Is the Product of Sigma Algebras Associative?

[haljordan45]

Homework Statement

Given 3 measure spaces (X,A,$\mu$), (Y,B,$\zeta$), (Z,C,$\gamma$), show that the product of the three sigma algebras A, B, and C is associative, meaning that:

AxBxC=(AxB)xC=Ax(BxC)

Homework Equations

We can make use of the fact that XxYxZ=(XxY)xZ=Xx(YxZ)

The Attempt at a Solution

I've tried looking at showing the generating set for AxBxC lies in the other two sigma algebras but am having trouble drawing the direct connection. Any help would be greatly appreciated.

 

[mighty2000]

Hello,

Thank you for your question. To show that the product of the three sigma algebras A, B, and C is associative, we can use the fact that XxYxZ=(XxY)xZ=Xx(YxZ). This means that the product of three measure spaces can be expressed as the product of two measure spaces, and then the product of the resulting measure space with the third measure space. We can use this fact to show that AxBxC=(AxB)xC=Ax(BxC).

First, let's consider the product of A and B, denoted by AxB. This product is defined as the sigma algebra generated by the product of the sets in A and B, denoted by AxB. Similarly, the product of B and C, denoted by BxC, is defined as the sigma algebra generated by the product of the sets in B and C, denoted by BxC. Now, the product of AxB and C, denoted by (AxB)xC, is defined as the sigma algebra generated by the product of the sets in AxB and C, denoted by (AxB)xC.

To show that AxBxC=(AxB)xC, we need to show that the generating set for AxBxC lies in (AxB)xC, and vice versa. This means that any set that can be formed by taking the product of sets in A, B, and C, can also be formed by taking the product of sets in AxB and C, and vice versa.

Let's take a set in AxBxC, denoted by D. This set can be expressed as the product of three sets, denoted by D=AxBxC. Using the fact that XxYxZ=(XxY)xZ=Xx(YxZ), we can rewrite this set as D=(AxB)xC. This means that D is also in (AxB)xC, and so the generating set for AxBxC lies in (AxB)xC.

Similarly, let's take a set in (AxB)xC, denoted by E. This set can be expressed as the product of two sets, denoted by E=(AxB)xC. Again, using the fact that XxYxZ=(XxY)xZ=Xx(YxZ), we can rewrite this set as E=AxBxC. This means that E is also in A