Confusion about conditionals: $X\perp Y$ but $E\{Y|X\}\neq Y$? That's not right... 
Protter Theorem 23.5. Let $Y$ be a positive or integrable r.v. on $(\Omega,\mathcal{F}, P)$. Let $\mathcal{G}$ be a sub $\sigma$-algebra. Then $E\{Y|\mathcal{G}\}=Y$ if and only if $Y$ is $\mathcal{G}$ measurable.
Proof. This is trivial from the definition of conditional expectation. $\square$

It's clear that the proof follows from the definition, but I think I am misunderstanding something: If $X=1 \ \forall \omega \in \Omega$, and $Y$ is some nontrivial RV, then clearly $X\perp Y$, so we would expect $E\{Y|X\}=Y$. 
But by Theorem 23.5, this can't be, since $\sigma(X)=\{\varnothing,\Omega\},$ and $Y$ is not $\sigma(X)$-measurable! 
What do I have wrong?

Conditional expectation $E\{Y|\mathcal{G}\}$, for integrable or positive $Y$ is defined: 

For an integrable, positive RV $Y:(\Omega,\mathcal{F})\rightarrow (\mathbb{R},\mathcal{B})$, and sub $\sigma$-algebra $\mathcal{G}\subseteq \mathcal{F},$ then a conditional expectation is a random variable $E\{Y|\mathcal{G}\}:(\Omega, \mathcal{G})\rightarrow (\mathbb{R},\mathcal{B})$ where for any RV $W$, measurable with respect to $\mathcal{G}$, then: $$E\{YW\}=E\{E\{Y|\mathcal{G}\}W\}.$$

I interpret this as 'If we know that $Y$ is going to be $\mathcal{G}$-measurable, then we construct an RV $E\{Y|\mathcal{G}\}$ that preserves $Y$'s behavior within all $\mathcal{G}$-measurable RVs'

If $X$ and $Y$ are independent, $\Bbb{E}(Y\mid X)=\Bbb{E}(Y)=c$ (not $Y$) which is surely measurable with respect to $\{\emptyset, \Omega\}$.
This should be obvious from elementary probability. If $X$ and $Y$ are independent, the expectation of $Y$ is just the expectation of $Y$, it doesn't care what $X$ does.
To show that this is the right random variable from the measure theoretic definition:
Of course the RV $c$ is measurable wrt $\{\emptyset, \Omega\}$. And for any $W$ that is measurable wrt $\{\emptyset,\Omega\}$, that is $W=d$, another constant, then:
$$\Bbb{E} (Yd)=cd=\Bbb{E}(cd)$$
