I'm really stuck.

If $\mathcal{A}$ is a sigma-algebra in $X$ and $C\subset X$ does not belong to $\mathcal{A}$, how do I need to show that the sigma-algebra $\sigma(\mathcal{A}\cup \{C\})$ (so generated by $\mathcal{A}$ and $C$) is equal to the collection of sets of the form $K=(A \cap C) \cup (B \cap C^c) $, with $A,B \in \mathcal{A}$.

Hope somebody can help me how to solve this problem.

  • $\begingroup$ You probably mean $$\sigma(\mathcal{A}\cup\{C\})$$ instead of $$\sigma(\mathcal{A}\cup C)$$ What did you try "to solve this problem"? $\endgroup$ – Did Jan 8 '18 at 18:30
  • $\begingroup$ Yes, that's what I mean. I will edit the original post. I already tried to write down all possibilities of the sigma-algebra generated by the set, but that are to much possibilities. After that I tried to show that $\sigma(\mathcal{A} \cup \{C\})$ is a subset of K and the other inclusion. But that also didn't work. I think drhab gives the most direct way to prove it. $\endgroup$ – MathI Jan 8 '18 at 19:06

$$[(A\cap C)\cup(B\cap C^{\complement})]^{\complement}=(A^{\complement}\cap C)\cup(B^{\complement}\cap C^{\complement})$$ so the collection of sets of that form is closed under complements.

$$\bigcup_{n=1}^{\infty}[(A_n\cap C)\cup(B_n\cap C^{\complement})]=((\bigcup_{n=1}^{\infty}A_n)\cap C)\cup ((\bigcup_{n=1}^{\infty}B_n)\cap C^{\complement})$$

so the collection of sets of that form is closed under countable unions.

Then it must be a $\sigma$-algebra that evidently contains $\mathcal A$ as a subcollection and has $C$ as one of its elements.

Further it is evident that any $\sigma$-algebra with these properties will contain every element of the form $(A\cap C)\cup(B\cap C^{\complement})$ where $A,B\in\mathcal A$.


$\begin{aligned}\left[\left(A\cap C\right)\cup\left(B\cap C^{\complement}\right)\right]^{\complement} & =\left(A\cap C\right)^{\complement}\cap\left(B\cap C^{\complement}\right)^{\complement}\\ & =\left(A^{\complement}\cup C^{\complement}\right)\cap\left(B^{\complement}\cup C\right)\\ & =\left(A^{\complement}\cap B^{\complement}\right)\cup\left(A^{\complement}\cap C\right)\cup\left(B^{\complement}\cap C^{\complement}\right)\\ & =\left(A^{\complement}\cap B^{\complement}\cap C\right)\cup\left(A^{\complement}\cap C\right)\cup\left(A^{\complement}\cap B^{\complement}\cap C^{\complement}\right)\cup\left(B^{\complement}\cap C^{\complement}\right)\\ & =\left(A^{\complement}\cap C\right)\cup\left(B^{\complement}\cap C^{\complement}\right) \end{aligned} $

  • $\begingroup$ Thanks for your reaction. I just have problems to see why $[(A\cap C)\cup(B\cap C^{\complement})]^{\complement}=(A^{\complement}\cap C)\cup(B^{\complement}\cap C^{\complement})$. Even when I paint the diagrams it's not the same. Did you change intersection and union or is it me who see this wrong? $\endgroup$ – MathI Jan 8 '18 at 17:35
  • $\begingroup$ Do you mean (one of) the 2 equalities that I mention, or is it something else? $\endgroup$ – drhab Jan 8 '18 at 17:37
  • $\begingroup$ I now complete my reaction. I cannot use Enter here, because then I add a comment directly. I mean the $[(A\cap C)\cup(B\cap C^{\complement})]^{\complement}=(A^{\complement}\cap C)\cup(B^{\complement}\cap C^{\complement})$. $\endgroup$ – MathI Jan 8 '18 at 17:40
  • $\begingroup$ I have edited to make things more clear. $\endgroup$ – drhab Jan 8 '18 at 18:24
  • $\begingroup$ Thank you, now it's clear! I just needed a few more steps. $\endgroup$ – MathI Jan 8 '18 at 19:03

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