[proofplan]
Fix a positive-probability treatment value $a$ and a measurable outcome event $B$. Consistency says that, on the arm event $\{A=a\}$, the observed outcome event $\{Y\in B\}$ is exactly the same event as the potential-outcome event $\{Y_a\in B\}$. Randomization then removes the conditioning on $\{A=a\}$ from the law of $Y_a$. Dividing by the positive probability $\mathbb P(A=a)$ gives the desired identification formula.
[/proofplan]
[step:Replace the observed arm event by the potential-outcome event using consistency]
Fix $a\in\mathcal A$ with $\{a\}\in\mathcal E_{\mathcal A}$ and $\mathbb P(A=a)>0$, and fix $B\in\mathcal E_{\mathcal Y}$. Define the treatment-arm event
\begin{align*}
E_a:=\{\omega\in\Omega:A(\omega)=a\}.
\end{align*}
Since $A$ is measurable and $\{a\}\in\mathcal E_{\mathcal A}$, one has $E_a\in\mathcal F$.
Define the observed outcome event
\begin{align*}
F_B:=\{\omega\in\Omega:Y(\omega)\in B\}
\end{align*}
and the potential-outcome event
\begin{align*}
G_B:=\{\omega\in\Omega:Y_a(\omega)\in B\}.
\end{align*}
Since $Y$ and $Y_a$ are measurable and $B\in\mathcal E_{\mathcal Y}$, both $F_B$ and $G_B$ belong to $\mathcal F$.
By consistency, $Y(\omega)=Y_a(\omega)$ for every $\omega\in E_a$. Therefore, for every $\omega\in E_a$,
\begin{align*}
\omega\in F_B \iff \omega\in G_B.
\end{align*}
Hence
\begin{align*}
E_a\cap F_B=E_a\cap G_B.
\end{align*}
[/step]
[step:Compute the conditional observed probability on the treatment arm]
Because $\mathbb P(E_a)>0$, the elementary [conditional probability](/page/Conditional%20Probability) $\mathbb P(F_B\mid E_a)$ is defined. Using the definition of conditional probability and the event equality from the previous step gives
\begin{align*}
\mathbb P(Y\in B\mid A=a)=\mathbb P(F_B\mid E_a).
\end{align*}
Thus
\begin{align*}
\mathbb P(Y\in B\mid A=a)=\frac{\mathbb P(E_a\cap F_B)}{\mathbb P(E_a)}.
\end{align*}
Since $E_a\cap F_B=E_a\cap G_B$, this becomes
\begin{align*}
\mathbb P(Y\in B\mid A=a)=\frac{\mathbb P(E_a\cap G_B)}{\mathbb P(E_a)}.
\end{align*}
[/step]
[step:Use randomization to remove the conditioning on treatment assignment]
Randomization in the fixed-arm sense applies with $c=a$, $C=\{a\}\in\mathcal E_{\mathcal A}$, and $D=B\in\mathcal E_{\mathcal Y}$. Since $E_a=\{A\in\{a\}\}$ and $G_B=\{Y_a\in B\}$, it gives
\begin{align*}
\mathbb P(E_a\cap G_B)=\mathbb P(E_a)\mathbb P(G_B).
\end{align*}
Substituting this into the preceding identity and using $\mathbb P(E_a)>0$, we obtain
\begin{align*}
\mathbb P(Y\in B\mid A=a)=\mathbb P(G_B).
\end{align*}
Since $G_B=\{Y_a\in B\}$, this is
\begin{align*}
\mathbb P(Y\in B\mid A=a)=\mathbb P(Y_a\in B).
\end{align*}
[guided]
The only probabilistic input in this step is independence from randomization. Randomization is stated in fixed-arm event form: for every treatment value $c\in\mathcal A$, every treatment-measurable set $C\in\mathcal E_{\mathcal A}$, and every outcome-measurable set $D\in\mathcal E_{\mathcal Y}$,
\begin{align*}
\mathbb P(A\in C,\,Y_c\in D)=\mathbb P(A\in C)\mathbb P(Y_c\in D).
\end{align*}
For the fixed treatment value $a$ and the fixed outcome event $B$, take $c=a$, $C=\{a\}$, and $D=B$. The treatment-arm event is
\begin{align*}
E_a=\{\omega\in\Omega:A(\omega)=a\},
\end{align*}
and the potential-outcome event is
\begin{align*}
G_B=\{\omega\in\Omega:Y_a(\omega)\in B\}.
\end{align*}
The randomization hypothesis therefore gives the product formula
\begin{align*}
\mathbb P(E_a\cap G_B)=\mathbb P(E_a)\mathbb P(G_B).
\end{align*}
We also derive the conditional-probability expression used here. By definition of elementary conditional probability and because $\mathbb P(E_a)>0$,
\begin{align*}
\mathbb P(Y\in B\mid A=a)=\frac{\mathbb P(E_a\cap F_B)}{\mathbb P(E_a)}.
\end{align*}
Consistency gave $E_a\cap F_B=E_a\cap G_B$, so
\begin{align*}
\mathbb P(Y\in B\mid A=a)=\frac{\mathbb P(E_a\cap G_B)}{\mathbb P(E_a)}.
\end{align*}
This gives
\begin{align*}
\mathbb P(Y\in B\mid A=a)=\frac{\mathbb P(E_a)\mathbb P(G_B)}{\mathbb P(E_a)}.
\end{align*}
The positivity assumption $\mathbb P(A=a)>0$ is exactly the assertion $\mathbb P(E_a)>0$, so division by $\mathbb P(E_a)$ is legitimate. Hence
\begin{align*}
\mathbb P(Y\in B\mid A=a)=\mathbb P(G_B).
\end{align*}
Finally, by the definition of $G_B$,
\begin{align*}
\mathbb P(G_B)=\mathbb P(Y_a\in B).
\end{align*}
Thus
\begin{align*}
\mathbb P(Y\in B\mid A=a)=\mathbb P(Y_a\in B).
\end{align*}
[/guided]
[/step]
[step:Conclude the identification formula]
The preceding equality holds for the fixed measurable outcome set $B\in\mathcal E_{\mathcal Y}$. Reversing the sides gives
\begin{align*}
\mathbb P(Y_a\in B)=\mathbb P(Y\in B\mid A=a).
\end{align*}
Since $B\in\mathcal E_{\mathcal Y}$ was arbitrary, this proves the claimed equality of the intervention distribution under treatment $a$ with the observed outcome distribution in the randomized treatment arm $A=a$.
[/step]
