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I'm trying to prove Proposition 1.17 from this note, i.e.,

Let $(\Omega, \mathcal F)$ be a measurable space and $X_1, X_2$ real-valued random variables. Let $g: \mathbb R^2 \to \mathbb R$ be Borel measurable and $Y = g(X_1, X_2)$. Then $Y$ is $\sigma(X_1, X_2)$-measurable, i.e., $\sigma(Y) \subset \sigma(X_1, X_2)$.

Could you have a check on my below attempt?

Let $B \in \mathcal B (\mathbb R)$. Then $g^{-1} (B) \in \mathcal B (\mathbb R^2)$ and $$ \{Y \in B\} = \{(X_1, X_2) \in g^{-1} (B) \}. $$

It suffices to prove that $$ \mathcal A :=\{A \in \mathcal B (\mathbb R^2) \mid \{(X_1, X_2) \in A \} \in \sigma(X_1, X_2) \}. $$ coincides with $\mathcal B (\mathbb R^2)$. However, this is true because $\mathcal A$ is a $\sigma$-algebra on $\mathbb R^2$ and $\mathcal A$ contains the subbase $\{A \times B \mid A, B \text{ open in } \mathbb R\}$ of the product topology of $\mathbb R^2$.

Akira
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  • What do you mean by $\sigma(X_1,X_2)$? – copper.hat Sep 22 '22 at 02:42
  • @copper.hat I mean by $\sigma(X_1,X_2)$ the $\sigma$-algebra generated by $X_1, X_2$, i.e., the smallest $\sigma$-algebra $\mathcal G$ on $\Omega$ such that $X_1, X_2$ are $\mathcal G$-measurable. – Akira Sep 22 '22 at 06:43

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