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Lagrange fanboy

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- TL;DR Summary
- Given a set from a product sigma algebra, how do I prove that it's projection is measurable?

I was reading page 33 of https://staff.fnwi.uva.nl/p.j.c.spreij/onderwijs/TI/mtpTI.pdf when I saw this claim:

Given measurable spaces ##(\Omega_1,\Sigma_1), (\Omega_2,\Sigma_2)## and the product space ##(\Omega_1\times \Omega_2, \Sigma)## where ##\Sigma## is the product sigma algebra, the embedding ##e_b:\Omega_1\rightarrow \Omega_1\times\Omega_2, x\mapsto (x,b)## is measurable.

After some reduction, I notice that ##e^{-1}_b(S)=\emptyset## if ##S\not = \{(x,b)|x\in A\}## for some ##A\subseteq \Omega_1##, otherwise ##e^{-1}_b(S)=\pi_1(S)##, the projection of ##S## onto the first component. Here I'm stuck. I can't show that if ##S\in \Sigma## then ##\pi_1(S)\in \Sigma_1##. I thought about using structural induction, taking advantage of the fact that ##\Sigma## is the smallest sigma algebra generated, but I think for structural induction to work I need to have a well-founded/well-ordered relation, which I cannot find.

Given measurable spaces ##(\Omega_1,\Sigma_1), (\Omega_2,\Sigma_2)## and the product space ##(\Omega_1\times \Omega_2, \Sigma)## where ##\Sigma## is the product sigma algebra, the embedding ##e_b:\Omega_1\rightarrow \Omega_1\times\Omega_2, x\mapsto (x,b)## is measurable.

After some reduction, I notice that ##e^{-1}_b(S)=\emptyset## if ##S\not = \{(x,b)|x\in A\}## for some ##A\subseteq \Omega_1##, otherwise ##e^{-1}_b(S)=\pi_1(S)##, the projection of ##S## onto the first component. Here I'm stuck. I can't show that if ##S\in \Sigma## then ##\pi_1(S)\in \Sigma_1##. I thought about using structural induction, taking advantage of the fact that ##\Sigma## is the smallest sigma algebra generated, but I think for structural induction to work I need to have a well-founded/well-ordered relation, which I cannot find.

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