Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
3  
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 3:35
1  
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 3:37
    
@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 4:38
    
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 4:43
    
@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
add comment

1 Answer

up vote 2 down vote accepted

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.

share|improve this answer
    
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 0:54
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.