I know the following characterization of conditional expectation:
Let $(\Omega, \mathbb F, P)$ be a probability room, $\mathbb G \subseteq \mathbb F$ be a $\sigma$-algebra and $X$ a random variable in terms of $\mathbb F $ such that $E(X) < \infty$. Then there exists almost surely a random variable $X_0$ on $(\Omega, \mathbb F, P)$ with
(i) $X_0$ is $\mathbb G$-measurable.
(ii) $\int_A X_0 dP = \int_A X dP \ $ for every $A \in \mathbb G$.
If $X_0$ fullfills both of these properties, then we call it the conditional expectation of $X$ given $\mathbb G$.
What confuses me is: Aren't we always assuming that some random variable $X$ is measurable? So if I wanted to find the conditional expectation of $X$ given $\mathbb G$, then it should be given by $X$ itself if $X$ is $G$-measurable, right? The equality of the integrals would be obvious in this case. So
$$E(X|\mathbb G) = X$$
when $X$ is $\mathbb G$-measurable. But that seems kinda arbitrary to me, since I always assumed that the measurability of a random variable would be given by premise.
Best Answer
Your conclusion is right: if $X$ is $\mathbb G$ measurable then $E(X|\mathbb G)=X$. The whole point here is $\mathbb G$ could be much smaller than the original sigma algebra $\mathbb F$ and $X$ may not be measurable w.r.t. $\mathbb G$