Show that if $Z\cup E$ is measurable then so is $E.$ Problem: 

Let $E,Z \in\mathbb{R}^{n}$ and $Z$ be a set of measure zero. Show that if $Z\cup E$ is measurable then so is $E.$

What I have done:

Since $Z\cup E$ is measurable then there exists 
  $G$ 
  such that $Z\cup E\subset G$ consequently $ E\subset G$.
  Also,
  $\vert G-(Z\cup E) \vert_{e}<\epsilon $
  But 
  $Z$ is also zero measure so there exists $G'$ such that
  $\vert G'-(Z) \vert_{e}<\epsilon $ and $Z\subset G'$
  But I can't say that 
  $\vert G-E\vert<\epsilon'$
  Actually, I have to find an open set like $G$ whcih cantains $E$(which I have done) and $\vert G-E\vert<\epsilon'$

I need just a hint, not the whole solution. Maybe my approach is wrong entirely. 

Hint
I suppose you are talking about the Lebesgue measure (tell me if that is not the case). Since $Z$ is a null set, it is measurable. Define the following set:
\begin{align} 
A:=[Z\cup E] \cap [Z\cap E^c]^c
\end{align} 
Now where is the set $A$ equal to? And what can you say about it? 
Edit: Thanks to @Bungo for pointing out a set which is more useful than the one I had in mind. 
