I am doing a measure theory problem which says:
Let $V$ be the Vitali set in $\mathbb{R}$. Deduce that $E=V \times \lbrace0\rbrace$ is Lebesgue measurable in $\mathbb{R}^2$ but it's not a Borel set.
I think I did well the first part. Let $V$ be a Vitali set. We know that $V \subset [0,1]$. Let $\varepsilon>0$. We can take the rectangle $[0,1] \times (-\varepsilon,\varepsilon)$, which contains $E=V \times \lbrace0\rbrace$, and is measurable. Taking limit when $\varepsilon \rightarrow 0$, we have a measure $0$ set. Since the Lebesgue measure is complete, every subset of a $0$ measure set has measure $0$, so $E=V \times \lbrace0\rbrace$ is measuable.
My problem is with the second part. I have an idea, but I don't know if it's correct or not:
We know that the Vitali set is not Borel. We can see $E$ as the inverse image of the inclusion of $\mathbb{R}$ in the $X$ axis of $\mathbb{R}^2$, which is a measurable function. So, I tried the typical proof by contradiction:
Suppose $E=V \times \lbrace0\rbrace$ is Borel. Then, $g^{-1}(E)$ has to be borel. But that is $V$! So $E$ is not Borel.
Is this ok? Thanks in advance