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Let $M$ be an infinite $\sigma$ algebra on X. Show that if $A \in M$, then the collection $M_A$={$B \cap A: B \in M$} is a $\sigma$ algebra on A.

My thoughts/attempt (Really struggling)

To show $M_A$ is a $\sigma$-algebra, I need to show that $M_A$ is closed under complements and countable unions.
I'm really confused on how to show either but here is my attempt anyways.

First I show finite unions:

Let $B_1 \cap A$ $\in M_A$ and $B_2 \cap A$ $\in M_A$ (thus $A$, $B_1$ and $B_2 \in M$).

$(B_1 \cap A)$ $\cup$ $(B_2 \cap A)$=($B_1 \cup B_2)$ $\cap$ $A$ $\in M_A$ $(B_1 \cup B_2 \in M$ as M is a $\sigma$ algebra). Thus we closed under finite union. Can do the same process to show for countable union.

Second: I show closure under complements:

Let $B_1 \cap A$ $\in M_A$. Need to show that $(B_1 \cap A)$$^{c}$ $\in$ $M_A$.

I can write $(B_1 \cap A)$$^{c}$ = ($B_1 \cap A^C$) $\cup$ ($B_1^C \cap A$).

$B_1 \cap A^C \in M_A$ as $A^c \in M$ as $M$ is a $\sigma$ algebra. $B_1^c \cap A \in M_A$ as $B_1^c \in M$ as $M$ is a $\sigma$ algebra. Thus, $(B_1 \cap A)$$^{c}$ $\in M_A$ as it is the union of two things in $M_A$ and we previously showed that $M_A$ is closed under finite union.
Is my proof correct? Or am I very off in my approach? Thank you.

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  • $\begingroup$ I think your approach is sound, you just need to make sure you get the countable union case right. But as you say, it's essentially the same as the finite case. $\endgroup$ – nathan.j.mcdougall Sep 5 '18 at 4:15
  • $\begingroup$ $(B_1 \cap A^C) \cap A = \emptyset$ $\endgroup$ – dEmigOd Sep 5 '18 at 6:14
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The main problem is that $B_1 \cap A^C \not\subset A$ !, so $B_1 \cap A^C \notin M_A$. The complement should be found in $A$, so the complement in $A$ for $B_1 \cap A$ is $B_1^C \cap A$.

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  • $\begingroup$ So the complement of B1 intersection A is B1 complement intersection A and this entity is in Ma as B1 complement is in M as M is a sigma algebra? $\endgroup$ – kemb Sep 5 '18 at 7:32
  • $\begingroup$ Also is the union part ok? or is it just the complements I need to fix? How do I show that the complement of ($B_1$ $\cap$ A is $B_1^c \cap A$ or is this just by construction (what I'm thinking)? $\endgroup$ – kemb Sep 5 '18 at 7:33
  • $\begingroup$ It is by construction. you are looking for complements in $A$, not in $X$ $\endgroup$ – dEmigOd Sep 5 '18 at 13:21

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