Thread: Fubini-Tonelli with complete measure space

1. Fubini-Tonelli with complete measure space

So let $(X,M,\mu), (Y,N,\nu)$ be complete $\sigma$-finite measure spaces. Then consider $(X\times Y,L,\lambda)$, the completion of $(X\times Y,M\times N,mu\times\nu)$.

This is the basic set up. I have to show that if $f$ is $L$-measurable and $f=0$ $\lambda$ almost everywhere, then $f_x,f^y$ are integrable and $\int f_xd\nu=\int f^yd\mu=0$ almost everywhere.

Note: $f_x(x,y)=f^y(x,y)=f(x,y)$ for fixed x,y.

I was told that for this part that I need to use the fact that $\mu,\nu$ are both complete but I don't see how.

I am starting out by assuming that $f=\chi_E$ (characteristic function). Then $f_x=E_x$ right? I assume that it is at this point I need to make use of completeness some how.

2. Originally Posted by putnam120
So let $(X,M,\mu), (Y,N,\nu)$ be complete $\sigma$-finite measure spaces. Then consider $(X\times Y,L,\lambda)$, the completion of $(X\times Y,M\times N,mu\times\nu)$.

This is the basic set up. I have to show that if $f$ is $L$-measurable and $f=0$ $\lambda$ almost everywhere, then $f_x,f^y$ are integrable and $\int f_xd\nu=\int f^yd\mu=0$ almost everywhere.

Note: $f_x(x,y)=f^y(x,y)=f(x,y)$ for fixed x,y.

I was told that for this part that I need to use the fact that $\mu,\nu$ are both complete but I don't see how.

I am starting out by assuming that $f=\chi_E$ (characteristic function). Then $f_x=E_x$ right? I assume that it is at this point I need to make use of completeness some how.
I don't see that completeness should be needed for this. According to Halmos, the definition of $\lambda(E)$ is $\lambda(E) = \textstyle\int\nu(E_x)\,d\mu(x) = \int\mu(E^y)\,d\nu(y)$ (of course, he has to show that those two integrals are equal). So the result for $f = \chi_E$ follows straight from that definition. See §36 of Halmos's book, pp.145–148. His results for product measures assume throughout that the measures on the component spaces are σ-finite, but not that they are complete.