Consider $\left(\Omega, \mathcal F, \mathbb P\right)$, where $\Omega = [0,1]$, $\mathcal F$ is the Borel $\sigma$-field on $\Omega$, and $\mathbb P$ is the Lesbesgue measure. Consider the random variables $X$ and $Y$ defined by $X(w)=2\omega ^2$ and
$$Y=\begin{cases}1, &\omega\in[0,\frac 1 3]\\ 2 & \omega \in(\frac 1 3, \frac 2 3) \\ 1 & \omega \in[\frac 2 3, 1]. \end{cases}$$
Compute $\mathbb E[X|Y].$
I tried to do this by saying that,
$$ \mathbb E (X | Y) (x) = \begin{cases} \mathbb E (X | Y=1), & x = 1\\ \mathbb E (X | Y=2), & x = 2.\end{cases} $$
From here I calculated the second case by using
$$\begin{align}\mathbb E(X|Y=2) = \frac{\mathbb E(X\; \cap\; Y=2)}{\mathbb P(Y=2)} &= 3\operatorname{\int}_{\frac 1 3}^{\frac 2 3} X\;d\mathbb P \\ &= 3\operatorname{\int}_{\frac 1 3}^{\frac 2 3} 2\omega^2\;d\omega =\frac{14}{27}.\end{align}$$
Is this correct so far, or is there a better way of doing this?