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.

Let $(\Omega,\mathscr A)$ be a measurable space. If $\varnothing \subset X \subset \Omega$, let $$\mathscr F = \{ F \subseteq \Omega, F = X \cap Y, Y \in \mathscr A\} \;. $$

I need to prove that $\mathscr F$ is a $ \sigma$-Algebra on $X$.

So, I have to show that

  1. $\varnothing \in \mathscr F$
  2. If $F \in \mathscr F$, then $F^C \in \mathscr F $
  3. If $F_i \in \mathscr F$, then $\bigcup_{i=1}^\infty F_i \in \mathscr F $

I have trouble in showing 2 and 3 conditions.

share|improve this question
    
In condition 2, one asks that $X\setminus F\in\mathscr F$ for every $F\in\mathscr F$, not that $\Omega\setminus F\in\mathscr F$. –  Did Sep 15 '12 at 20:50
    
Yes you are right. New Sigma Algebra should be on X. Does it make us go further? –  mathguy Sep 15 '12 at 21:10
    
Yes: for example, you could try to write down $X\setminus F$ using $\Omega\setminus F$. –  Did Sep 15 '12 at 21:16
    
Ok, I am trying to understand this: So we have to show that $X \setminus F \in \mathscr F$. I can't figure out how to use Y in this case. –  mathguy Sep 15 '12 at 22:44
    
So... you assume that $F=X\cap Y$ with $Y$ in $\mathscr A$ and you want to find $Z$ in $\mathscr A$ such that $X\setminus F=X\cap Z$. Any idea? –  Did Sep 15 '12 at 23:22

1 Answer 1

HINTS: For both (2) and (3), note that $F\in\mathscr{F}$ iff there is a $Y_F\in\mathscr{A}$ such that $F=X\cap Y_F$.

(2) What is $X\cap(\Omega\setminus Y_F)$?

(3) What is $X\cap\bigcup_iY_{F_i}$?

share|improve this answer
    
@ Brian, Thank you for the hints. As far as I can see, $ X\cap(\Omega\setminus Y_F) = X $ and $ X\cap\bigcup_iY_{F_i} = \mathscr F $. I am sorry but where does this lead to? –  mathguy Sep 16 '12 at 0:44
    
@mathguy: No, $X\cap(\Omega\setminus Y_F)=\{x:x\in X\text{ and }x\in\Omega\text{ and }x\notin Y_F\}=\{x:x\in X\text{ and }x\notin Y_F\}=X\setminus Y_F=X\setminus F$. The claim that $X\cap\bigcup_iY_{F_i}$ doesn’t even make sense: the lefthand side is a subset of $X$, and the righthand side is a collection of subsets of $X$. They don’t even have the same kind of objects as members. In fact $X\cap\bigcup_iY_{F_i}=\bigcup_i(X\cap Y_{F_i})$ by de Morgan’s law, and this is $\bigcup_iF_i$. –  Brian M. Scott Sep 16 '12 at 5:52

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.