# Sigma Algebra Proof

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• Sep 2nd 2010, 05:19 AM
AdvoTV
Sigma Algebra Proof
Hello, I would be grateful if I could receive some assistance with the following question:

Let $\mathcal{C}$ be a $\sigma-field$, and let $A \in \mathcal{C}$.

Define $\mathcal{C}_{\mathcal{A}}=\{C:C=A \cap B \text{ for some } B \in\mathcal{C}\}$

Show that $\mathcal{C}_{\mathcal{A}}$ is a $\sigma-field$. [Note: Complements of sets are formed with respect to $A$]

Thank you so much in advance to anybody who helps me with this. This material is so confusing for me, and I'm sure I can get it, but it will take time, practice, and guidance.
• Sep 2nd 2010, 06:11 AM
Plato
You do understand that this is not a homework service?
Please at least show what you have tried.
• Sep 2nd 2010, 06:19 AM
AdvoTV
Of course
I understand so far that if I take B to be the null set (since it is an element of the sigma algebra we are given), that it shows the null set is part of the described set since the intersection of the null set and A would be the null set.

When I start with the intersection of A and B, however, I'm having trouble showing that the complement can be written in the form that satisfies the described set. I'm not sure I'm supposed to go towards the use of DeMorgan's laws to write things as an intersection? So, I'm essentially having a problem showing closure under complements.

I'm also not quite sure why it was necessary that the problem states that complementation is with respect to A?
• Sep 2nd 2010, 06:39 AM
Plato
Quote:

Originally Posted by AdvoTV
why it was necessary that the problem states that complementation is with respect to A?

In terms of the ‘new’ $\sigma$-algebra it means that $C'=A\setminus C$.
Therefore in the given $\sigma$-algebra $C'=A'\cup B'$ in new the $\sigma$-algebra $C'=A\cap B'$.