How does one check the integral condition of Fubini's?

Question

I.e. that

$$\int_{A \times B} |f(x,y)|d(x,y) < \infty$$

Isn't the use of Fubini's to avoid having to calculate this?

I don't understand how to calculate such integral w/o iterated.

@LeanderTilstedKristensen Okay so, Tonelli requires that $f$ from $X \times Y$ to $[0, \infty]$ is non-negative measurable. And then the integral equals the iterated. So this reason's "backwards" to the assumption of Fubini's? Or you suggest to replace Fubini's with Tonelli's. — mavavilj, Oct 22 '20 at 17:23
$|f|$ is non-negative and measurable (if $f$ is measurable). — Leander Tilsted Kristensen, Oct 22 '20 at 17:24
What i'm saying is. Use Tonellis theorem on $|f|$ to check that the condition $$\int_{A\times B} |f(x,y)| d(x,y) < \infty$$ is satisfied. Afterwards you can use fubinis to compute $$\int_{A\times B} f(x,y) d(x,y)$$ — Leander Tilsted Kristensen, Oct 22 '20 at 17:29
@LeanderTilstedKristensen But I don't understand how to compute that integral w/o being able to split it to iterated ones. — mavavilj, Oct 22 '20 at 18:57
What is your objective here? Determine if $f$ is absolutely integrable (thereby meeting the conditions of Fubini's theorem) so that you can then do ... what? Do you have a specific example? — RRL, Oct 22 '20 at 19:15
@RRL Fubini's says that one must check that the prior integral is finite, before being able to apply Fubini's. The problem may be, how to check it (w/o being able to resort to iterated integrals). — mavavilj, Oct 22 '20 at 19:18
OK so to ascertain that $\int_{A \times B} |f| < \infty$ without actually evaluating. Often that can be done by a comparison test. — RRL, Oct 22 '20 at 19:19
@RRL I saw an example that did that by fixing or "parametrizing along" the other variable so that a $\mathbb{R}^2$ integral becomes $\int_{min_y}^{max_y} |f| dy$. — mavavilj, Oct 22 '20 at 19:22
It really depends on the specific structure. An example: https://math.stackexchange.com/a/2456906/148510 — RRL, Oct 22 '20 at 19:31

score 1 · Answer 1 · answered Oct 22 '20 at 19:14

Suppose we have two $\sigma$-finite measure spaces $(A,\mathcal{M},\mu)$ and $(B,\mathcal{N},\nu)$, and a measurable function $f:A\times B \to \Bbb{C}$. To check the hypothesis of Fubini's theorem, we usually apply Tonelli's theorem to guarantee that $f$ is integrable (with repsect to the product measure).

Tonelli's theorem tells us that if $\phi:A\times B \to [0,\infty]$ is measurable (relative to the appropriate $\sigma$-algebras) then you can always iterate the integrals and the result is always the same: \begin{align} \int_{A\times B}\phi(x,y)\, d(x,y) = \int_A \int_B \phi(x,y)\, dy \, dx = \int_B \int_A \phi(x,y)\, dx \, dy \end{align} Just to be more explicit, this says all three terms are finite and equal, or they are all equal to $\infty$.

So, with this in mind, to check the hypothesis $\int_{A\times B}|f(x,y)| \,d(x,y)< \infty$ which is required for Fubini's theorem, we first apply Tonelli's theorem: \begin{align} \int_{A\times B}|f(x,y)|\, d(x,y) = \int_A \int_B |f(x,y)|\, dy \, dx = \int_B \int_A |f(x,y)|\, dx \, dy \tag{$*$} \end{align} Out of these three, calculate whichever integral is the easiest to do (after all, they are all equal). If it is finite, great! Fubini's theorem's hypotheses are satisfied so you can remove absolute values and say that \begin{align} \int_{A\times B}f(x,y)\, d(x,y) = \int_A \int_B f(x,y)\, dy \, dx = \int_B \int_A f(x,y)\, dx \, dy \end{align} If however $(*)$ results in $\infty$, then that is very unfortunate, and you have to be very careful in your subsequent analysis.

How does one check the integral condition of Fubini's?

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