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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.

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    @$A$rturo I see. I have trouble in showing that M is closed under countable unions. Can you give some hint on this?2011-09-16

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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)

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

  2. 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 $\bigcup_{i=1}^{\infty}(A_i\triangle B_i) = \left(\bigcup_{i=1}^{\infty}(A_i-B_i)\right) \cup \left(\bigcup_{i=1}^{\infty}(B_i-A_i)\right).$

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

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

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

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

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

  8. 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.

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    @Jean: It is actually easier to do the following (thought of it later): the union $\cup (A_i\triangle B_i)$ is contained in $(\cup A_i)\cup V$, and contains $(\cup A_i)-V$. As both of these lie in $M$ and their difference is a subset of $V$, it follows that the original union lies in $M$ as well.2011-09-17