I have a small computation to do and I am not able to prove it:
Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space. Let $X$ be an integrable random variable and $A\in\mathcal{F}$. Let $\mathcal{A}=\{\emptyset,\Omega,A,A^c\}$.
What is $\mathbb{E}[X \mid \mathcal{A}]$? I do not how to proceed, is there anybody who can give a detailed proof of the expression of $\mathbb{E}[X \mid \mathcal{A}]$ in this particular situation?
Thanks in advance!
By the definition of conditional expectation, $Y:= \mathbb{E}(X \mid \mathcal{A})$ is $\mathcal{A}$-measurable. So, it's a good start to think about the question how $\mathcal{A}$-measurable random variables look like. In this case, you can easily show that they are of the form $$c_1 \cdot 1_A + c_2 \cdot 1_{A^c}$$ where $c_1, c_2$ are constants. Thus we conclude $$Y= c_1 \cdot 1_A + c_2 \cdot 1_{A^c} \tag{1}$$ for some suitable constants $c_1$, $c_2$. Obviously, we still have to determine $c_1, c_2$.
Again by the definition of conditional expectation, we know that $Y$ has to fulfill $$\forall B \in \mathcal{A}: \int_B X \, d\mathbb{P} = \int_B Y \, d\mathbb{P} \stackrel{(1)}{=} \int_B (c_1 \cdot 1_A + c_2 \cdot 1_{A^c}) \, d\mathbb{P}$$
First, we choose $B=A$ and obtain $$\int_A X \, d\mathbb{P} = \int_A (c_1 \cdot 1_A + c_2 \cdot 1_{A^c}) \, d\mathbb{P} = \int (c_1 \cdot 1_A + c_2 \cdot \underbrace{1_A \cdot 1_{A^c}}_{0}) \, d\mathbb{P} = c_1 \cdot \mathbb{P}(A) \\ \Rightarrow c_1 = \begin{cases} \frac{\mathbb{E}(X \cdot 1_A)}{\mathbb{P}(A)} & \mathbb{P}(A) \not= 0 \\ 0 & \mathbb{P}(A) = 0 \end{cases} $$
A similar calculation for $B=A^c$ yields $$c_2 = \begin{cases} \frac{\mathbb{E}(X \cdot 1_{A^c})}{\mathbb{P}(A^c)} & \mathbb{P}(A^c) \not= 0 \\ 0 & \mathbb{P}(A^c) = 0 \end{cases} $$