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


  • $(\Omega, \mathcal F)$, $(\Omega_1, \mathcal F_1)$, 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.

  • 2
    $\begingroup$ Seems okay to me. Is there any particular step you're not sure about? $\endgroup$
    – Alex Ortiz
    Oct 16, 2021 at 14:30
  • 1
    $\begingroup$ It looks correct to me. $\endgroup$
    – Mason
    Oct 16, 2021 at 16:37

1 Answer 1


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 $$f^{-1}(\sigma(\mathcal{F}_1 \times \mathcal{F}_2)) \subset \mathcal{F}.$$

  • $\begingroup$ How coincidental it is! I proved $f^{-1}(\sigma(\mathcal{G})) = \sigma(f^{-1}(\mathcal{G}))$ 3 months ago here. $\endgroup$
    – Akira
    Oct 16, 2021 at 18:50
  • $\begingroup$ 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. $\endgroup$
    – Akira
    Oct 16, 2021 at 20:14
  • $\begingroup$ @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. $\endgroup$ Oct 16, 2021 at 21:13
  • $\begingroup$ Ah $\mathcal{X} \subset \sigma(\mathcal{C})$ is a typo. I've deleted it. $\endgroup$
    – Akira
    Oct 16, 2021 at 21:15
  • $\begingroup$ @Akira: Your $\mathcal{X}$ is just the biggest $\sigma$-algebra for which $f$ is measurable. It is probably called: final $\sigma$-algebra. $\endgroup$ Oct 16, 2021 at 21:15

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