Let, $f,g : \Bbb R^2 \to \Bbb R$ be two functions defined by, $f(x,y) := \max\{ x,y \}$ and $g(x,y) := \min\{ x,y \}$ . Then what is the sigma field generated by $f$ and $g$ i.e. What is $\sigma(f,g)$ ?
My attempt :
The pre-image of an open interval $(a,b)$ under $f$ is sets of the form :
And, The pre-image of an open interval $(a,b)$ under $g$ is sets of the form :
Thus, $\sigma(f,g)$ contains rectangles of the form : $$\{(x,y)\in \Bbb R^2 : a<x<b,a<y<b , a,b \in \Bbb R\}$$ (as intersection of the above two figures) and then combining the facts that :
(a) Bounded open rectangles generate $\beta(\Bbb R^2)$.
(b) If $C$ is a generating class of $\mathscr{S}$, then $\sigma(f^{-1}(C)) = f^{-1}(\sigma(C))= f^{-1}(\mathscr{S})$,
I deduce that $\sigma(f,g) = \beta(\Bbb R^2)$.
Please point out mistakes if any. Thanks in advance for help!
Your answer is correct but the proof needs some improvements. You started by proving that $(a,b)\times (a,b)$ is in the sigma algebra whenever $a<b$. These don't exhaust all open rectangles. But it is true that any open set in $\mathbb R^{2}$ is a countable union of these special rectangles. Hence the sigma algebra contains all open sets which implies it contains all Borel sets. Also, $f$ and $g$ are continuous functions, so the sigma algebra generated by them is contained in the Borel sigma algebra. This completes the proof.