Given a probability space $(\Omega, \mathcal{F}, P)$, let $\mathcal{G}$ be a $\sigma$-subalgebra of $\mathcal{F}$. Does there exist a random variable $X$ such that $\sigma(X) = \mathcal{G}$, and if so, is there a simple way to construct such an $X$?
I am asking this question because I find the definition of conditional expectation much more intuitive when the conditioning is on a random variable, because then we can just think of the conditional expectation $E[Y|X]$ as the function of $X$ that minimizes the expected mean square error.
The answer is NO. $\sigma (X)$ is generated by the countable family of sets $\{X^{-1}(a,b): a,b \in \mathbb Q, a<b\}$. Not every $\sigma$ algebra is countably generated. For example if you take $X=\mathbb R, \mathcal F=\mathcal G=\mathcal P(\mathbb R)$ then no such r.v. $X$ exists.