# Find the least $\sigma$- algebra generated by $C=\{A,B\}$, subsets of $\Omega$ where $A\cup B= \Omega$?

Basically i tried to prove that $A\cup B= \Omega \Rightarrow B= A^{C}$. so then the $\sigma(C)= \{\emptyset, \Omega, A, A^{C}\}$. is right?.

• $B=A^c$ just if they are disjoint (and its immediate if they really are) – Robson Sep 13 '18 at 16:04
• $A\cup B=\Omega$ does not imply $B=A^C$, so you will not be able to prove this. – Mike Earnest Sep 13 '18 at 16:09

how about $\{A,B,A^c,B^c,A^c\cup B^c, A\cap B, \Omega,\emptyset\}$
The $\sigma$-álgebra generated by a set $C$ is by definition the least $\sigma$-álgebra that contains $C$.
Observe that the following set is a $\sigma$-álgebra and is in fact the least one: $$\mathcal{A}=\{\emptyset,\Omega,A,B,A^c,B^c,A\cap B,A\cap B^c,B\cap A^cA^c\cap B^c\}$$ Note that here it isn't of our preocupation if some of the elements are write twice or more.
You can see that $\mathcal{A}$ satisfy all properties for being a $\sigma$-álgebra.
And it is the least one because if $\mathcal{A'}$ is another $\sigma$-álgebra such that $A,B\in \mathcal{A'}$, then by the properties of a $\sigma$-álgebra we have that all elements of $\mathcal{A}$ are elements of $\mathcal{A'}$ too, i.e., $\mathcal{A}\subset\mathcal{A'}$.