I am working on a problem in which we have a random variable $Y$ on $([0,1], \mathcal{B}([0,1]), \Lambda)$ defined by $Y(\omega) = \omega^2$.
We also have a given $\sigma$-algebra $\mathcal{F} = \sigma\left([0,\frac{1}{4}], (\frac{1}{4}, \frac{2}{3}], (\frac{2}{3}, 1]\right)$
I have written out the explicit elements of $\mathcal{F}$, however I am unsure how to proceed in determining $\mathbb{E}(Y|\mathcal{F})$.
I was inclined to integrate $\omega^2$ and divide this by Lebesgue measure, but I am not convinced this is valid as this conditioning is for a $\sigma$-algebra rather than individual events...
Any hints would be greatly appreciated!
We can use the definition of conditional expectation $$Y(\omega) = \mathbb{E}(X|\mathcal{F})(\omega)=\frac{\mathbb{E}(X\mathbb{1}_{B_i}(\omega))}{\Lambda(B_i)}$$ where $B_1 = [0, 1/4], B_2 = (1/4, 2/3], B_3 = (2/3, 1]$ are pairwise disjoint and partition our probability space $\Omega$.
Thus we calculate for each $B_i$:
$$Y(\omega) = \begin{cases}\frac{1}{\Lambda([0,1/4])}\int_{[0,1/4]}\omega^2d\omega\quad :\omega \in [0,1/4] \\ \frac{1}{\Lambda((1/4,2/3])}\int_{[1/4,2/3]}\omega^2d\omega \quad : \omega \in (1/4, 2/3] \\ \frac{1}{\Lambda((2/3, 1])}\int_{(2/3,1]}\omega^2d\omega \quad :\omega \in (2/3,1] \end{cases}$$
Explicitly integrating gives the exact values the random variable takes.