The question is from a textbook in Chinese that I use for supplementary reading.
Let $f_{i}(\omega_{i})$ be measurable on $\sigma$ -finite measure spaces $(\Omega_{i}, \mathcal{A}_{i}, \mu_{i}), i=1,2 .$ Show
- $g_{i}\left(\omega_{1}, \omega_{2}\right)=f_{i}\left(\omega_{i}\right), i=1,2$ is $\mathcal{A}_{1} \times \mathcal{A}_{2}$ -measurable.
- $h(\omega_{1}, \omega_{2})=f_{1}(\omega_{1}) f_{2}(\omega_{2}).$ is $\mathcal{A}_{1} \times \mathcal{A}_{2}$ -measurable.
For the first part of the question, the definition looks very similar to how random vector is defined, so I think showing
$$\{g_i(\omega_1,\omega_2)<x_i,i=1,2\}\\=\{f_i(\omega_i)<x_i,i=1,2\}\\=\{(\omega_1,\omega_2):f_i(\omega_i)<x_i,i=1,2\}\\=\{\omega_1:f_1(\omega_1)<x_1\}\times\{\omega_2:f_2(\omega_2)<x_2\}\in\mathcal{A_1\times\mathcal{A_2}}$$
would be suffice. It is the second part of the question I am unsure about.
At first I tried to show by cases, and the outcome turns out to be very nusty. I tried to show $\{0<h(\omega_1,\omega_2)<x\}$ when $x>0$ is a union of some $A_1\times A_2$ that is supposed to be in $\mathcal{A_1}\times\mathcal{A_2}$,similar for $\{h(\omega_1,\omega_2)=x\}$, when $x=0$ and $\{h(\omega_1,\omega_2)\leq x\}$ when $x<0$. The written is very massive and I don't know if doing so make sense. (it would be nice if I can find expression for $\{h(\omega_1,\omega_2)<x\}$ but I kept on confusing myself.)
However, I do know in the one-dimensional case, the product of two measurable functions is still measurable. I wonder if I could simply say $f_1(\omega_1)=g_1(\omega_1,\omega_2)$ and $f_2(\omega_2)=g_2(\omega_1,\omega_2)$ is $\mathcal{A_1}\times\mathcal{A_2}$-measurable, so will their products. The problem is when I show the first part of the question, I thought $(g_1,g_2)$ as a whole, each representing a component. They are only $\mathcal{A_1}\times\mathcal{A_2}$-measurable when considering them together. In other words, I am not sure if my reasoning makes sense or not.
If either reasonings are not correct I am not very sure how to show the second part of the question. Could someone please explain to me how should I show the second part of the question?
Also, an additional question: most of the theorems stated in the book work in the other direction: provided a function is $\mathcal{A_1}\times\mathcal{A_2}$-measurable, its section or integration over $\Omega_i$ is also measurable provided certain condition hold. Are there any theorems that state the other direction? i.e. if $f_i$ is $\mathcal{A_i}$-measurable, through operations satisfies certain condition will result in a $A_1\times A_2$-measurable, or it is rather case by case.
Best Answer
Just do the first part separately.
1.1. Let $g_1: \Omega_1 \times \Omega_2 \rightarrow \Bbb R$ be defined by $g_{1}\left(\omega_{1}, \omega_{2}\right)=f_{1}\left(\omega_{1}\right)$. Then, for any $x_1 \in \Bbb R$, we have $$ g_1^{-1}((-\infty, x_1)) =\{(\omega_{1}, \omega_{2}) \in \Omega_1 \times \Omega_2 : f_1(\omega_1)< x_1\} = f_1^{-1}((-\infty, x_1)) \times \Omega_2 \in \mathcal{A}_{1} \times \mathcal{A}_{2}$$ So $g_1$ is $\mathcal{A}_{1} \times \mathcal{A}_{2}$ -measurable.
1.2. Let $g_2: \Omega_1 \times \Omega_2 \rightarrow \Bbb R$ be defined by $g_2\left(\omega_{1}, \omega_{2}\right)=f_2\left(\omega_{2}\right)$. Then, for any $x_2 \in \Bbb R$, we have $$ g_2^{-1}((-\infty, x_2)) =\{(\omega_{1}, \omega_{2}) \in \Omega_1 \times \Omega_2 : f_2(\omega_2)< x_2\} = \Omega_1 \times f_2^{-1}((-\infty, x_2)) \in \mathcal{A}_{1} \times \mathcal{A}_{2}$$ So $g_2$ is $\mathcal{A}_{1} \times \mathcal{A}_{2}$ -measurable.
Now, for the second part, just note that $$h(\omega_{1}, \omega_{2})=f_{1}(\omega_{1}) f_{2}(\omega_{2})=g_{1}\left(\omega_{1}, \omega_{2}\right)g_{2}\left(\omega_{1}, \omega_{2}\right)$$ and then use the fact that the product of two measurable functions (from the same domain) is still measurable.
Remark: Regarding your additional question. Yes, there are theorem that goes from the measurability of the "component" functions to the measurability of the "combined" function, but such theorems depend of how exactly the functions will be combined into the resulting function.