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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,

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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.

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  • $\begingroup$ Thanks a ton... :) .. The example was a "bingo".. (Y) $\endgroup$ – user92360 May 17 '17 at 11:53

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