Why do $m^*(A-x_0\cap E)=m^*(A\cap E+x_0)$?

Question

During a proof a an equation was used: $$m^*((A-x_0)\cap E)+m^*((A-x_0)\cap E^C)=m^*(A\cap (E+x_0))+m^*(A\cap (E^C+x_0))$$

$E$ is measurable and the outer measure is invariant to shifting, which property have been used for this equation?

Answer 1

This is basic set theory in that $(A-x) \cap E = A \cap (E+x) - x$. To proof this:

1. $y\in (A-x) \cap E$. Thus $y+x \in A$ and $y+x\in E+x$ thus $y+x \in A \cap (E+x)$, thus $y\in A \cap (E+x)-x$, thus $(A-x) \cap E \subseteq A \cap (E+x) - x$.

2. $y\in A \cap (E+x) - x$ implies $y+x \in A$ and $y+x\in E+x$ which implies $y\in A-x$ and $y\in E$, thus $y\in (A-x)\cap E$. Thus $A \cap (E+x) - x \subseteq (A-x) \cap E$.