# Is projection of a measurable subset in product $\sigma$-algebra onto a component space measurable?

$(\Omega_i, \mathcal{F}_i), i \in I$ are measurable spaces. $\prod_{i \in I} \mathcal{F}_i$ is the product $\sigma$-algebra of $\mathcal{F}_i, i \in I$.

For any $A \in \prod_{i \in I} \mathcal{F}_i$ and $k \in I$, is $\{\omega_k \in \Omega_k: \exists \omega_i \in I/\{k\}, (\omega_i)_{i \in I} \in A\}$ measurable relative to $\mathcal{F}_k$? If not, how about when $I$ is countable or finite?

For any $A \in \prod_{i \in I} \Omega_i$, if its projection onto any component space defined as above is measurable, will $A \in \prod_{i\in I} \mathcal{F}_i$?

Thanks!

Added: For any $A \in \prod_{i \in I} \Omega_i$, if all of its sections onto the component spaces are measurable, will $A \in \prod_{i\in I} \mathcal{F}_i$? A section of $A$ determined by $(\omega_i)_{i \in I/\{k\}}$ is defined as $\{\omega_k \in \Omega_k: (\omega_i)_{i \in I} \in A \}$.

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For the first question, see the discussion here: mathoverflow.net/questions/34142/… – Byron Schmuland Nov 3 '11 at 18:05

The answer to the second question is "no". Take $\Omega_1=\Omega_2=\{0,1\}$ and let ${\cal F}_1=\{\emptyset,\{0\},\{1\},\{0,1\}\}$ and ${\cal F}_2=\{\emptyset,\{0,1\}\}$. The diagonal in $\Omega_1\times\Omega_2$ is not measurable with respect to the product $\sigma$-algebra ${\cal F}_1\times {\cal F}_2$, but its projection either way is the whole space.

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Byron: Thanks! What is "the diagonal in $\Omega_1 \times \Omega_2$? – Ethan Nov 3 '11 at 18:10
The diagonal is $\{(0,0),(1,1)\}$. – Byron Schmuland Nov 3 '11 at 18:11
Thanks! What about "for any $A \in \prod_{i \in I} \Omega_i$, if all of its sections onto the component spaces are measurable, will $A \in \prod_{i\in I} \mathcal{F}_i$?" A section of $A$ determined by $(\omega_i)_{i \in I/\{k\}}$ is defined as $\{\omega_k \in \Omega_k: (\omega_i)_{i \in I} \in A \}$. – Ethan Nov 3 '11 at 18:17
Again, no. See Gerald Edgar's answer here: mathoverflow.net/questions/72922/… – Byron Schmuland Nov 3 '11 at 18:26

The answer to the first question is no in general. Let $V\subset\mathbb{R}$ be non-Lebesgue measurable and $W\subset\mathbb{R}$ of measure zero. Then $A=V\times W\subset\mathbb{R^2}$ is Lebesgue measurable in $\mathbb{R^2}$ (since has outer measure $0$ and Lebesgue measure is complete), and the projection of $A$ on the first component of $\mathbb{R^2}$ is $V$.

## Edit

As noted in the comments, Lebesgue measure on $\mathbb{R^2}$ is not the product of Lebesgue measure on $\mathbb{R}$, but its completion. So the answer above is not correct.

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Borel $\sigma$-algebra on $\mathbb{R}^2$ is the product $\sigma$-algebra of Borel $\sigma$-algebras on $\mathbb{R}$. Is Lebesgue $\sigma$-algebra on $\mathbb{R}^2$ the product $\sigma$-algebra of Lebesgue $\sigma$-algebras on $\mathbb{R}$? – Ethan Nov 3 '11 at 18:03
Does that example really work? I mean: in the question no measures are involved and the product $\sigma$-algebra of Lebesgue measure is not the Lebesgue $\sigma$-algebra on the product and I don't see how you can conclude that your set is measurable with respect to the product $\sigma$-algebra. I think one has to work a bit harder to come up with a counterexample. See analytic sets – t.b. Nov 3 '11 at 18:05
@Ethan I have edited my answer. – Julián Aguirre Nov 3 '11 at 20:13