Let $E$ be a Lebesgue measurable subset of $\mathbb{R} $ and let $A \subset \mathbb{R} $.Show that $m^*(E \cap A) + m^*(E \cup A)=m^*(E)+ m^*(A)$.

Question

If $A$ was given to be measurable then I can do it by breaking the union into disjoint set. How to do it in general case?

Answer 1

For all $B \subset \mathbb{R}$, $m^*(B) = m^*(B \cap E) + m^*(B \cap E^c)$ because $E$ is measurable.

Let $B = A \cup E$. Then, \begin{align} m^*(A \cup E) &= m^*((A \cup E) \cap E) + m^*((A \cup E) \cap E^c)\\ &= m^*(E) + m^*(A \cap E^c). \end{align}

Add $m^*(A \cap E)$ to both sides and use the measurability of $E$ once more to get \begin{align} m^*(A \cup E) + m^*(A \cap E) &= m^*(E) + m^*(A \cap E^c) + m^*(A \cap E)\\ &= m^*(E) + m^*(A). \end{align}