# Sigma-Algebra

• Aug 7th 2009, 07:01 PM
MrJack1990
Sigma-Algebra
Let $\displaystyle A,B,C$ be a partition of $\displaystyle \Omega$. Write out the smallest $\displaystyle \sigma$-algebra containing the sets $\displaystyle A,B,$ and $\displaystyle C$
• Aug 7th 2009, 07:43 PM
MrJack1990
I have no idea if this is correct, but this is my best estimation of what the question is asking.

A partition is such that
$\displaystyle A \cap B = \emptyset$
$\displaystyle A \cap C = \emptyset$
$\displaystyle B \cap C = \emptyset$

and $\displaystyle A \cup B \cup C = \Omega$ i.e. is exhaustive

Then by definition of borel-algebra we have
(1) $\displaystyle \Omega \ \in \ \mathcal{F}$
(2) As $\displaystyle A,B,C \ \in \mathcal{F}, A^{c}, B^{c}, C^{c} \in \mathcal{F}$
(3) As $\displaystyle A,B,C \in \mathcal{F}, A \cup B \cup C \in \mathcal{F}$

Is this right?
• Aug 7th 2009, 11:18 PM
Moo
Hello,

(you're asked for a sigma algebra, not a Borel algebra :))

So yup, the empty set and $\displaystyle \Omega$ belong to $\displaystyle \mathcal{F}$ (first+second axiom)

A,B,C belong to $\displaystyle \mathcal{F}$

Their complement belong to $\displaystyle \mathcal{F}$.
But... you can have a more precise formula for these complements :

$\displaystyle A^c=B\cup C$
$\displaystyle B^c=A\cup C$
$\displaystyle C^c=A\cup B$

Do you agree ?
We don't need to include $\displaystyle A\cup B\cup C$ because it's $\displaystyle \Omega$...

And thus the third axiom is automatically checked.

So the smallest $\displaystyle \sigma$-algebra containing A,B,C (also called the generated $\displaystyle \sigma$-algebra by A,B,C) is :
$\displaystyle \mathcal{F}=\{\emptyset,A,B,C,A\cup B,A\cup C,B\cup C,\Omega\}$