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

Question

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.

Answer 1

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.

Answer 2

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.