My teacher gave me an example of performing the subject:
Example
Let $\Omega = \Bbb R$ and $\mathcal R = \{(-\infty,-1),(1,+\infty)\}$.
Then $\sigma(\mathcal R) = \{\emptyset, \Bbb R, (-\infty,-1),
(1,+\infty), [-1, \infty), (-\infty,1],
(-\infty,-1)\cup(1,+\infty),[-1,1]\}$.
There was a different example, where she also generated the smallest $\sigma$-algebra for the family of sets $\mathcal A = \{A,B\} \subset 2^\Omega$ in the same way: $\sigma(\mathcal R) =
\{\emptyset, \Omega, A, B, A^C,B^C,A\cup B, (A\cup B)^C\}$.
I certainly understand why $\emptyset$, $\Omega $ and a family of sets itself are there, and I certainly know that $\sigma$-algebra is closed under the operations of complement and union. What I don't understand about the other elements of generated $\sigma$-algebras: why we are taking exactly them? Does this work in a general case?
Best Answer
You start with a set of sets, in your example, $\{A,B\}$. To obtain the smallest $\sigma$-algebra containing it, all you need to do is add the missing sets that make it a $\sigma$-algebra (instead of just being a set).
What this means is that you want to add all sets so that the resulting set is closed with respect to taking complements and union. To make $\{A,B\}$ closed with respect to taking complements you need to add $A^c, B^c$. To make it closed with respect to union you need to add $A \cup B$. Now you have added new elements and again you need to add all elements so that the new set of sets is closed under complement and union. Hence you need to add $A^c \cup B^c = (A \cap B)^c$ and $(A \cup B)^c = A^c \cap B^c$. Next you need to add $A \cap B$ (to make it closed with respect to complements).
So far we have $ \{ A, B, A^c , B^c, A \cup B, (A \cup B)^c, A^c \cup B^c , A \cap B \}$. Do we need to add any more sets? Or does this set constitute a $\sigma$-algebra?