Three questions about $E[X \mid Y,g(Y)] = E[X|Y]$

Question

QUESTION 1

$$\text{Show:} \quad E[X \mid Y,g(Y)] = E[X|Y]$$

My attempt:

To show this first I will condition on the event $Y=y$ and then at the end replace $y$ with $Y$...

$$E[X \mid Y=y,g(Y=y)] = \sum_x x P(X=x \mid Y=y, g(Y=y))$$

And the trick is to note that the two events $(Y=y) \cap (g(Y=y))$ is redundant and equal to just the event that $Y=y$, thus

$$= \sum_x x P(X=x \mid Y=y)$$

which looks familiar in its way to $E[X|Y]$ which completes the proof. Is this correct?

QUESTION 2

Following the same reasoning, since $E[X|Y]$ and $Var(X|Y)$ are just functions of $Y$, can you confirm the following are true:

$$E[X \mid Y,E[X|Y], Var(X|Y)] = E[X|Y]$$

QUESTION 3

I also believe

$$E[X \mid g(Y)] = E[X \mid Y]$$

is not always true... because given you know $g(Y) = k$ is different from knowing that $Y=y$, is that correct? Thank you.

Answer 1

1) and 2) are correct. For 3) let $X$ take values $\pm 1$ with probability $\frac 1 2 $ each, $Y=X$ and $g(y)=y^{2}$. In this case $E(X|g(Y))=0$ and $E(X|Y)=X$.