# What does "$f(X_1, ... , X_k) \in \sigma(X_1, ... , X_k)$" mean?

(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.

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

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.