# Prove projection of a measurable set from product space is measurable

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• Lagrange fanboy
Lagrange fanboy
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.

Last edited:
I think you need a lemma from p.289
S Lang

Lagrange fanboy
Just an ignorant suggestion without doing the work: a measurable product set is presumably a union of rectangles with both sides measurable, and projection of a rectangle is a side. so projection of a measurable set is apparently a union of measurable sets, hence measurable. does something like this work?

You need the following simple fact/lemma:

Let a function ##f:X\to Y## be fixed, and let ##\Sigma## be a sigma-algebra on ##X##, and ##\mathcal C## be some collection of subsets in ##Y##. Assume that for any ##A\in\mathcal C## the inverse image ##f^{-1}(A)## belongs to ##\Sigma##. Then ##f^{-1}(B)\in\Sigma## for all ##B## in ##\Sigma(A)##, where ##\Sigma(A)## is the sigma-algebra generated by ##\mathcal C##.

One of the corollaries of the above lemma is the fact that if for a real-valued function ##f## the sets ##f^{-1}([a, \infty))## are measurable for all ##a\in\mathbb R##, then ##f^{-1}(B)## is measurable for any Borel set ##B\subset\mathbb R##. (This is also true for rays of form ##(a, \infty)##, ##(-\infty, a]##, ##(-\infty, a)##).

The proof of the lemma is quite elementary: consider the collection ##\mathcal B## of all ##B\subset Y## such that ##f^{-1}(B)\in\Sigma##. It is easy to show, using the fact that ##\Sigma## is a sigma-algebra, that ##\mathcal B## is also a sigma-algebra. Therefore ##\Sigma(\mathcal C) \subset\mathcal B##, which proves the statement.

Returning to the original question, it is easy to see that if ##A\in\Sigma_1##, ##B\in\Sigma_2##, then ##\pi_b^{-1}(A\times B)## is measurable (it is ##A## if ##b\in B##, and ##\varnothing## othervise). But such sets ##A\times B## generate ##\Sigma##, so the embedding ##\pi_b## is indeed measurable.

Finally, projections of measurable sets are not always measurable. For example, a coordinate projection of a Borel measurable set in ##\mathbb R^2## is not necessarily Borel. It is a so-called Souslin (a.k.a. analytic) set, but generally it is not Borel.

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