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August 29th, 2017, 07:13 PM  #1 
Senior Member Joined: Oct 2015 From: Antarctica Posts: 128 Thanks: 0  Show the the intersection of two sigma fields is a sigma field
In probability, a sigma field (algebra) is defined as a set F whose elements are sets and that satisfies the following properties: 1. The null set and the sample space are members of F. 2. If an event A is a member of F, then the complement of A is a member of F 3. If events A1, A2, … An are members of F, then the event A1 U A2 … U An is a member of F. So I have to show that the intersection of ANY two sigma fields is a sigma field. The first property is easy to show. By definition, any any two sigma fields will contain the null set and the sample space, so the intersection between any two sigma fields will inevitably include both the null set and the sample space. Done. The second two properties are where I'm getting tripped up. I'm not sure how I would show, with mathematical notation, that for the intersection of any two sigma fields in which the resulting set is different from the trivial sigma field ({Null Set, Sample Space}) will always contain at least one event and its complement. I understand it in my head, but don't know how to write it down. Finally, I have no idea where to start with the proof for the unions. Would somebody please offer some advice on how to write this proof in mathematical notation? Thank you. 
August 29th, 2017, 07:22 PM  #2 
Senior Member Joined: Aug 2012 Posts: 2,010 Thanks: 574 
Doesn't the "sigma" in sigma field indicate that the unions and intersections may be countable? (Yes). Your notation has them as finite. The proof won't be much different either way but if you intend countability you should indicate that. If X is in the intersection then it's in each of the fields hence its complement is in each field hence ... For the unions, same kind of reasoning. If X and Y are in the intersection then X and Y are each in both fields therefore X union Y is in each field therefore ... Last edited by Maschke; August 29th, 2017 at 07:57 PM. 
August 29th, 2017, 08:04 PM  #3 
Senior Member Joined: Oct 2015 From: Antarctica Posts: 128 Thanks: 0 
Okay, I greatly appreciate the help. I believe that I fully understand where to go from here. Thank you.

September 3rd, 2017, 04:36 AM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
A set is in the intersection of fields F and G if and only if it is in both F and G. Since the set is in F, its complement is in F. Since it is in G, its complement is in G. Therefore, its complement is in the intersection of F and G. Similarly for (3). If all of sets A1, A2, ... are in F intersection G then each is in F and in G. Since they are all in F, their union is in F. Since they are all in G, their union is G. Therefor their union is in the intersection of F and G. Do you see why, if you take the union of F and G it is not necessarily a field? 

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