1
$\begingroup$

I have been getting confused by multiple source's versions of Fubini-Tonelli's theorem and I would like to simply make sure I am getting the definition straight.

It seems to me that Fubini-Tonelli's theorem has 2 versions:

  1. If $\Omega = \Omega_1 \times \Omega_2$, $F = F_1 \times F_2$ and $\mu = \mu_1 \times \mu_2$ where $\mu_1,\mu_2$ are sigma finite measures, $f$ is measurable with respect to $(\Omega, F)$ and $f$ is integrable with respect to $(\Omega, F)$, then:

    $$\int_\Omega f(x,y) d(\mu_1 \times \mu_2) = \int_{\Omega_1} \int_{\Omega_2} f(x,y) d \mu_1 d \mu_2 = \int_{\Omega_1} \int_{\Omega_2} f(x,y) d \mu_2 d \mu_1 $$

  2. If $\Omega = \Omega_1 \times \Omega_2$, $F = F_1 \times F_2$ and $\mu = \mu_1 \times \mu_2$ where $\mu_1,\mu_2$ are sigma finite measures, $f$ is measurable with respect to $(\Omega, F)$ and if either $\int_{\Omega_1} \int_{\Omega_2} |f(x,y)| d \mu_1 d \mu_2 < \infty$ or $ \int_{\Omega_1} \int_{\Omega_2} |f(x,y)| d \mu_2 d \mu_1 < \infty$, then:

    $$\int_\Omega f(x,y) d(\mu_1 \times \mu_2) = \int_{\Omega_1} \int_{\Omega_2} f(x,y) d \mu_1 d \mu_2 = \int_{\Omega_1} \int_{\Omega_2} f(x,y) d \mu_2 d \mu_1 $$

Is this correct? Are there any conditions I have missed?

$\endgroup$
  • $\begingroup$ One assumption implies the others, so you may assume any of the 3 possible integrability assumptions. Also, Tonelli's theorem tells that we have the same conclusion if the integrability condition is replaced by the non-negativity condition $f \geq 0$. $\endgroup$ – Sangchul Lee Mar 12 '18 at 3:45
0
$\begingroup$

The two "versions" of the theorem can be seen as the two sides of an "if and only if". Namely, you can write what you wrote in the following way:

If $\Omega=\Omega_1\times\Omega_2$, $F=F_1\times F_2$ and $μ=μ_1\times μ_2$, where $μ_1,μ_2$ are sigma-finite measures, and $f$ is measurable with respect to $(Ω,F)$, the following statements are equivalent:

  • $\displaystyle \int_\Omega |f(x,y)| d(\mu_1 \times \mu_2)<\infty$

  • $\displaystyle \int_{\Omega_1} \int_{\Omega_2} |f(x,y)| d \mu_1 d \mu_2 < \infty$

  • $\displaystyle \int_{\Omega_1} \int_{\Omega_2} |f(x,y)| d \mu_2 d \mu_1 < \infty$

When the above conditions hold, we have $$ \int_\Omega f(x,y) d(\mu_1 \times \mu_2) = \int_{\Omega_1} \int_{\Omega_2} f(x,y) d \mu_1 d \mu_2 = \int_{\Omega_1} \int_{\Omega_2} f(x,y) d \mu_2 d \mu_1. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.