Symmetric difference of a Vitali subset and a Lebesgue measurable sets Could anyone help to show the following? Thanks!
Let $V \subset [0,1]$ be a Vitali set. Let 
$$
M= \{ A \Delta B \,:\, A \subset [0,1] \text{ is Lebesgue measurable}, B\subset V \}
,$$ 
where $\Delta$ denote the symmetric difference. Prove that $M$ is a sigma-algebra. 
 A: So you are having trouble showing that $M$ is closed under countable unions. Here's a walkthrough of how I proved it (other solutions are likely possible): (Added: Indeed, there is a much simpler way of doing it which I thought of much later, added at the bottom)


*

*Show that if $A\in M$ and $B\subseteq V$, then $A$, $B$, $A\cup B$, and $A-B$ are all in $M$.

*Let $\{A_i\triangle B_i\}_{i=1}^{\infty}$ be a countable family of elements of $M$. Writing the symmetric difference as $X\triangle Y = (X-Y)\cup (Y-X)$, note that
$$\bigcup_{i=1}^{\infty}(A_i\triangle B_i) = \left(\bigcup_{i=1}^{\infty}(A_i-B_i)\right) \cup \left(\bigcup_{i=1}^{\infty}(B_i-A_i)\right).$$

*Show that $$\bigcup_{i=1}^{\infty}(B_i-A_i) = \mathcal{B}\subseteq V,$$
hence $\mathcal{B}\in M$. 

*Prove that
$$\left(\bigcup_{i=1}^{\infty}A_i\right) - \left(\bigcup_{i=1}^{\infty} B_i\right) \subseteq \bigcup_{i=1}^{\infty}(A_i-B_i) \subseteq \left(\bigcup_{i=1}^{\infty}A_i\right) - \left(\bigcap_{i=1}^{\infty}B_i\right).$$

*Show that
$$\left(\bigcup_{i=1}^{\infty}A_i\right) - \left(\bigcup_{i=1}^{\infty} B_i\right) \in M.$$

*Think about what kind of elements can lie in
$$\left(\left(\bigcup_{i=1}^{\infty}A_i\right) - \left(\bigcap_{i=1}^{\infty}B_i\right)\right) - \left(\left(\bigcup_{i=1}^{\infty}A_i\right) - \left(\bigcup_{i=1}^{\infty} B_i\right)\right).$$

*Conclude that
$$\bigcup_{i=1}^{\infty}(A_i-B_i) \in M.$$

*Conclude that $M$ is closed under countable unions.

Added. In fact, much simpler is to note that
$$\left(\bigcup_{i=1}^{\infty}A_i\right) - V \subseteq \bigcup_{i=1}^{\infty}(A_i\triangle B_i) \subseteq \left(\bigcup_{i=1}^{\infty} A_i\right)\cup V.$$
Now, both the smallest and largest of the three sets lie in $M$, and their difference is a subset of $V$; therefore, the middle set will lie in $M$ by point 1.
