \setminus B) + 0 = m^*(A \setminus B),$$ completing the remaining half of the question.

Answer 2

For any $r>0$ let $U_r$ be open with $U_r\supset A\setminus B$ and $m(U_r)\leq m^*(A\setminus B)+r/2.$ And let $V_r$ be open with $V_r\supset B$ and $m(V_r)\leq r/2.$ Then $U_r\cup V_r$ is open and covers $A$ so $m^*(A)\leq m(U_r\cup V_r).$ $$\text {So we have }\quad m^*(A\setminus B)\leq m^*(A)\leq m(U_r\cup V_r)\leq$$ $$\leq m(U_r)+m(V_r)\leq$$ $$\leq m^*(A\setminus B)+r/2+r/2=m^*(A\setminus B)+r.$$ $$\text {Therefore }\quad m^*(A\setminus B)\leq m^*(A)\leq m^*(A\setminus B)+r.$$ This is not possible for $all $ $r>0$ unless $m^*(A\setminus B)=m^*(A).$