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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.

t.b.
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Jean Carr
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    It should really be "a" Vitali set. There are lots of Vitali sets, and they can have wildly different properties! See for example this answer. – Arturo Magidin Sep 16 '11 at 03:35
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    Well, show it has the necessary properties: it contains $\mathbb{R}$; it's closed under complements (what is the complement of a symmetric difference? ); and it's closed under countable unions... Which one is giving you trouble? – Arturo Magidin Sep 16 '11 at 03:37
  • @Arturo Yes the original question states "a vitali set", so I think the conclusion holds for arbitrary vitali sets. – Jean Carr Sep 16 '11 at 04:38
  • Indeed, it will hold for any of them; my point is that "the Vitali set" is incorrect, since there are so many of them and they can be very different from one another. – Arturo Magidin Sep 16 '11 at 04:43
  • @Arturo I see. I have trouble in showing that M is closed under countable unions. Can you give some hint on this? – Jean Carr Sep 16 '11 at 14:56

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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.

Arturo Magidin
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  • I have trouble in showing step 5 and 7. Can you give some more details? I appreciate a lot. – Jean Carr Sep 16 '11 at 18:24
  • @Jean: For step 5, note that the $A_i$ are Lebesgue measurable, and the countable union of Lebesgue measurable sets is Lebesgue measurable. The $B_i$ are each contained in $V$, so their union is contained in $V$. Is there a prior step that gives the desired conclusion now? – Arturo Magidin Sep 16 '11 at 18:26
  • @Jean: For step 7: if you can show that the difference in step 5 is completely contained inside $V$, then that means that there exists a set $\mathcal{B}\subseteq V$ such that $((\cup A_i)-(\cap B_i))-\mathcal{B}=\cup(A_i-B_i)$; or a set $\mathcal{C}\subseteq V$ such that $((\cup A_i)-(\cup B_i))\cup \mathcal{C} = \cup(A_i-B_i)$. That should allow you to reach 7. – Arturo Magidin Sep 16 '11 at 18:28
  • @Jean: Oops; there was a typo in Step 1. Step 1 should have $A\in M$, not $A\subseteq V$. – Arturo Magidin Sep 16 '11 at 18:46
  • This is really helpful, thank you very much! – Jean Carr Sep 17 '11 at 00:54
  • @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. – Arturo Magidin Sep 17 '11 at 21:29