# $f: \Omega \to \Omega_1 \times \Omega_2$ is measurable if and only if $f_1,f_2$ are measurable

Following this question, I was suggested to give a try on this theorem. Could you check if my proof is fine?

Let

• $$(\Omega, \mathcal F)$$, $$(\Omega_1, \mathcal F_1)$$, and $$(\Omega_2, \mathcal F_2)$$ be measurable spaces.

• $$\mathcal F_1 \otimes \mathcal F_2$$ the product $$\sigma$$-algebra of $$\mathcal F_1$$ and $$\mathcal F_2$$.

• $$\rho_1$$ and $$\rho_2$$ the projection maps from $$\Omega_1 \times \Omega_2$$ to $$\Omega_1$$ and $$\Omega_2$$ respectively.

• $$f_1 = \rho_1 \circ f,f_2 = \rho_2 \circ f$$ the first and second coordinates of $$f$$.

Then $$f: \Omega \to \Omega_1 \times \Omega_2$$ is measurable if and only if $$f_1,f_2$$ are measurable.

My attempt:

By construction of $$\mathcal F_1 \otimes \mathcal F_2$$, $$\rho_1$$ and $$\rho_2$$ are measurable. If $$f$$ is measurable, then $$f_1$$ is the composition of $$2$$ measurable functions and thus is measurable. With similar argument, $$f_2$$ is measurable. Now we prove the converse direction.

Assume $$f_1$$ and $$f_2$$ are measurable. Because $$\mathcal F_1 \otimes \mathcal F_2 = \sigma (\mathcal F_1 \times \mathcal F_2)$$, we just have to verify that $$f^{-1}(A \times B) \in \mathcal F$$ for all $$A \in \mathcal F_1$$ and $$B \in \mathcal F_2$$. In fact, \begin{aligned}f^{-1}(A \times B) &= \{ \omega \mid f(\omega) \in A \times B\} \\ &= \{ \omega \mid f_1(\omega) \in A \text{ and } f_2 (\omega) \in B\} \\ &= f_1^{-1} (A) \cap f_2^{-1} (B) \in \mathcal F. \end{aligned}

This completes the proof.

Update: I added a lemma to justify that it's enough to prove $$f^{-1}(\mathcal{F}_1 \times \mathcal{F}_2) \subseteq \mathcal{F}$$.

Lemma: Let $$\left(\Omega_{1}, \mathcal{F}\right)$$ and $$\left(\Omega_{2}, \sigma(\mathcal{C})\right)$$ with $$C \subseteq \Omega_2$$ be two measurable spaces. A function $$f: \Omega_{1} \rightarrow \Omega_{2}$$ is measurable if $$A \in \mathcal{C} \quad \text {implies} \quad f^{-1}(A) \in \mathcal{F}.$$

Proof: Consider $$\mathcal X = \{A \subseteq \Omega_2 \mid f^{-1}(A) \in \mathcal{F}\}$$. We have $$f^{-1}(A^c) = (f^{-1}(A))^c$$ and $$f^{-1}(\cup_n A_n)=\cup_n f^{-1}(A_n)$$. This means $$\mathcal X$$ is a $$\sigma$$-algebra over $$\Omega_2$$. Moreover, $$\mathcal C \subseteq \mathcal X$$. By definition of $$\sigma(\mathcal{C})$$, we obtain $$\sigma(\mathcal{C}) \subseteq \mathcal X$$. Hence $$f$$ is measurable.

• Seems okay to me. Is there any particular step you're not sure about? Oct 16, 2021 at 14:30
• It looks correct to me. Oct 16, 2021 at 16:37

In my opinion, the difficult part is showing that, in general, $$f^{-1}(\sigma(\mathcal{G})) = \sigma(f^{-1}(\mathcal{G})).$$
You seem to be assuming this fact. You have proven that $$f^{-1}(\mathcal{F}_1 \times \mathcal{F}_2) \subset \mathcal{F}.$$
But you want to prove that $$f^{-1}(\sigma(\mathcal{F}_1 \times \mathcal{F}_2)) \subset \mathcal{F}.$$
• How coincidental it is! I proved $f^{-1}(\sigma(\mathcal{G})) = \sigma(f^{-1}(\mathcal{G}))$ 3 months ago here. Oct 16, 2021 at 18:50
• After looking closer to the problem, I found that we don't need $f^{-1}(\sigma(\mathcal{G})) = \sigma(f^{-1}(\mathcal{G}))$. Please see my update. Oct 16, 2021 at 20:14
• @Akira: Yes! You just have to show that $f^{-1}(\sigma(\mathcal{G})) \subset \sigma(f^{-1}(\mathcal{G}))$. Your statement that $\mathcal{X} \subset \sigma(\mathcal{C})$ is wrong, though. What you can conclude is only the opposite inclusion. Oct 16, 2021 at 21:13
• Ah $\mathcal{X} \subset \sigma(\mathcal{C})$ is a typo. I've deleted it. Oct 16, 2021 at 21:15
• @Akira: Your $\mathcal{X}$ is just the biggest $\sigma$-algebra for which $f$ is measurable. It is probably called: final $\sigma$-algebra. Oct 16, 2021 at 21:15