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What is a measurable set defined in terms of random variables? I was studying Stochastic Approximation and trying to understand the proof of Dvoretzky Stochastic Approximation Theorem when this expression appeared.

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The paper can be found here: link

The sentence I am confused about is

$A$ any set in $\mathcal{F}$ the definition of which can be made in terms of $X_1, ..., X_m$.

$A$ is a measurable set and $X_1, ..., X_m$ are all random variables. Is this maybe an outdated expression? Because I have never read or heard it before.

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  • $\begingroup$ may be it's just $A\in\sigma(X_1,X_2,\ldots, X_m)$? $\endgroup$
    – PNDas
    Mar 19 at 16:54
  • $\begingroup$ @PNDas It makes sense to me, I was thinking it could be something like that, but I was not at all sure. Just to clarify myself, $\sigma(X_1,X_2, ..., X_m)$ is the smallest $\sigma$-algebra that makes $X_1,...,X_m$ measurable right? $\endgroup$
    – Kareit
    Mar 19 at 17:14
  • $\begingroup$ Yes @Kareit. More explicitly, if $X_i:(\Omega,\mathcal F)\to (\mathbb R,\mathcal B(\mathbb R))$ is measurable, then $$\sigma(X_1) = \{ \{X_1 \in B\}: B\in \mathcal B(\mathbb R))\}$$ and $$\sigma(X_1,X_2) = \sigma\left( \sigma(X_1)\cup \sigma(X_2)\right)$$ $\endgroup$
    – Andrew Zhang
    Mar 19 at 18:50

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