Is the Diagonal Set D Measurable in the Product Space [0,1] x [0,1]?

[fantispug]

[SOLVED] Using the product measure

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

Homework Statement

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)

Homework 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
$$\mu\times\nu(B)=\inf\{\sum_{i=1}^{\infty}\mu(A_i)\nu(B_i)|B\subseteq\bigcup_{i=1}^{\infty} A_i \times B_i | A_i \in X | B_i \in Y\}$$

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.

 

[mighty2000]

Thank you for sharing your solution to this problem. It seems like you have made good progress in solving it. However, I would like to offer some additional thoughts and suggestions.

Firstly, I would like to clarify that the Catheodory extension theorem is used to extend a measure on a smaller sigma-algebra to a larger one. In this case, we are dealing with a product measure, which is already defined on the complete sigma-algebra. So, we do not need to use the Catheodory extension theorem in this case.

Secondly, I believe your approach is correct. If the intersection B∩D is countable, then it is easy to show that it is measurable with respect to the product measure. If it is uncountable, then we need to consider the cases where it is contained in a diagonalized interval and where it is not. However, I think you may have overlooked the fact that the product measure is complete, which means that every subset of a null set is measurable. So, even if the intersection is not contained in a diagonalized interval, it will still be measurable.

Finally, I would like to suggest another approach to this problem. Instead of directly showing that D is measurable, we can show that the complement of D is measurable. Since D is the diagonal of the square [0,1]×[0,1], its complement is the union of the two triangles above and below the diagonal. These triangles are measurable because they can be written as the product of a measurable set in X and a null set in Y. Therefore, the complement of D is measurable, and hence, D itself is also measurable.

I hope this helps and good luck with your studies.