Given $\mathcal{X}=\{X_1,\ldots,X_n\}$ and some collection $J_1,\ldots,J_k\in \mathcal{P}(\mathcal{X})$, with PDF's $F_{J_i}$ for all $i=1,\ldots,k$, such that whenever $J\subset J_i\cap J_j$ we have that $F_{J_i}$ and $F_{J_j}$ induce the same marginal PDF $F_J$, we don't necessarily have $F_{\mathcal{X}}$ inducing $F_{J_i}$ for all $i=1,\ldots,k$.
Consider, e.g., $X,Y,Z$ where
$$F_{X,Y}(x,y) = \begin{cases}0 & \min(x,y)<0\\ \frac{1}{2} & 0\leq\min(x,y)<1\\ 1 & \min(x,y)\geq 1\end{cases}$$
(which amounts to having $X\stackrel{a.s.}{=}Y\sim B(0.5)$),
$$F_{Y,Z}(y,z) = \begin{cases}0 & \min(y,z)<0\\ \frac{1}{2} & 0\leq\min(y,z)<1\\ 1 & \min(y,z)\geq 1\end{cases}$$
($Y\stackrel{a.s.}{=}Z\sim B(0.5)$) and
$$F_{X,Z}(x,z) = \begin{cases}0 & \min(x,z)<0 \vee \max(x,z)<1\\ \frac{1}{2} & 0\leq\min(x,z)<1\leq\max(x,z)\\ 1 & \min(x,z)\geq 1\end{cases}$$
($(1-X)\stackrel{a.s.}{=}Z\sim B(0.5)$).
Then from any pair of PDF's we consistently get
$$F_X(t) = F_Y(t) = F_Z(t) = \begin{cases}0 & t<0\\ \frac{1}{2} & 0\leq t<1\\ 1 & t\geq 1\end{cases}$$
but no joint PDF $F_{X,Y,Z}$ generates all three marginal distribution functions (since that would imply $X\stackrel{a.s.}{=}(1-X)$).
However, if the maximal elements of $J_1,\ldots,J_k$ (w.r.t. inclusion) are all pairwise-disjoint, then we can construct $F_{\mathcal{X}}$ by assuming independence.
