I have a problem and don't understand a lot of stuff in there.
First, it is given that if there are random variable X and Y where X has a well defined finite expectation, the, $\mathbb{E}(X|Y)$ is defined as the unique random variable that satisfies
a) $F_{\mathbb{E}(X|Y)} \subseteq F_{Y}$ b) for any $A \in F_{Y}, \mathbb{E}(\mathbb{I}_{A}\mathbb{E}(X|Y)) = \mathbb{E}(\mathbb{I}_{A}X).$
Then, the question is :
Let $\Omega =[0,1], F$ be the Borel $\sigma$-field in [0,1] and let $\mathbb{P}$ be the Lebesgue measurement.
Let $X(\omega)=2\omega$ for all $\omega \in[0,1]$ and let $Y(\omega)=2$ for all $\omega \in [0,1]$.i) Compute $\mathbb{E}(X|Y)$
ii) Show that $\mathbb{E}(X|Y)$ satisfies the two conditions above.
First, I don't understand the notation $\mathbb{I}_{A}$.
Also, just in general, how to go about solving conditional expectation.
The trick here is that the $\sigma$-algebra generated by a degenerate random variable is trivial. Because $\mathbb P(Y=2)=1$, we have $Y^{-1}(B)=\varnothing$ if $2\notin B$ and $Y^{-1}(B)=[0,1]$ if $2\in B$, and hence $$\sigma(Y) := \left\{Y^{-1}(B): B\in\mathcal B_{[0,1]} \right\}=\{\varnothing,[0,1]\}. $$ We compute $$\mathbb E[X\mathsf 1_{\varnothing}]=0 $$ and $$\mathbb E[X] = \int_0^1 2x\ \mathsf dx = 1. $$ It follows that $$\mathbb E[X\mid Y]=1.$$