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Suppose two $X,Y$ are random variables defined on $(\Omega,\mathscr{F},\mathbf{P})$ and $f:\mathbb{R}^2\rightarrow\mathbb{R}$ is a measurable function. Does this $[f(X,Y)]^{-1}(\mathscr{B}(\mathbb{R}))\subset \sigma(Y) $ hold true?

Since $f$ is a measurable,$$\forall B\in\mathscr{B}(\mathbb{R}),f^{-1}(B):= \left\{(x,y):f(x,y)\in B \right\}\in \mathscr{B}(\mathbb{R^2}).\Longrightarrow C:=\left\{\omega:(X,Y)\in f^{-1}(B) \right\}\in \mathscr{F}.$$

Let $\pi_{y}:\mathbb{R}^2\mapsto \mathbb{R},(x,y)\mapsto y.$ $$C \subset \left\{\omega:(X,Y)\in \pi_{y}^{-1}(\pi_{y}(f^{-1}(B)))\right\}.$$ But we can't guarantee that $\pi_{y}(f^{-1}(B))\in \mathscr{B(\mathbb{R}}).$

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user250236
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    False. Take $Y=0, f(x,y)=x$ and see what happens. – geetha290krm Oct 06 '23 at 05:35
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    If $Y$ is constant with value $y$ then $\sigma(Y)$ is the trivial $\sigma$-algebra and the condition in question holds if and only if $f(X,y)$ is constant. – Jochen Oct 06 '23 at 05:36

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