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. 

I proceed in this way:
According to the Hint by @shashi, 
$E=(E\cup Z)\cap (Z\cap E^{c})^{c} $
we know that 
$E\cup Z$ 
is measurable. 
But we know that $(Z\cap E^{c})\subset Z$ and $Z$ is zero measure and our space is Lebesgue measure and consequently being sigma algebra and having completion(every subset of zero measure is of measure zero). 
Therefore, $ Z-E $ is measurable and then $(Z-E)^{c}$ is measurable. Then $E$ is measurable.     
