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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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    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?2012-09-15

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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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    @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 $\bigcup_iF_i$.2012-09-16