Ex3-34

Let E and Z be two sets in \mathbb{R}^n such that Z has measure 0. If E \cup Z is measurable, show that E is measurable and |E\cup Z|=|E| .
Solution:
# In order to get E from E \cup Z , we can remove the elements of Z from E \cup Z first. Since this operation will also remove the elements of E that are in Z , we need to add the elements of E \cap Z back. Hence, we get\begin{align}
E&= ((E \cup Z) - Z ) \cup (E \cap Z).
\end{align}By using some set operations, \begin{align}
E&= ((E \cup Z) - Z ) \cup (E \cap Z)\\
&= ((E\cup Z)\cup (E \cap Z)) \cap (Z^C \cup (E\cap Z))\\
&=(E\cup Z) \cap (Z^C \cup E)\\
&=(E\cup Z)\cap (Z\cap E^C)^C.
\end{align}Now, since Z\cap E^C is a subset of Z , it has measure 0, hence measurable. This implies that the complement of Z\cap E^C is measurable. Thus, E being the intersection of two measurable sets is measurable. 
#To prove |E\cup Z|=|E| , since E and Z are measurable,\begin{align}
|E \cup Z| &=|E|+|Z|-|E \cap Z|\\
&= |E| && \text{(since }E \text{ and } E\cap Z \text{ have mesaure zero.)}
\end{align}