[proofplan] Fix a measurable point $x$ with positive treatment event $\{X=x\}$ and fix a measurable outcome set $A$. The consistency assumption gives equality of $Y$ and $Y_x$ outside a null subset of the conditioning event. Therefore the two outcome events $\{Y\in A\}$ and $\{Y_x\in A\}$ have the same intersection with $\{X=x\}$ up to a null set. The definition of [conditional probability](/page/Conditional%20Probability) given a positive-probability event then gives the desired equality after division by $\mathbb P(X=x)$. [/proofplan] [step:Reduce consistency to equality of the two conditioned outcome events] Fix $x\in\mathcal X$ such that $\{x\}\in\mathcal E$ and $\mathbb P(X=x)>0$, and fix $A\in\mathcal G$. Define the conditioning event \begin{align*} B:=X^{-1}(\{x\})=\{\omega\in\Omega:X(\omega)=x\}. \end{align*} Since $X$ is $(\mathcal F,\mathcal E)$-measurable and $\{x\}\in\mathcal E$, one has $B\in\mathcal F$. Define the observed-outcome event \begin{align*} E:=Y^{-1}(A)=\{\omega\in\Omega:Y(\omega)\in A\}, \end{align*} and define the potential-outcome event \begin{align*} E_x:=Y_x^{-1}(A)=\{\omega\in\Omega:Y_x(\omega)\in A\}. \end{align*} Since $Y$ and $Y_x$ are $(\mathcal F,\mathcal G)$-measurable and $A\in\mathcal G$, both $E$ and $E_x$ belong to $\mathcal F$. By consistency on $B$, there exists $N_x\in\mathcal F$ with $\mathbb P(N_x)=0$ such that $Y(\omega)=Y_x(\omega)$ for every $\omega\in B\setminus N_x$. Hence, for every $\omega\in B\setminus N_x$, \begin{align*} \omega\in E \iff \omega\in E_x. \end{align*} Therefore \begin{align*} (B\cap E)\triangle(B\cap E_x)\subset N_x. \end{align*} Since $N_x$ is null, the two events $B\cap E$ and $B\cap E_x$ have equal probability: \begin{align*} \mathbb P(B\cap E)=\mathbb P(B\cap E_x). \end{align*} [guided] We first isolate the event on which we are conditioning. Define \begin{align*} B:=X^{-1}(\{x\})=\{\omega\in\Omega:X(\omega)=x\}. \end{align*} This is an event because the singleton $\{x\}$ is assumed to lie in $\mathcal E$ and $X:(\Omega,\mathcal F)\to(\mathcal X,\mathcal E)$ is measurable. The positivity assumption says $\mathbb P(B)>0$. Next define the two outcome events whose conditional probabilities we want to compare: \begin{align*} E:=Y^{-1}(A)=\{\omega\in\Omega:Y(\omega)\in A\}, \end{align*} and \begin{align*} E_x:=Y_x^{-1}(A)=\{\omega\in\Omega:Y_x(\omega)\in A\}. \end{align*} Both are elements of $\mathcal F$ because $A\in\mathcal G$ and both $Y$ and $Y_x$ are random variables from $(\Omega,\mathcal F)$ to $(\mathcal Y,\mathcal G)$. The consistency assumption is used only on the conditioning event $B$. It gives an event $N_x\in\mathcal F$ with $\mathbb P(N_x)=0$ such that \begin{align*} Y(\omega)=Y_x(\omega) \end{align*} for every $\omega\in B\setminus N_x$. For such an $\omega$, membership in $A$ is the same for $Y(\omega)$ and $Y_x(\omega)$, so \begin{align*} \omega\in E \iff \omega\in E_x. \end{align*} Thus the only points where $B\cap E$ and $B\cap E_x$ can differ lie inside the null event $N_x$: \begin{align*} (B\cap E)\triangle(B\cap E_x)\subset N_x. \end{align*} Since a subset of a null event has probability zero, the two measurable events $B\cap E$ and $B\cap E_x$ have the same probability: \begin{align*} \mathbb P(B\cap E)=\mathbb P(B\cap E_x). \end{align*} [/guided] [/step] [step:Divide by the positive probability of the conditioning event] By the definition of conditional probability given the positive-probability event $B$, \begin{align*} \mathbb P(Y\in A\mid X=x)=\frac{\mathbb P(B\cap E)}{\mathbb P(B)}. \end{align*} Similarly, \begin{align*} \mathbb P(Y_x\in A\mid X=x)=\frac{\mathbb P(B\cap E_x)}{\mathbb P(B)}. \end{align*} The denominator is positive by hypothesis, and the numerators are equal by the previous step. Therefore \begin{align*} \mathbb P(Y\in A\mid X=x)=\mathbb P(Y_x\in A\mid X=x). \end{align*} This proves the asserted equality for the fixed admissible $x$ and fixed $A\in\mathcal G$, and hence for all such $x$ and $A$. [/step]