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(Probability path by Rensick, P.85, Q.3) Suppose $f : \mathbb{R}^k \to \mathbb{R}$ and $f \in \mathcal{B}(\mathbb{R}^k)/\mathcal{B}(\mathbb{R})$. Let $X_1, ... , X_k$ be random variables on $(\Omega, \mathcal{B})$. Then $f(X_1, ... , X_k) \in \sigma(X_1, ... , X_k)$.

I know that $f \circ X : \Omega \to \mathbb{R}$ is a random variable where $ X(\omega): = (X_1(\omega), ... , X_k(\omega))$ since it is the composition of measurable functions. Then, by definition, $(f \circ X)^{-1} (\mathcal{B}(\mathbb{R})) \subset \mathcal{B}$, but I don't understand how to get $f(X_1, ... , X_k) \in \sigma(X_1, ... , X_k)$. Does this mean a measruable function is an element of sigma algebra generated by all random variables? If yes, this interpretation does not make sense to me. I would appreciate if you give some help.

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  • $\begingroup$ I think it probably means that $f(X_1,\ldots,X_k)$ is measureable w.r.t. the sigma-algebra generated by the $X_i$. $\endgroup$ Sep 19, 2020 at 16:12

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The set $\sigma(X_1,\ldots,X_k)$ is a set of subsets of $\Omega.$

To say that a certain random variable whose domain is $\Omega$ is a member of that set is false if taken literally.

I am tentatively presuming that what is meant is that for every Borel subset $B$ of $\mathbb R,$ the set $\{ \omega\in\Omega : f(X_1(\omega), \ldots,X_k(\omega) \in B \}$ is a member of $\sigma(X_1,\ldots,X_k),$ i.e. the function $f(X_1,\ldots,X_k):\Omega\to \mathbb R$ is a measurable function with respect to that sigma-algebra.

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