# Prove the set is a $\sigma$-algebra

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.

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

• 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 Apr 30, 2022 at 20:18
• 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) Apr 30, 2022 at 20:33
• 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. Apr 30, 2022 at 20:37