$(\Omega_i, \mathcal{F}_i), i \in I$ 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 $\{\omega_k \in \Omega_k: (\omega_i)_{i \in I} \in A \}$.

up vote 4 down vote accepted

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.

  • Byron: Thanks! What is "the diagonal in $\Omega_1 \times \Omega_2$? – Ethan Nov 3 '11 at 18:10
  • 1
    The diagonal is $\{(0,0),(1,1)\}$. – user940 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
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    Again, no. See Gerald Edgar's answer here: mathoverflow.net/questions/72922/… – user940 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.

  • 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
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    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

The answer to Ethan's Question 1 is no. I n descriptive set theory, a subset $A$ of a Polish space $X$ is an analytic set if it is a continuous image of a Polish space. These sets were first defined by Luzin (1917) and his student Souslin (1917)(see, http://en.wikipedia.org/wiki/Analytic_set ).

We need well known facts from the descriptive set theory.

Fact 1. The following conditions on a subset $A$ of a Polish space $X$ are equivalent:

(a) $A$ is analytic;

(b) There is a Polish space $Y$ and a Borel set $B \subseteq X \times Y$ such that $A$ is projection of $B$., that is $A=\{ x \in X | (\exists y)(x,y) \in B\}$.

Fact 2. There exists an analytic set $A_0$ in a Polish space $X$ which is not Borel.

Let $A_0$ becomes from Fact 2. For $A_0$, the condition (b) of Fact 1 implies that there exists a Polish space $Y_0$ and a Borel set $B_0 \subset X \times Y_0 $ such that $A_0=\{ x \in X | (\exists y)(x,y) \in B\}$. Obviously, $B_0$ stands an example of Borel subset of $X \times Y_0$ whose projection on $X$ is a non-Borel set $A_0$.
\medskip

The answer to Ethan's Question 2 is no. Let $Y$ be a non-Borel subset of $[0.1]$.

Let consider a set

$C:=([0,1]\times [0,1]) \setminus ((\{1/2\}\times Y) \cup (Y \times \{1/2\}))$.

Then its projection onto any component space defined as above is Borel measurable(more precisely, coincides with corresponding component space) but $C$ is not Borel measurable.

Remark on Julián Aguirre example.

Let $\cal{F}_i$ be the $\sigma$-algebra of Lebesgue measurable subsets of $R$ for $i=1,2$.

If $V \subset R$ is non-Lebesgue measurable and $ W \subset R$ is of measure zero, then $A =V \times W \subset R^2$ is not measurable with respect to the product of $\sigma$-algebras $(\cal{F}_i)_{1 \le i \le 2}$. Indeed, assume the contrary. For $x \in W$, we have $B:= R \times \{x\} \in \cal{F}_1 \times \cal{F}_2$ which (under our assumption) implies that $A \cap B \in \cal{F}_1 \times \cal{F}_2$, but $A\cap B = V \times \{x\} $ and we claim that $A \cap B$ is not element of the $\sigma$-algebra $\cal{F}_1 \times \cal{F}_2$. This is a contradiction.

Julián Aguirre's set $V \times W$ belongs to the $\sigma$-algebra $\cal{F}(R^2)$ of Lebesgue measurable subsets of $R^2$ but it does not belong the $\sigma$-algebra $\cal{F}_1 \times \cal{F}_2$. Hence the product of Lebesgue $\sigma$-algebras differs from the $\sigma$-algebra of Lebesgue measurable subsets of the product space.

Thus the answer to the Ethan's Question 3 is no.

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