A translated conditional density factorizes after subtraction

For each fixed $a,c$, the change $b=d+c$ is the translation in [F1]; the full map $S$ is invertible linear with determinant one. Let $E$ be a Lebesgue null set outside which the given factorization holds. Applying [F3] to $1_E$ shows that $S^{-1}E$ is null, so $q(S(a,d,c))=p_A(a)f(d)p_C(c)$ almost everywhere. For a nonnegative Borel test $\mathrm{\Phi }$, the function $(a,b,c)\mapsto \mathrm{\Phi }(a,b-c,c)q(a,b,c)$ is Lebesgue measurable. Change of variables [F3] gives $E\mathrm{\Phi }(A,B-C,C)=\int \mathrm{\Phi }(a,d,c)q(S(a,d,c))dadddc=\int \mathrm{\Phi }(a,d,c)p_A(a)f(d)p_C(c)dadddc$. In this last integral the factors depend on separate coordinates, and $\mathrm{\Phi }$ is measurable for the product Borel sigma-algebra. The integrand is therefore measurable for the product of the factor Lebesgue sigma-algebras; the completed-product identification in [F3] identifies its Euclidean integral with the product integral to which [F2] applies. This establishes the transformed joint density without assuming product measurability of $f(b-c)$.

Integrating this joint density over $c$ and using $\int p_C(c)dc=1$, Tonelli [F2] yields $p_{A,D}(a,d)=p_A(a)f(d)$. Its marginals are $p_A$ and $f$, so the product identity proves independence and identifies the density of $D$.