A problem on outer measure and Lebesgue measure I need some help with

Question

Let $X$ be a set that is Lebesgue measurable. Let $Y$ be another set, and define $X \triangle Y$ to be their symmetric difference. Given this information we also have that $\mu^{\ast}(X \triangle Y) = 0$. I want to show that $Y$ is also Lebesgue measurable and $\mu(X) = \mu(Y)$. Here the $\mu^{\ast}$ is the outer measure. The help would be greatly appreciated for this problem!!

Answer 1

We have $Y=Y\cap X \cup (Y \setminus X)$ and $X=Y\cap X \cup (X \setminus Y)$. Since $\mu^*(X\Delta Y)=0$ we claim that $\mu^*(Y \setminus X)=0$ and $\mu^*(X \setminus Y)=0$ which follows $\mu(Y \setminus X)=0$ and $\mu(X \setminus Y)=0$. Hence $Y \setminus X$ and $X \setminus Y$ are Lebesgue measurable. Since $X$ is Lebesgue measurable we deduce that $X \setminus (X \setminus Y)=X \cap Y$ is Lebesgue measurable which implies that $(X \cap Y) \cup (Y \setminus X)=Y$ also is Lebesgue measurable. Finally we get $\mu(Y)=\mu((Y \cap X) \cup (Y \setminus X))=\mu((X \cap Y))+\mu(Y \setminus X)=\mu((X \cap Y))=\mu((X \cap Y))+\mu(X \setminus Y)=\mu(X).$