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Let $X_1, \dots, X_n$ be a set of random variables defined on a probability space $(\Omega, \mathcal{F}, P)$ and denote by $\mathcal{S} = \sigma(X_1, \dots, X_n)$ the $\sigma$-algebra generated by the random variables. Take an arbitrary function $f: \mathbf{R}^n \to \mathbf{R}$. What conditions should be imposed on $f$ such that one can claim that $f(X_1, \dots, X_n)$ is $\mathcal{S}$-measurable? I have Durrett 2010 at hand, and there is the following theorem there:

Theorem 1.3.3. If $X_1, \dots, X_n$ are random variables and $f: (\mathbf{R}^n, \mathcal{R}^n) \to (\mathbf{R}, \mathcal{R})$ is measurable, then $f(X_1, \dots, X_n)$ is a random variable.

So, $f(X_1, \dots, X_n)$ is $\mathcal{F}$-measurable. Can we also conclude that it is $\mathcal{S}$-measurable?

Thank you.

Regards, Ivan

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2 Answers 2

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Let $g:=f(X_1,\dots,X_n)$. If $B\subset\mathbb R$ is a Borel set, then $g^{-1}(B)=\{\omega,(X_1(\omega),\dots,X_n(\omega))\in f^{-1}(B)\}$. Since $f^{-1}(B)$ is a Borel subset of $\mathbb R^n$, $g^{-1}(B)\in\mathcal S$.

Indeed, take $\mathcal B$ the class of Borel subsets $S$ of $\mathbb R^n$ such that $\{\omega,(X_1(\omega),\dots,X_n(\omega))\in S\}\in \mathcal S$. The class $\mathcal B$ certainly contains the sets of the form $S_1\times\dots\times S_n$, with $S_i\subset\mathbb R$ a Borel subsets, and is a $\sigma$-algebra.

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Yes, we can. Note that the $X_i$ are random variable also on the measure space $(\Omega, \mathcal S, P)$ (as they are trivially $\mathcal S$-measurable). Now apply 1.3.3.

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