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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.

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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

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 $X\cap\bigcup_iY_{F_i}$?

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@ 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

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