Product Integral: Integrability

Question

Given measure spaces $X$ and $Y$.

Then it holds: $$\int_Y\int_X|\eta(x,y)|\mathrm{d}\mu(x)\mathrm{d}\nu(y)<\infty\implies\int_X|\eta(x,y)|\mathrm{d}\mu(x)<\infty\quad(y\in Y)$$

Can this actually fail?

Answer 1

I think I got it:

For Lebesgue measures: $$\mu,\nu:\mathcal{B}([0,1])\to\mathbb{R}_+:\quad\mu,\nu:=\lambda$$

Consider the function: $$\eta:(0,1]\times[0,1]\to\mathbb{C}:(x,y)\mapsto\frac{1}{\sqrt{x^2+y}}$$

Then one has: $$\int_{[0,1]}\eta(x,y)\mathrm{d}\lambda(y)=\int_0^1\int_0^1\frac{1}{\sqrt{x^2+y}}\mathrm{d}y\mathrm{d}x\\=\int_0^1\bigg\{2\sqrt{x^2+1}-2\sqrt{x^2}\mathrm{d}\bigg\}\mathrm{d}x\leq{\int_0^1}2\sqrt{2}\mathrm{d}x=2\sqrt{2}<\infty$$

But one obtains: $$\int_{[0,1]}\eta(x,0)\mathrm{d}\lambda(x)=\int_0^1\frac{1}{x}\mathrm{d}x=\infty$$

That shows failure.

Answer 2

No, it doesn't hold. In general this looks like Fubini or Tonelli, but both times we integrate always with respect to each measure, which we don't want to do here in this case.

As already pointed out by Freeze_S, we could apply Fubini/Tonelli in the following example and we therefore would then achieve $0$, but this only holds if we integrate with respect to both measures:

So let's assume we have $$ \eta(x,y)=f(x)g(y) $$ with $f$ integrable with respect to $\mu(x)$ and $g$ integrable with respect to $\nu(x)$ while we assume $g$ is almost $\nu$-everywhere $0$ but on a nullset $\mathcal{N}$ infinity, so it still holds $\int g(y)\mathrm{d}(\nu(y)=0 $

So now take take a $y^\ast$ from $\mathcal{N}$ and see, that $$ \int_Y\int_X|\eta(x,y)|\mathrm{d}\mu(x)\mathrm{d}\nu(y)=0 $$ but $\eta(x,y^\ast)=\infty$, so $$ \int_X|\eta(x,y^\ast)|\mathrm{d}\mu(x)\not<\infty $$ because the integral is not defined.