If I have a joint distribution $\mu$ on $X\times Y$ can I write is as $(F_X,F_{Y \mid X}$) where $F_X$ is the marginal for $X$ and $F_{Y \mid X}$ is the conditional distribution of $Y$ conditioned on $X$?
If so, then in general could I write a joint distribution on $W\times X\times Y\times \cdots \times Z$ as $(F_W,F_{X \mid W}, F_{Y \mid W,X},\cdots, F_{Z \mid W,X,Y,\dots})$?
$\def\d{\mathrm{d}}\def\peq{\mathrel{\phantom{=}}{}}$It suffices to prove that $F_{X_1}, F_{X_2 \mid X_1}, \cdots, F_{X_n \mid X_1, \cdots, X_{n - 1}}$ uniquely determine $F_{X_1, \cdots, X_n}$. Below $μ_X$ means the probability measure of $X$ and assume that $X_1, \cdots, X_n$ are real-valued random variables.
Since $F_{X_1}$ is known, $μ_{X_1}$ is uniquely determined. For $1 \leqslant k \leqslant n - 1$, since $F_{X_{k + 1} \mid X_1, \cdots, X_k}$ is known, define$$ φ_k(x_1, \cdots, x_k; B) = \int\limits_B \d F_{X_{k + 1} \mid X_1, \cdots, X_k}(x; x_1, \cdots, x_k), \quad \forall B \in \mathscr{B} $$ then $μ_{X_{k + 1} \mid X_1, \cdots, X_k}$ is uniquely determined by$$ μ_{X_{k + 1} \mid X_1, \cdots, X_k}(B) = φ_k(X_1, \cdots, X_k; B). \quad \forall B \in \mathscr{B} $$ Next, to prove by induction that $μ_{X_1, \cdots, X_k}$ is uniquely determined for $1 \leqslant k \leqslant n$, the base case $k = 1$ is already proved. Now assume that the proposition holds for $k$. For $k + 1$, because for any $B_1, \cdots, B_{k + 1} \in \mathscr{B}$,\begin{align*} &\peq μ_{X_1, \cdots, X_{k + 1}}(B_1, \cdots, B_{k + 1}) = P(X_1 \in B_1, \cdots, X_{k + 1} \in B_{k + 1})\\ &= E(I_{\{ X_1 \in B_1, \cdots, X_{k + 1} \in B_{k + 1} \}}) = E(I_{\{X_1 \in B_1\}} \cdots I_{\{X_{k + 1} \in B_{k + 1}\}})\\ &= E(E(I_{\{X_1 \in B_1\}} \cdots I_{\{X_{k + 1} \in B_{k + 1}\}} \mid X_1, \cdots, X_k))\\ &= E(I_{\{X_1 \in B_1\}} \cdots I_{\{X_k \in B_k\}} E(I_{\{X_{k + 1} \in B_{k + 1}\}} \mid X_1, \cdots, X_k))\\ &= E(I_{\{X_1 \in B_1\}} \cdots I_{\{X_k \in B_k\}} μ_{X_{k + 1} \mid X_1, \cdots, X_k}(B_{k + 1}))\\ &= E(I_{\{X_1 \in B_1\}} \cdots I_{\{X_k \in B_k\}} φ_k(X_1, \cdots, X_k; B_{k + 1}))\\ &= \mathop{\intop\cdots\intop}\limits_{B_1 × \cdots × B_k} φ_k(x_1, \cdots, x_k; B) \,μ_{X_1, \cdots, X_k}(\d x_1, \cdots, \d x_k), \end{align*} then $μ_{X_1, \cdots, X_{k + 1}}$ is uniquely determined on$$ \mathscr{A}_{k + 1} = \{B_1 × \cdots × B_{k + 1} \mid B_1, \cdots, B_{k + 1} \in \mathscr{B}\}. $$ Note that $\mathscr{B}^{k + 1} = σ(\mathscr{A}_{k + 1})$, thus $μ_{X_1, \cdots, X_{k + 1}}$ is uniquely determined on $\mathscr{B}^{k + 1}$. End of induction.
Finally, since $μ_{X_1, \cdots, X_n}$ is uniquely determined, so is $F_{X_1, \cdots, X_n}$.