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## Homework Statement

Let [tex]\Omega = [0,1)[/tex]

Let G be the collection of all subsets of [tex]\Omega[/tex] of the form

[a

_{1},b

_{1}),[tex]\cup[/tex][a

_{2},b

_{2}),[tex]\cup[/tex]...[tex]\cup[/tex][a

_{r},b

_{r})

For r any non-negative integer and 0<=a

_{1}

and a

_{1}<=b

_{1}<= a

_{2}....

Show that G is a field

Show that G is not a [tex]\sigma[/tex]-field

## Homework Equations

Definition:

Let [tex]\Omega[/tex] be any set. A collection F of subsets of [tex]\Omega[/tex]

is a field if:

1. [tex]\emptyset[/tex][tex]\in[/tex]F

2. given A in F, then A

^{c}=[tex]\Omega[/tex]\A

3. given A and B in F, A[tex]\cup[/tex]B is in F

In addition, the collection F is a [tex]\sigma[/tex]-field if

given A

_{1}, A

_{2}, A

_{3}.... are all in F, so is there union

[tex]\bigcup[/tex]A

_{i}

## 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=[a

_{i},b

_{i})

So A

^{c}= [tex]\Omega[/tex]/A

3. This is where I have the problem,

Set A in G as [a

_{i},b

_{i})

And B in G as [a

_{j},b

_{j})

So we want to show A U B in G

But if b

_{i}= a

_{j}

Then we will have A U B =[a

_{i},b

_{j})

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.

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