If $X_1$, $X_2$ and $X_3$ are mutually independent, I'd like to show $$ P(X_1 < X_3, X_2 < X_3 \mid X_3) = P(X_1 < X_3 \mid X_3)\cdot P(X_2 < X_3 \mid X_3), \quad a.s. $$ Here's what I have:
The conditional probabilities are functions on the sample space $\Omega$ that are defined to satisfy, for any $C \in \sigma(X_3)$, \begin{align*} \int_C P(X_1 < X_3 \mid X_3)\, dP & = P(X_1 < X_3, C) \\ \int_C P(X_2 <X_3 \mid X_3)\, dP & = P(X_2 < X_3, C) \\ \int_C P(X_1 < X_3, X_2 <X_3 \mid X_3)\, dP & = P(X_1 < X_3, X_2 < X_3, C) \end{align*}
I can't seem to get anywhere from here, though, as it's not clear how the original independence can be used.
Let $F_i$, for $i=1, \ldots, 3$, be the respective distribution function of $X_i$. Then, for any Borel set $B$, \begin{align*} \int_{X_3 \in B} P(X_1 < X_3\mid X_3) dP &=\int_{\Omega}1_{X_1<X_3} 1_{X_3 \in B} dP \\ &=\iint_{\mathbb{R}^2} 1_{x< z} 1_{z\in B} dF_1(x) dF_3(z)\\ &=\int_{\mathbb{R}} F_1(z) 1_{z\in B} dF_3(z)\\ &=\int_{\Omega} F_1(X_3)1_{X_3 \in B} dP \\ &=\int_{X_3 \in B}F_1(X_3) dP. \end{align*} That is, $$E(X_1 < X_3 \mid X_3) = F_1(X_3).$$ Similarly, $$E(X_2 < X_3 \mid X_3) = F_2(X_3).$$ Then, \begin{align*} \int_{X_3 \in B} 1_{X_1 < X_3} 1_{X_2 < X_3} dP &= \iiint_{\mathbb{R}^3}1_{x<z}1_{y<z}1_{z\in B} dF_1(x)dF_2(y)dF_3(z)\\ &=\int_{\mathbb{R}} F_1(z)F_2(z) 1_{z\in B} dF_3(z)\\ &=\int_{\Omega}F_1(X_3) F_2(X_3) 1_{X_3 \in B} dP \\ &=\int_{X_3 \in B}F_1(X_3) F_2(X_3) dP\\ &=\int_{X_3 \in B} E(X_1 < X_3 \mid X_3) E(X_2 < X_3 \mid X_3) dP. \end{align*} Therefore, $$E(X_1 < X_3, X_2 < X_3 \mid X_3) = E(X_1 < X_3 \mid X_3) E(X_2 < X_3 \mid X_3). $$