Precise Defintion of Fubini-Tonelli's Theorm

137 Views Asked by At

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?

1

There are 1 best solutions below

0
On BEST ANSWER

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