Let $\Omega=[0,1]$, $X(x)=x^2$ and $Y(x)=1-|2x-1|$ with Lebesgue measure. I want to calculate $E[X\mid Y]$
My work so far
Notice that :
For $x \in [0,\frac12)$ we have $Y(x)=2x$
For $x \in [\frac12,1]$ we have $Y(x)=2-2x$
Let's consider first case
$$E[X\mid Y=2x]=\frac{1}{P(Y=2x)}\int_{\{Y=2x\}}x^2=2\cdot \int_0^\frac{1}{2}x^2=\frac{1}{12}$$
Using analogous way of thinking :
$$E[X\mid Y=2-2x]=\frac{1}{P(Y=2-2x)}\int_{\{Y=2-2x\}} x^2 = 2 \cdot \int_{\frac{1}{2}}^1x^2 = \frac{2}{3}$$
Am I correct with such justification ?