So I was asked to determine $E[X|Y]$ on the probability space $([-1,1], B[1,-1],\mathbb{P})$ where $\mathbb{P}$ is the uniform probability.
Where $ X(\omega) = \omega^2 $ for $ \omega \in [-1,1] $ and,
$ Y(\omega) =\left\{ \begin{array}{lr} 0, & \text{for } \omega < 0\\ x(n-1), & \text{for } \omega \geq 0 \end{array} \right\}. $
So by using the conditional expectation given a partition in $Y$ I obtained for ${Y = 0}$:
$E[X|{Y=0}] = \frac{E[X{1_{[-1,0)}]}}{\mathbb{P}([-1,0])} = 2\int_{[-1,0)}{} X d\mathbb{P}$.
The solutions say that this can be transformed into the following integral:
$\int_{[-1,0)} X dm$,
by using a radon nikodym density of $f = \frac{1}{2}$ and then this integral can obviously be easily computed. What I don't understand is how to obtain the Radon density to make this transformation?