$E[X|X=Y]=E[Y|X=Y]=E[X]=E[Y]$? The last piece to comprehend this answer is to establish that
$$E[X|X=Y]=E[Y|X=Y].$$
I assume
$$E[X|X=Y]=E[X|\{\omega : X(\omega)=Y(\omega)\}]=E[X|A].$$ 
If we consider this the expectation of $X$ in $A$, then clearly the equality above holds because it must be the same as the expectation of $Y$ in $A$, as we've assumed that these two RVs are equal in $A$. 
Finally, isn't it the case that $\{\omega : X(\omega)=Y(\omega)\}=\Omega,$ and therefore that $E[X|X=Y]=E[Y|X=Y]=E[X]=E[Y], $ or more generally $E[X|\Omega]=E[X]?$
So essentially the information $X=Y$ is like no information at all?
 A: 
Finally, isn't it the case that  {ω:X(ω)=Y(ω)}=Ω [...] So essentially the information X=Y is like no information at all?

The random variable $Y$ doesn't have to map every element of the state space into the same real number as the random variable $X$. If it does, you're just giving an additional label, $Y$ to the random variable $X$. That gives you no information about $X$, conditioning on $X=Y$ is just conditioning on the state space as you note. But if $X$ and $Y$ are different random variables (but possibly with the same distribution), then $X=Y$ does provide information about $X$ or $Y$. 
Here's a simple example. The state space $\Omega$ is $\{\omega_0, \omega_1, \omega_2 \}$ all equally likely. Specify the random variable $X$ as  $X(\omega_0) = 0$ and  $X(\omega_1) = 1$, $X(\omega_2) = 2$. Specify $Y$ as $Y(\omega_0) = 0$ and $Y(\omega_1) = 1$, $Y(\omega_2) = 3$.
Then the event $X=Y$ is the set $\{\omega_0, \omega_1\}$. Conditional on that event, the expected value of both variables is of course the same, $0.5$
