How do you reconcile the definitions of multiple mutually independent events and multiple mutually independent random variables? 1) If $A, B, C$ are events then to prove mutual independence it's not enough to show $P(A,B,C)=P(A)P(B)P(C)$. You also have to show $P(A,B)=P(A)P(B)$ etc.
2) For random variables, the definition of mutual independence looks different. If $X,Y,Z$ are random variables, then it's enough to show $P(X,Y,Z)=P(X)P(Y)P(Z)$, for every possible value of $X,Y,Z$
The situation is different. In the random variable case each assignment of values to $X,Y,Z$ is a collection of events for which you'd have to show independence as in case 1).
case 2) of course follows from case 1). I am looking for a simple proof, since it seems like it should follow in a straightforward way. Most texts/PDFs simply state it without proof. For example, here
 A: Your statement about the definition of independent random variables is
applicable only to discrete random variables and not for continuous
random variables since in the continuous case what A. Webb writes
as
$$\Pr(X=x,Y=y,Z=z)= \Pr(X=x) \Pr(Y=y) \Pr(Z=z)  \qquad \text{for all} ~x,y,z$$
is trivially satisfied with both sides having value $0$ for all 
choices of $x,y,z$.
A more verbose alternative definition of the mutual stochastic independence of three events $A,B, C$ is that all eight
of the following identities must hold:
\begin{align}
\Pr(A\cap B\cap C) &= \Pr(A)\Pr(B)\Pr(C) \tag{1}\\
\Pr(A\cap B\cap C^c) &= \Pr(A)\Pr(B)\Pr(C^c)\tag{2}\\
\Pr(A\cap B^c\cap C) &= \Pr(A)\Pr(B^c)\Pr(C) \tag{3}\\
\Pr(A\cap B^c\cap C^c) &= \Pr(A)\Pr(B^c)\Pr(C^c) \tag{4}\\
\Pr(A^c\cap B\cap C) &= \Pr(A^c)\Pr(B)\Pr(C) \tag{5}\\
\Pr(A^c\cap B\cap C^c) &= \Pr(A^c)\Pr(B)\Pr(C^c)\tag{6}\\
\Pr(A^c\cap B^c\cap C) &= \Pr(A^c)\Pr(B^c)\Pr(C) \tag{7}\\
\Pr(A^c\cap B^c\cap C^c) &= \Pr(A^c)\Pr(B^c)\Pr(C^c) \tag{8}\\
\end{align}
which I hope will seem to you to be eerily akin to checking whether
$$\Pr(X=x,Y=y,Z=z)= \Pr(X=x) \Pr(Y=y) \Pr(Z=z)$$
holds for all $x,y,z$!
These $8$ equalities are a little redundant in that you don't
need to check on all of them; checking just a few suffices. 
Note that $(1)$ is an equality that you need to check (you knew
that already!) but if you also verify that $(2)$ holds, then
you can add $(1)$ and $(2)$ to get that
$$\Pr(A\cap B) = \Pr(A)\Pr(B)\tag{9}$$ holds. Similarly,
after checking $(3)$, adding $(1)$ and $(3)$ shows that
$$\Pr(A\cap C) = \Pr(A)\Pr(C)\tag{10}$$ holds, and the sum
of $(1)$ and $(5)$ is $$\Pr(B\cap C) = \Pr(B)\Pr(C) \tag{11}$$
at which point you can stop.  But, $(1)$-$(8)$ are useful
to keep in mind when dealing with events and their complements:

the probability of an intersection is the product of the
  probabilities, regardless of whether we are dealing with
  mutually independent events or their complements or both.

A: Case 2 should read
\begin{equation}
\Pr(X=x,Y=y,Z=z)= \Pr(X=x) \Pr(Y=y) \Pr(Z=z)  \qquad \text{for all x,y,z}
\end{equation}
That this holds for any combination of $x,y,z$ is a strong condition. Recall that we get the marginal distribution of $X,Y$ by summing over $Z$
\begin{align}
\Pr(X=x,Y=y) &= \sum_z \Pr(X=x,Y=y,Z=z) &\qquad\text{marginal distribution}\\
&= \sum_z \Pr(X=x) \Pr(Y=y) \Pr(Z=z) &\qquad\text{assumed condition} \\
&= \Pr(X=x) \Pr(Y=y) \cdot \sum_z  \Pr(Z=z) \\
&= \Pr(X=x) \Pr(Y=y)
\end{align}
so pairwise independence follows. The continuous case is analagous.
