[proofplan] Write $S=e(X)$ and prove conditional independence through bounded test functions, so that all null-set issues are absorbed into conditional-expectation identities. First compute the conditional law of the binary treatment given $S$ and $X$: because $S$ is a version of $\mathbb P(T=1\mid X)$ and $S$ is $\sigma(X)$-measurable, conditioning further on $S$ changes nothing. This gives the balancing identity $T\perp X\mid S$. For the ignorability conclusion, test against bounded functions of the potential-outcome pair and use conditional ignorability given $X$ to factor the [conditional expectation](/page/Conditional%20Expectation); the remaining factor depends on treatment only through $S$, so the same calculation gives independence given $S$. [/proofplan]

[step:Record the measurability relations generated by the propensity score] Define the score [random variable](/page/Random%20Variable) \begin{align*} S:(\Omega,\mathcal F)\to([0,1],\mathcal B([0,1])) \end{align*} by \begin{align*} S=e(X). \end{align*} Since $e$ is $\mathcal E/\mathcal B([0,1])$-measurable and $X$ is $\mathcal F/\mathcal E$-measurable, $S$ is $\sigma(X)$-measurable. Hence \begin{align*} \sigma(S)\subset \sigma(X). \end{align*} The propensity-score hypothesis says that $S$ is a chosen version of the conditional expectation \begin{align*} \mathbb E[\mathbb 1_{\{T=1\}}\mid \sigma(X)]. \end{align*} Therefore \begin{align*} \mathbb E[\mathbb 1_{\{T=0\}}\mid \sigma(X)] = 1-S \end{align*} almost surely, because $\mathbb 1_{\{T=0\}}=1-\mathbb 1_{\{T=1\}}$. [/step]

[step:Compute the treatment law after conditioning on the score]Let $t\in\{0,1\}$. Define the score-level treatment probability \begin{align*} p_t:\Omega\to[0,1] \end{align*} by \begin{align*} p_t=\mathbb E[\mathbb 1_{\{T=t\}}\mid \sigma(S)]. \end{align*} Using $\sigma(S)\subset\sigma(X)$ and the defining consistency of iterated conditional expectation, we obtain \begin{align*} p_1 = \mathbb E[\mathbb E[\mathbb 1_{\{T=1\}}\mid \sigma(X)]\mid \sigma(S)]. \end{align*} Substituting the propensity-score version gives \begin{align*} p_1 = \mathbb E[S\mid \sigma(S)]. \end{align*} Since $S$ is $\sigma(S)$-measurable, \begin{align*} p_1=S. \end{align*} Similarly, \begin{align*} p_0=1-S. \end{align*}[/step]

[guided]We want the treatment probability inside a score stratum. For $t\in\{0,1\}$, define \begin{align*} p_t:\Omega\to[0,1] \end{align*} by \begin{align*} p_t=\mathbb E[\mathbb 1_{\{T=t\}}\mid \sigma(S)]. \end{align*} The key point is that conditioning on $X$ is finer than conditioning on $S$, because $S=e(X)$ is a [measurable function](/page/Measurable%20Function) of $X$. Thus $\sigma(S)\subset\sigma(X)$. For $t=1$, the defining property of the propensity score gives \begin{align*} \mathbb E[\mathbb 1_{\{T=1\}}\mid \sigma(X)]=S \end{align*} almost surely. Taking conditional expectation of both sides with respect to $\sigma(S)$ gives \begin{align*} \mathbb E[\mathbb E[\mathbb 1_{\{T=1\}}\mid \sigma(X)]\mid \sigma(S)] = \mathbb E[S\mid \sigma(S)]. \end{align*} The left-hand side is $\mathbb E[\mathbb 1_{\{T=1\}}\mid \sigma(S)]$ by iterated conditioning along the inclusion $\sigma(S)\subset\sigma(X)$. The right-hand side is $S$, because $S$ is $\sigma(S)$-measurable. Therefore \begin{align*} \mathbb E[\mathbb 1_{\{T=1\}}\mid \sigma(S)]=S. \end{align*} Since $\mathbb 1_{\{T=0\}}=1-\mathbb 1_{\{T=1\}}$, linearity of conditional expectation gives \begin{align*} \mathbb E[\mathbb 1_{\{T=0\}}\mid \sigma(S)]=1-S. \end{align*} Thus, inside each score stratum, the conditional treatment law is Bernoulli with success probability $S$.[/guided]

[step:Prove that the treatment is conditionally independent of covariates given the score] Let \begin{align*} h:(E,\mathcal E)\to(\mathbb R,\mathcal B(\mathbb R)) \end{align*} be a bounded measurable map. For $t=1$, the product $h(X)S$ is integrable and $\sigma(X)$-measurable, so \begin{align*} \mathbb E[h(X)\mathbb 1_{\{T=1\}}\mid \sigma(S)] = \mathbb E[\mathbb E[h(X)\mathbb 1_{\{T=1\}}\mid \sigma(X)]\mid \sigma(S)]. \end{align*} Because $h(X)$ is bounded and $\sigma(X)$-measurable, \begin{align*} \mathbb E[h(X)\mathbb 1_{\{T=1\}}\mid \sigma(X)] = h(X)\mathbb E[\mathbb 1_{\{T=1\}}\mid \sigma(X)]. \end{align*} Using the propensity-score version, \begin{align*} \mathbb E[h(X)\mathbb 1_{\{T=1\}}\mid \sigma(S)] = \mathbb E[h(X)S\mid \sigma(S)]. \end{align*} Since $S$ is $\sigma(S)$-measurable, \begin{align*} \mathbb E[h(X)\mathbb 1_{\{T=1\}}\mid \sigma(S)] = S\mathbb E[h(X)\mid \sigma(S)]. \end{align*} By the preceding step, $S=\mathbb E[\mathbb 1_{\{T=1\}}\mid \sigma(S)]$, hence \begin{align*} \mathbb E[h(X)\mathbb 1_{\{T=1\}}\mid \sigma(S)] = \mathbb E[h(X)\mid \sigma(S)]\mathbb E[\mathbb 1_{\{T=1\}}\mid \sigma(S)]. \end{align*} For $t=0$, the same computation with $\mathbb 1_{\{T=0\}}$ and $1-S$ gives \begin{align*} \mathbb E[h(X)\mathbb 1_{\{T=0\}}\mid \sigma(S)] = \mathbb E[h(X)\mid \sigma(S)]\mathbb E[\mathbb 1_{\{T=0\}}\mid \sigma(S)]. \end{align*} Every bounded map $k:\{0,1\}\to\mathbb R$ has the representation \begin{align*} k(T)=k(1)\mathbb 1_{\{T=1\}}+k(0)\mathbb 1_{\{T=0\}}. \end{align*} Linearity of conditional expectation therefore yields \begin{align*} \mathbb E[h(X)k(T)\mid \sigma(S)] = \mathbb E[h(X)\mid \sigma(S)]\mathbb E[k(T)\mid \sigma(S)]. \end{align*} This is precisely $T\perp X\mid S$ in the bounded-test-function formulation. [/step]

[step:Use ignorability given covariates to factor outcome and treatment tests] Assume now that $(Y(1),Y(0))\perp T\mid X$. Define the potential-outcome pair \begin{align*} U:(\Omega,\mathcal F)\to(\mathbb R^2,\mathcal B(\mathbb R^2)) \end{align*} by \begin{align*} U=(Y(1),Y(0)). \end{align*} Let \begin{align*} g:(\mathbb R^2,\mathcal B(\mathbb R^2))\to(\mathbb R,\mathcal B(\mathbb R)) \end{align*} be bounded and measurable. Choose a bounded $\sigma(X)$-measurable version \begin{align*} M_g:\Omega\to\mathbb R \end{align*} of \begin{align*} \mathbb E[g(U)\mid \sigma(X)]. \end{align*} Conditional ignorability gives, for each $t\in\{0,1\}$, \begin{align*} \mathbb E[g(U)\mathbb 1_{\{T=t\}}\mid \sigma(X)] = M_g\mathbb E[\mathbb 1_{\{T=t\}}\mid \sigma(X)] \end{align*} almost surely. [/step]

[step:Average the covariate-level factorization over score strata] For $t=1$, iterated conditioning and the factorization from the previous step give \begin{align*} \mathbb E[g(U)\mathbb 1_{\{T=1\}}\mid \sigma(S)] = \mathbb E[M_g S\mid \sigma(S)]. \end{align*} Since $S$ is $\sigma(S)$-measurable, \begin{align*} \mathbb E[g(U)\mathbb 1_{\{T=1\}}\mid \sigma(S)] = S\mathbb E[M_g\mid \sigma(S)]. \end{align*} Again by iterated conditioning, \begin{align*} \mathbb E[M_g\mid \sigma(S)] = \mathbb E[g(U)\mid \sigma(S)]. \end{align*} Using $\mathbb E[\mathbb 1_{\{T=1\}}\mid \sigma(S)]=S$, we obtain \begin{align*} \mathbb E[g(U)\mathbb 1_{\{T=1\}}\mid \sigma(S)] = \mathbb E[g(U)\mid \sigma(S)]\mathbb E[\mathbb 1_{\{T=1\}}\mid \sigma(S)]. \end{align*} For $t=0$, replacing $S$ by $1-S$ gives \begin{align*} \mathbb E[g(U)\mathbb 1_{\{T=0\}}\mid \sigma(S)] = \mathbb E[g(U)\mid \sigma(S)]\mathbb E[\mathbb 1_{\{T=0\}}\mid \sigma(S)]. \end{align*} Therefore, for every bounded map $k:\{0,1\}\to\mathbb R$, \begin{align*} \mathbb E[g(U)k(T)\mid \sigma(S)] = \mathbb E[g(U)\mid \sigma(S)]\mathbb E[k(T)\mid \sigma(S)]. \end{align*} This is the bounded-test-function formulation of \begin{align*} U\perp T\mid S. \end{align*} Since $U=(Y(1),Y(0))$, this proves \begin{align*} (Y(1),Y(0))\perp T\mid e(X). \end{align*} Together with the balancing result, the theorem follows. [/step]