Can a Counterexample Disprove a Conditional Expected Value Statement?

[Linux]

I know that the following is true and I've already proven it.

Let $Y$ be a random variable and $\varphi$ a measurable function.

Let $A$ be a $\Sigma_Y$ measurable set.

If $ X (\omega) = \varphi(Y (\omega))$ for all $\omega\in A $ , then $\mathbb{E}(X|Y )(\omega) = \varphi(Y (\omega))$ for almost all $\omega\in A $.

I cannot, however, come up with an counterexample which proves that the above statement can be false if $A \not\in \Sigma_Y$.

Could you help me with that?

That is, I'm looking for a random variable $Y$ sych that for $A \not\in \sigma(Y)$ , $\omega \in A$ $$ X (\omega) = \varphi(Y (\omega))$$ but $$\mathbb{E}(X|Y )(\omega) \neq \varphi(Y (\omega)).$$

 

[Valkarie]

A counterexample could be the following. Let $Y$ be a uniform random variable on $[0,1]$ and $\varphi(x) = x^2$. Let $A = \{ \omega \in [0,1] : \omega \not\in \mathbb{Q} \}$ so that $A \not\in \sigma(Y)$. Then for $\omega \in A$, $X(\omega) = \varphi(Y(\omega)) = Y(\omega)^2$. However, $\mathbb{E}(X|Y )(\omega) = \frac{1}{2}Y(\omega)^2 \neq \varphi(Y (\omega))$.