Positive joint density of a finite additive-noise model

First suppose every $q_j$ is Borel measurable. For each $j$, let $Q_j=q_{j}d\lambda$ on Borel sets; [F1, F3] makes $Q_j$ a probability measure. The function $g(n)={\prod }_{r=1}^p{q}_{v_r}(n_{v_r})$ is Borel, hence product-measurable, by [F7, F8]. Repeated Tonelli gives $\int gd\rho =1$ and, on every Borel rectangle, ${\int }_{\prod B_j}gd\rho ={\prod }_j{Q}_j(B_j)$. Uniqueness of product measures identifies this Borel density measure with the joint law of the independent noises.

Recursion in topological order defines a Borel map $T:{\mathbb{R}}^p\to {\mathbb{R}}^p$ from the noise vector to $X$. Its inverse has coordinates $n_j=x_j-f_j(x_{\mathrm{pa}(j)})$, so the observational law is the Borel pushforward under $T$.

Start from step 1.1 and change one coordinate at a time in topological order. With earlier endogenous coordinates fixed, $x_{v_r}=n_{v_r}+f_{v_r}(x_{\mathrm{pa}(v_r)})$ is a fixed translation. Translation invariance changes its density factor to $q_{v_r}(x_{v_r}-f_{v_r}(x_{\mathrm{pa}(v_r)}))$. Every intermediate integrand for a nonnegative Borel test function is product-measurable by [F7, F8], so Tonelli justifies the successive one-dimensional substitutions. The resulting identity $\mathbb{E}[h(X)]=\int h(x)p_X(x)d\rho (x)$ for every nonnegative Borel $h$ proves the density assertion on Borel sets. The product is Borel and strictly positive. No choice principle was used in this case.

Now assume Countable Choice and allow arbitrary Lebesgue-measurable $q_j$. By [F9] choose Borel representatives, replace their values outside $(0,\mathrm{\infty })$ by $1$, and call the resulting functions ${\bar{q}}_j$. Since each original density is positive and finite almost everywhere, this changes it only on a null set and preserves its integral and noise law. Choose a Borel null hull $Z_j$ of $\{q_j\mathrm{\ne }{\bar{q}}_j\}$. These are only finitely many applications of [F9]. Step 2.1 applied to ${\bar{q}}_j$ gives the positive Borel density ${\bar{p}}_X$ relative to $\rho$.

For each $j$, let $E_j=\{x:x_j-f_j(x_{\mathrm{pa}(j)})\in Z_j\}$. It is Borel by [F8]. With all coordinates except $x_j$ fixed, its $x_j$-section is a translate of $Z_j$ and has measure zero by [F6]; for $p>1$, Tonelli gives $\rho (E_j)=0$, and for $p=1$ this follows directly from translation invariance. Off the finite union $E={\bigcup }_j{E}_j$, the products $p_X$ and ${\bar{p}}_X$ agree. Completeness [F10] therefore makes $p_X$ measurable for $\bar{\rho }$, with the same integral on every completed measurable set as ${\bar{p}}_X$. Every original factor is positive, proving strict positivity as well.

In the Countable Choice case, the density measures in steps 2.1 and 4.1 extend the Borel observational law. To see uniqueness among absolutely continuous extensions, first note that every product-measurable set differs from a Borel set by a subset of a Borel $\rho$-null set: the sets with this property form a sigma-algebra under Countable Choice and contain every Lebesgue rectangle by [F9]; Borel null coordinate slabs have product measure zero by Tonelli. A product-null set has a Borel null hull by the same representation, so every $\bar{\rho }$-measurable set has such a representation too. Any absolutely continuous extension assigns zero to the exceptional set and is determined by the Borel core. Thus the stated extensions and both density conclusions hold.