Probability Spaces: Showing G is Field and not Sigma Field

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The discussion focuses on demonstrating that the collection G of subsets of the interval [0,1) is a field but not a sigma-field. It establishes that G contains the empty set and is closed under complementation and finite unions, fulfilling the criteria for a field. However, it struggles with showing that G is not a sigma-field, as the union of infinitely many intervals can lead to sets not contained in G, such as the interval (0,1). The participants clarify that while finite unions of intervals are included in G, the requirement for sigma-fields necessitates closure under countable unions, which G does not satisfy. Overall, the distinction between field and sigma-field properties is crucial for understanding the limitations of G.
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Homework Statement


Let \Omega = [0,1)

Let G be the collection of all subsets of \Omega of the form
[a1,b1),\cup[a2,b2),\cup...\cup[ar,br)
For r any non-negative integer and 0<=a1
and a1 <=b1 <= a2 ...

Show that G is a field

Show that G is not a \sigma-field



Homework Equations



Definition:
Let \Omega be any set. A collection F of subsets of \Omega
is a field if:

1. \emptyset\inF
2. given A in F, then Ac=\Omega\A
3. given A and B in F, A\cupB is in F

In addition, the collection F is a \sigma-field if

given A1, A2, A3... are all in F, so is there union

\bigcupAi


The Attempt at a Solution



To show G is a field:

1.
Assume empty set is not in G, then there must be an element in empty which is not in the G - a contradiction since there are no elements in the empty set.

2.
If we have A in G then A=[ai,bi)
So Ac= \Omega/A

3. This is where I have the problem,
Set A in G as [ai,bi)
And B in G as [aj,bj)
So we want to show A U B in G
But if bi= aj
Then we will have A U B =[ai,bj)
Which is not in G

To show G is not a sigma field:
(0,1)= the union of intervals [1/n,1)
[1/n,1) is in G for all n
(0,1) is not in G
So G is not a sigma field.

So essentially the problem is with part 3 of the definition.
 
Last edited:
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Can you say "If we have A in G then A=[ai,bi)"? Isn't each element of G a union of r intervals? What would A look like when r > 1? Same comment applies to both part 2 and part 3 of your answer.
 
EnumaElish said:
Can you say "If we have A in G then A=[ai,bi)"? Isn't each element of G a union of r intervals? What would A look like when r > 1? Same comment applies to both part 2 and part 3 of your answer.

Thanks yes! OK so we say

A=[ai,bi)U[a(i+1),b(i+1)U...U[ar,br)

Ac = omega/A

Then 3 is now easy

A=[ai,bi)U[a(i+1),b(i+1)U...U[ar,br)
B=[aj,bj)U[a(j+1),b(j+1)U...U[as,bs)

so AUB = [ai,bi)U[a(i+1),b(i+1)U...U[ar,br)U[aj,bj)U[a(j+1),b(j+1)U...U[as,bs)

AUB=[ak,bk)U[a(k+1),b(k+1)U...U[at,bt)
where k=min{i,j} and t=max{r,s}
 
What is the difference between 3 and the sigma-field property?
 
For 3 we need the union of any two intervals to be in G
For the sigma field property we want all possible unions of all possible intervals to be in GPoint 3 basically requires [0,x) be in G for x<=1, which we do have.
In order for that sigma to hold we would need the interval (0,1) to be in the collection G.

Though I'm still not sure i fully understand the difference :S
 
OK well I am working with Probability and random processes By Geoffrey Grimmett, David Stirzaker, the google books preview covers the section i am dealing with (1.1 and 1.2 right at the start). This uses the definitions given in the exercise.

http://books.google.com/books?id=G3...rontcover&dq=probability+and+random+processes

In the link you have given i can't distinguish any definition of a field or sigma field, or any treatment of an infinite omega.
 
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