Conditional expectation should be $\cal F$ a measurable.

Question

I was reading this answer that gives some intuition about what conditional expectation is. I can more or less understand it now but at some point in the response it is written :

A single numerical response is not enough because the particular piece of information that I will give you is itself random. In fact, your response is necessarily a function of this particular piece of information. Mathematically, this is reflected in the requirement that ${\mathbb E}(X\ |\ {\cal F})$ must be $\cal F$ measurable.

How can I see that this mathematically translate to ${\mathbb E}(X\ |\ {\cal F})$ is $\cal F$ measurable?

Answer 1

Let's assume that $\mathcal{F}$ is the sigma algebra generated by event $A$. Then $$\mathbb{E}(X|\mathcal{F})(\omega) = C_1\cdot \mathbb{1}_A(\omega) + C_2\cdot \mathbb{1}_{A^C}(\omega)$$ Hence, if we know whether or not event $A$ happened, we know this value. That is essentially what the measurability means intuitively.

Does that make sense?

Answer 2

In the context of that answer, $\mathbf E(X|\mathcal F)$ is the best guess of a random variable $X$ in view of the information contained in $\mathcal F$.

In some cases (as in the parity example in the answer) the best estimate of $X$ given $\mathcal F$ is not a fixed number but it depends on the specific value in $\mathcal F$. In the die example

$$\mathbf E(Die|Parity) = \begin{cases}4 & \text{if $Parity=Even$},\\ 3& \text{if $Parity = Odd$},\end{cases}$$

This means that $\mathbf{E}(X|\mathcal F)$ is a random variable, however its value can be uniquely determined using information in $\mathcal F$ (that is, wether the die rolled is odd or even); hence, $\mathbf{E}(X|\mathcal F)$ has to be $\mathcal F$-measurable.