Let $(A,\mathcal A), (B, \mathcal B), (C, \mathcal C)$ be measure spaces. Let $f:A\rightarrow C$ and $g :B\rightarrow C$ be mensurable functions. Let's also suppose that $D=\{(c,c): c \in C\} \in \mathcal C \otimes\mathcal C.$ Show that $$E=\{(a,b) \in A \times B : f(a)=g(b)\} \in \mathcal A \otimes \mathcal B.$$
$\{(a,b) \in A \times B : f(a)=g(b)\} \in \mathcal A \otimes \mathcal B$
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measure-theory
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0Okay so just to be clear, $\cal A\times B$ is literally just the cartesian product of the two $\sigma$-algebas, not the $\sigma$-algebra generated by that set? – 2017-01-04
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1@AlexMathers: Presumably not literally the Cartesian product (which is a set of ordered pairs of sets, not a set of sets of ordered pairs), but perhaps we mean $\mathcal{A} \times \mathcal{B} = \{ U \times V : U \in \mathcal{A}, V \in \mathcal{B}\}$? But if so the assumption $D \in \mathcal{C} \times \mathcal{C}$ can never hold, unless $C$ is empty or a singleton. – 2017-01-04
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0@NateEldredge:Yes, by $\mathcal A \times \mathcal B$ I mean $\{U\times V: U \in \mathcal A, V \in \mathcal B\}.$ But I see the problem. – 2017-01-04
1 Answers
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Since $D \in \mathcal C \otimes \mathcal C,$ it's easy to see that if we define $h:A\times B\rightarrow C \times C $ by $h(a,b)=(f(a),g(b))$, then $h$ is mensurable and since $h^{-1}(D)=E$ and $D$ is mensurable we get that $E$ is mensurable, that is, $E \in \mathcal A \otimes \mathcal B$.