Checking measurability of the diagonal Let me clarify my doubts. 
As I have understood from the following link:

Non-measurable set in product $\sigma$-algebra s.t. every section is measurable.

We can conclude that: For all measurable spaces $(\Omega , \mathcal F)$ it does NOT necessarily hold that: $\{(\omega,\omega) \in \Omega^{2} : \omega \in \Omega \} \in \mathcal F \otimes \mathcal F$ .
Now, along the same line my question is: 
For any probability space $(\Omega, \mathcal F , \mathbb P)$, any measurable space $(S, \mathcal S)$ and any $\mathcal F/ \mathcal S$ - measurable functions $X,Y : \Omega \to S$, does the set $\{X=Y\}:= \{\omega \in \Omega : X(\omega) = Y(\omega)\}$ belong to $\mathcal F$ ??
P.S.:- What I am thinking is: my question is somehow about the measurability of the "domain" whereas, the link provides non-measurability of the co-domain (under the mapping, in particular, for $X=Y=\mbox{Id}$). Isn't it?? And if it is the case, does it help anyway??
Thanks in advance,
 A: Your question: 

For any probability space $(\Omega, \mathcal F , \mathbb P)$, any measurable space $(S, \mathcal S)$ and any $\mathcal F/ \mathcal S$ - measurable functions $X,Y : \Omega \to S$, does the set $\{X=Y\}:= \{\omega \in \Omega : X(\omega) = Y(\omega)\}$ belong to $\mathcal F$ ??

The answer is no. 
Consider any probability space $(\Omega^2, {\mathcal F} \otimes {\mathcal F}, \mathbb P)$, such that $D= \{(\omega,\omega) : \omega \in \Omega \}$ is not ${\mathcal F} \otimes {\mathcal F}$-measurable. 
Consider the functions $X$ defined by $X (\omega_1,\omega_2) = \omega_1$ and $Y$ defined by  $Y (\omega_1,\omega_2) = \omega_2$. It is easy to prove that they are measurable functions and so they are random variables. However, 
 $\{X=Y\}:= \{(\omega_1,\omega_2) \in \Omega^2 : X(\omega_1,\omega_2) = Y(\omega_1,\omega_2)\}= \{(\omega_1,\omega_2) \in \Omega^2 : \omega_1 = \omega_2\}=D$. 
So,  $\{X=Y\}$ is not ${\mathcal F} \otimes {\mathcal F}$-measurable.
Here is a simple detailed example:
Consider $\Omega=[0,1]$ and $\mathcal F =\{E : E\subseteq [0,1] \textrm{ and } E  \textrm{ is countable or co-countable} \}$. ($E$ is co-countable, if $[0,1] \setminus E$ is countable.) 
It is easy to see that  $\mathcal F$ is a $\sigma$-algebra. Define $\mu$ on $\mathcal F$ as $\mu(E) = 0$ if $E$ is countable and  $\mu(E) = 1$ if $E$ is co-countable. It is easy to see that $\mu$ is a measure, in fact a probability. 
Now, consider $(\Omega^2, {\mathcal F} \otimes {\mathcal F}, \mu \otimes \mu)$. It is a probability space.  
Note that $D=\{(\omega,\omega) : \omega \in \Omega \}$ is not ${\mathcal F} \otimes {\mathcal F}$-measurable.  
The functions $X$ defined by $X (\omega_1,\omega_2) = \omega_1$ and $Y$ defined by  $Y (\omega_1,\omega_2) = \omega_2$ are measurable functions and so they are random variables. 
And we have
 $\{X=Y\}:= \{(\omega_1,\omega_2) \in \Omega^2 : X(\omega_1,\omega_2) = Y(\omega_1,\omega_2)\}= \{(\omega_1,\omega_2) \in \Omega^2 : \omega_1 = \omega_2\}=D$. 
So,  $\{X=Y\}$ is not ${\mathcal F} \otimes {\mathcal F}$-measurable.
