# Using the product measure

1. Oct 4, 2007

### fantispug

[SOLVED] Using the product measure

Edit: I think I've solved it. I don't know how to delete the thread.

1. The problem statement, all variables and given/known data
Let $$(X,A,\mu)$$ be the Lebesgue measure on [0,1] and $$(Y,B,\nu)=$$([0,1],Power set of [0,1],counting measure). If D=$$\{(x,y)\in X\timesY|x=y\}$$ and f be the characteristic function of D. Show that D is $$\mu\times\nu$$ measurable (i.e. with respect to the complete product measure obtained via the Catheodory extension theorem)

2. Relevant equations
In the Catheodory extension theorem the measurable sets are defined as the sets E that satisfy
$$\mu\times\nu(B)=\mu\times\nu(B\capE)+\mu\times\nu(B\capE^{c})$$
Where
$$\mu\times\nu(B)=\inf\{\sum_{i=1}^{\infty}\mu(A_i)\nu(B_i)|B\subseteq\bigcup_{i=1}^{\infty} A_i \times B_i | A_i \in X | B_i \in Y\}$$

3. The attempt at a solution
If $$B\capD$$ is countable then $$\mu\times\nu(B\capD)=0$$ and it is straight forward to show (via subadditivity and monotonicity of $$\mu\times\nu$$) that
$$\mu\times\nu(B)=\mu\times\nu(B\capE)+\mu\times\nu(B\capE^{c})$$

Similarly if $$\mu\times\nu(B\capD)=\infty$$ then the measurability condition holds. I'm not sure that this is necessarily the case if the intersection is uncountable though. It is certainly the case if the intersection contains a (diagonalised) interval.

So I'm stuck. I need to cover the case where the intersection is uncountable (and not containing a diagonalised interval e.g. the diagonalised cantor set), or approach this from a different road. I have the feeling there may be a better route of proof but I can't think of one.

Last edited: Oct 4, 2007