Probability distribution functions: factorization 3-way implies 2-way? I recently asked a question about pairwise versus mutual independence (also related to this and this q). 
However, 
(1) I inadvertently used incorrect terminology:

three events, A, B, C are mutually independent when:
P[A,B]=P[A]P[B], P[B,C]=P[B]P[C], P[A,C]=P[A]P[C],
  P[A,B,C]=P[A]P[B]P[C]

Did and others pointed out that

"Mutual independence means the four identities you copied, pairwise
  independence means the first three of these identities." -- Did

Note that the term mutual has varying definitions across math. For example, mutual information is a pairwise relation. 
(2) Going back to probability, GC Rota said the theory can be approached two ways: focusing on random variables (event algebra) or focusing on distributions. Here I am interested in distributions, where independence can be interpreted as factorization of the probability distribution function. The conditions are the same as above, where P is interpreted as the PDF function. 
The following graphic based on a standard example from Counterexamples in Probability and Statistics of a 3-dimensional binomial PDF that factorizes pairwise (ie, each of the 3 pairs of random variables are independent and the 2-dim joint distributions can all be written as the product of the respective marginals) but not 3-way independent (the joint distribution cannot be written as the product of the individual marginal distributions)

My question is whether the opposite can happen, ie if the 3-dim (or perhaps higher) joint distribution factorizes into the 1-dim marginals, does that imply the pairwise factorization of all 2-dim joint distributions into the marginals? 
 A: Indeed, assume that $\mu$ is the product of the probability measures $\mu_1$, $\mu_2$ and $\mu_3$, hence, for every $(A_1,A_2,A_3)$, $\mu(A_1\times A_2\times A_3)=\mu_1(A_1)\cdot\mu_2(A_2)\cdot\mu_3(A_3)$. 
Then, for example, the 2-marginal distribution $\nu$ corresponding to the two first coordinates is such that $\nu(A)=\mu(A\times\mathbb R)$ for every 2-dimensional $A$.
One sees that, for every $(A_1,A_2)$, $\nu(A_1\times A_2)=\mu(A_1\times A_2\times\mathbb R)=\mu_1(A_1)\cdot\mu_2(A_2)$, that is, that $\nu$ is indeed the product of $\mu_1$ and $\mu_2$.
A: If the joint density of $X$,$Y$, and $Z$ is $f_{X,Y,Z}(x,y,z)$, then for any 
pair, say $Y$ and $Z$,
$$f_{Y,Z}(y,z) = \int_{-\infty}^{\infty}f_{X,Y,Z}(x,y,z)\,\mathrm dx.$$
Now suppose that $f_{X,Y,Z}(x,y,z) = f_X(x)f_Y(y)f_Z(z)$ for all real numbers
$x,y$ and $z$. Substitution in the
integral gives
$$f_{Y,Z}(y,z) = \int_{-\infty}^{\infty}f_X(x)f_Y(y)f_Z(z) \mathrm dx
= f_Y(y)f_Z(z)\int_{-\infty}^{\infty}f_X(x)\mathrm dx = f_Y(y)f_Z(z).$$
I will leave it to you to verify that a similar result holds for the
other two pairs of random variables.
