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Let be $(X,F)$ a measurable space and $P \subset X$ a subset. Show that

$$T=\{(A \ \cap \ P)\ \cup (B\ \cup P^{c}): A,B \in F) \}$$ is a $\sigma$-algebra.

My thoughts to this exercise:

Well $X$ is an arbitrary set and $F$ is a $\sigma$-algebra. An intersection of multiple $\sigma$-algebras is also a $\sigma$ -algebra, but in the exercise $M$ is simply a subset of an arbitrary set. I know also the definition of $\sigma$-algebra but I have no idea how to prove these 3 points of the definition. I would appreciate a lot your help in advance.

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I mean first you need to show that $X\in T$. Since $F$ is a sigma algebra on $X$ we know that $X\in F$. Hence take $A=\emptyset\in F$ and $B=X$ then $X=(A\cap P)\cup (B\cup P^c)\in T$.

Now let $U\in T$ we need also to show that $U^c\in T$. Since $U\in T$ we know that $U=(A\cap P)\cup (B\cup P^c)$ for some $A,B\in F$ but then $U^c=\left((A\cap P)\cup (B\cup P^c)\right)^c=(A\cap P)^c\cap (B\cup P^c)^c=(A^c\cup P^c)\cap (B^c\cap P)$. But now let me remark that since $F$ is a sigma algebra $A^c, B^c\in F$. Hence $U^c\in T$.

Lastly take $U_1,...,U_n\in T$ and denote $U_i=(A_i\cap P)\cup (B_i\cup P^c)$. We want to show that $U:=\bigcup_{i=1}^n U_i\in T$. By definition $$U=\bigcup_{i=1}^n (A_i\cap P)\cup (B_i\cup P^c)=\bigcup_{i=1}^n (A_i\cap P)\cup \bigcup_{i=1}^n (B_i\cup P^c)=\left( \bigcup_{i=1}^nA_i\cap P\right) \cup\left( \bigcup_{i=1}^n B_i\cup P^c\right)$$

Now since $F$ is a sigma algebra and $A_i, B_i\in F$ we know that also $\bigcup_{i=1}^nA_i, \bigcup_{i=1}^nB_i\in F$. Hence $U\in T$

This shows by using the definition that $T$ is a $\sigma$-algebra

I hope this helps

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  • $\begingroup$ very nice solution. Could I use "the same proof" if for instance $T=\{(A \ \cap \ P)\ \cup (B\ \cap P^{c}): A,B \in F) \}$ or $T=\{(A \ \cup \ P)\ \cup (B\ \cap P^{c}): A,B \in F) \}$ just by changing the adequate $ \cap$ or $\cup$ signs? this question is purely for my interest. Thank you in advance $\endgroup$
    – Herrpeter
    Apr 30, 2022 at 20:18
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    $\begingroup$ I mean you can try it. It's a good exercise but I somehow don't think this will give you a $\sigma$-algebra. (Sorry I have no time to think about it yet) $\endgroup$
    – user123234
    Apr 30, 2022 at 20:33
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    $\begingroup$ Maybe one idea to find a counterexample is $X=\Bbb{R}$, $F=\{\emptyset, X\}$. Then define $T$ as you want. Assuming your $T$ is a $\sigma$-algebra $\emptyset \in T$. Now maybe you get a contradiction. But I'm not sure here. $\endgroup$
    – user123234
    Apr 30, 2022 at 20:37

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