My statement for Fubini theorem is:
{Let $(X, \mathcal{S}, \mu)$ and $(Y, \mathcal{T}, \lambda)$ be $\sigma$-finite measure spaces. and let $f$ be a $\mathcal{S} \times \mathcal{T}-$measurable function on $X \times Y.$ if $f$ is a real-valued and if $\psi^*(x) = \int_{Y}|f_{x}|d\lambda$ and if $\int_{X} \phi^* d\mu < \infty$ then $f\in L^1(\mu \times \lambda).$
My question is:
In the answer for this question Why are we allowed to replace the integral with respect to the product measure $\mu$ with iterated integrals? I do not understand why $f$ is $\mathcal{S} \times \mathcal{T}-$measurable function on $X \times Y.$ so, could someone please show me how to fulfill the assumptions of Fubini as I am confused a little bit ?
My $\phi(x) = \lambda (\phi_{x})= \int_{Y} \chi_{\phi_{x}}(yd\lambda(y)),$ therefore, $\int_{X} \phi^* d\mu = \int_{X}d\mu(x)\int_{Y}|f|d\lambda (y).$ I am not sure if my $\int_{X} \phi^* d\mu$ is correct or if it should be the $x-$ section of $f$? Could anyone check this for me please?
Best Answer
I think you are overthinking this. The integrand is measurable because it is continuous (redefining $\sin x/x$ at $0$ in the usual way.)
We have
$\psi^*(x) = \int_{E_Y}|f_{x}|dy=\int_0^{\sqrt x}\left|\frac{y}{x} e^{-x}\sin x\right|dy\le 2\int_0^{\sqrt x}y\ dy$
and
$\phi^*(y) = \int_{E_X}|f_{y}|dx=\int_{y^2}^{\infty}\left|\frac{y}{x} e^{-x}\sin x\right|dx\le 2y\int_{y^2}^{\infty}e^{-x}\ dx$
and each of these is finite.
On the other hand, your statement of Fubini is not clear as it stands. If you use the standard statement instead, then the answer to your question is a bit easier.