Let $X, Z$ be random variables and $f$ a measurable function. Let $X_{Z=z}$ denote a R.V. that is the counterfactual $X$ we would have seen had $Z$ been set to $z$ (i.e., $Z(\omega) \equiv z, \forall \omega \in \Omega$), leaving all else fixed. In a proof of some lemma in a paper I'm reading there's a statement that goes:
If $f(X)$ is $\{X_{Z=z}\}_z$-measurable then ...
My question is: How can $f(X)$ be NOT $\{X_{Z=z}\}_z$-measurable since by the observational consistency axiom $f(X(\omega)) = f(\sum_z \mathbf{1}_{Z=z}(\omega)X_{Z=z}(\omega)) = \sum_z \mathbf{1}_{Z=z}(\omega)f(X_{Z=z}(\omega))$, where $\mathbf{1}_{X=x}(\omega)$ is the indicator function defined for $\omega \in \Omega$ as: $$ \begin{equation} \mathbf{1}_{X=x}(\omega) = \begin{cases} 1, & X(\omega) = x \\ 0, & X(\omega) \ne x \end{cases} \end{equation} $$ and by the existence of potential outcomes axiom $X_{Z=z}$ exists for all $z \in Z(\Omega)$ (see "Causal Models on Probability Spaces").
Thank you, David
Fix a probability space $(\Omega, \mathcal{F}, P)$. Suppose you have two random variables $X:\Omega\rightarrow\mathbb{R}$ and $Z:\Omega\rightarrow\mathbb{R}$. I believe you mean to define a new random variable for each $z\in \mathbb{R}$ by $$ Y_z = X \cdot 1_{\{Z=z\}} = \left\{\begin{array}{cc} X & \mbox{ if $Z=z$} \\ 0 & \mbox{ else} \end{array}\right.$$ Then I think you are asing if $X$ is always $\sigma(\{Y_z\}_{z \in \mathbb{R}})$-measurable. The answer is no.
A case where it is true: It is true if $Z$ can take only a finite or countably infinite number of values: Define $S=\{Z(\omega) \in \mathbb{R}: \omega \in \Omega\}$. If $S$ is finite or countably infinite then indeed we have $$ X = \sum_{z \in S} Y_z$$ and so $X$ is $\sigma(\{Y_z\}_{z \in \mathbb{R}})$-measurable.
A case where it is not true is this: Let $X \in \{-1,1\}$ equally likely. Let $Z$ be independent of $X$ and suppose $Z$ has a continuous CDF (such as a $N(0,1)$ distribution). It can be shown that $X$ is $\sigma(\{Y_z\}_{z \in \mathbb{R}})$-measurable if and only if $X$ is a measurable function of a countably infinite number of the $Y_z$ variables. But for any countably infinite set of real numbers $\{z_1, z_2, z_3, ...\}$, we have $$P[\cap_{i=1}^{\infty}\{Y_{z_i}=0\}]=1$$ So, with prob 1, we obtain no useful information from the $Y_{z_i}$ variables and we cannot obtain $X$ from $(Y_{z_1}, Y_{z_2}, Y_{z_3}, ...)$. So $X$ is not $\sigma(\{Y_z\}_{z \in \mathbb{R}})$-measurable. For the same reason, $Z$ is not $\sigma(\{Y_z\}_{z \in \mathbb{R}})$-measurable.