# Thread: Fubini-Tonelli with complete measure space

1. ## Fubini-Tonelli with complete measure space

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

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

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

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

I am starting out by assuming that $\displaystyle f=\chi_E$ (characteristic function). Then $\displaystyle f_x=E_x$ 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 $\displaystyle (X,M,\mu), (Y,N,\nu)$ be complete $\displaystyle \sigma$-finite measure spaces. Then consider $\displaystyle (X\times Y,L,\lambda)$, the completion of $\displaystyle (X\times Y,M\times N,mu\times\nu)$.

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

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

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

I am starting out by assuming that $\displaystyle f=\chi_E$ (characteristic function). Then $\displaystyle f_x=E_x$ 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 $\displaystyle \lambda(E)$ is $\displaystyle \lambda(E) = \textstyle\int\nu(E_x)\,d\mu(x) = \int\mu(E^y)\,d\nu(y)$ (of course, he has to show that those two integrals are equal). So the result for $\displaystyle f = \chi_E$ 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.