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I'd like some help in an exercise that asks to show by an example that given two $\sigma$-algebras, the projection of a set in the product $\sigma$-algebra need not to be in the corresponding $\sigma$-algebra.

Is there any simple example of that? I only managed to realize that if the set is countable, then the projection has to be in the corresponding $\sigma$-algebra.

VSSF
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  • The third answer on http://math.stackexchange.com/questions/78628/is-projection-of-a-measurable-subset-in-product-sigma-algebra-onto-a-componen?answertab=active#tab-top might be useful to you. – Jeffrey Dawson Nov 04 '16 at 20:41
  • Thanks for the reply! I wonder if there is a more elementary example. – VSSF Nov 05 '16 at 00:34

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consider the lebesgue $\sigma$-Algebra on $\mathbb{R},\mathcal{L}(\mathbb{R})$ and the product $\sigma$-Algebra $\mathcal{L}(\mathbb{R}) \otimes \mathcal{L}(\mathbb{R})$.

Now take a set $A \not\in \mathcal{L}(\mathbb{R})$ and consider $A \times \{0\}$.

Then $A \times \{0\} \in \mathcal{L}(\mathbb{R}) \otimes \mathcal{L}(\mathbb{R})$ but the projection is A

Gono
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  • Are you sure $A \times {0} \in \mathcal{L}(\mathbb{R}) \otimes \mathcal{L}(\mathbb{R})$? Note that $\mathcal{L}(\mathbb{R}) \otimes \mathcal{L}(\mathbb{R}) \neq \mathcal{L}(\mathbb{R}^2)$; the second is the completion of the first. – Jeffrey Dawson Nov 04 '16 at 20:33