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 $\{\omega_k \in \Omega_k: (\omega_i)_{i \in I} \in A \}$.
 A: 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. 
A: 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, https://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 $A_0$.
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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.
A: 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.
A: 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.  
