Suppose we have a random variable $X$ bounded in expectation, i.e., $E[X] < c$. If $X \geq 0$, then is it correct to infer that the conditional expectation of $X$ given another r.v. $Y=y$ is bounded as well, i.e., $E[X|Y=y] = E[X|y] < c$?
I am trying to prove this using the idea that $p(X) = p(X|y)+ p(X|\bar{y})$, where $\bar{y}$ is the complement of event $Y=y$. So,
$$ E[X] = \int_x xp(x) dx = \int_x x \left( p(x|y) + p(x|\bar{y} \right)dx = \int_x x p(x|y) dx + \int_x xp(x|\bar{y})dx = E[X|y] + E[X|\bar{y}] $$
So if $E[X|y] < c$ and $ X \geq 0$, $E[X|y] + E[X|\bar{y}] < c$ implying $E[X|y] < c$.
Is this reasoning correct? It definitely seems counterintuitive, so I feel I am making some mistake that is leading up to this.
Let $X$ have uniform distribution on $(0,1)$ and $Y=X$. Then $EX=\frac 1 2<\frac 3 4$ but $E(X|Y)=X$ which is not bounded above by $\frac 3 4$.