On the consistency conditions for stochastic processes This result is due to Kolmogorov. Here's the standard version from Wikipedia

Let $T$ denote some interval (thought of as "time"), and let $n\in \mathbb {N}$. For each $k\in \mathbb {N}$ and finite sequence of distinct times
$t_{1},\dots ,t_{k}\in T$, let $\nu _{t_{1}\dots t_{k}}$ be a probability measure on $(\mathbb {R} ^{n})^{k}$. Suppose that these measures satisfy two consistency conditions:

*

*for all permutations $\pi$  of $\{1,\dots ,k\}$ and measurable sets $F_{i}\subseteq \mathbb {R} ^{n}$,
$\nu _{t_{\pi (1)}\dots t_{\pi (k)}}\left(F_{\pi (1)}\times \dots \times F_{\pi (k)}\right)=\nu _{t_{1}\dots t_{k}}\left(F_{1}\times \dots \times F_{k}\right);$

*for all measurable sets $F_{i}\subseteq \mathbb {R} ^{n}$, $m\in \mathbb {N}$ ${\displaystyle \nu _{t_{1}\dots t_{k}}\left(F_{1}\times \dots \times F_{k}\right)=\nu _{t_{1}\dots t_{k},t_{k+1},\dots ,t_{k+m}}\left(F_{1}\times \dots \times F_{k}\times \underbrace {\mathbb {R} ^{n}\times \dots \times \mathbb {R} ^{n}} _{m}\right).}$


Then there exists a probability space $(\Omega ,{\mathcal {F}},\mathbb {P} )$ and a stochastic process $X:T\times \Omega \to \mathbb {R} ^{n}$ such that $${\displaystyle \nu _{t_{1}\dots t_{k}}\left(F_{1}\times \dots \times F_{k}\right)=\mathbb {P} \left(X_{t_{1}}\in F_{1},\dots ,X_{t_{k}}\in F_{k}\right)}$$ for all $t_{i}\in T$, $k\in \mathbb {N}$  and measurable sets $F_{i}\subseteq \mathbb {R} ^{n}$; i.e. $X$ has $\nu _{t_{1}\dots t_{k}}$ as its finite-dimensional distributions relative to times $t_{1}\dots t_{k}$.

Since I'm not very knowledgeable about measure theory, I looked at some other sources. In the book "Stochastic processes in physics and chemistry" by Van Kampen, there seems to be an equivalent formulation of conditions $(1)$ and $(2)$, by means of finite-dimensional probability density functions:

*

*(symmetry) $\displaystyle p(\dots;x_i,t_i;\dots;x_j,t_j;\dots)=p(\dots;x_j,t_j;\dots;x_i, t_i;\dots)$
2.(completeness) $\displaystyle \int p(x_n, t_n;\dots;x_1, t_1)\mathrm{d}x_n=p(x_{n-1}, t_{n-1};\dots;x_1, t_1)$
On one hand, I don't see a link between these two ways of phrasing Kolmogorov's theorem, especially between point(s) 2. On the other hand, the symmetry and completeness conditions seem trivial.
The first one simply says that $$P(A\cap B)=P(B\cap A)$$
For what concerns the second, $$\sum_{x\in B}P(A\cap B\cap C)=P(A\cap C)$$ If we want to paraphrase in the simplest case, the probability of finding "head" after a single coin toss is equivalent to the joint probability of finding "head AND whatever" in a toss of N coins.
 A: 1.
The two formulations of conditions (1) and (2) are equivalent provided that the measures $\nu$ have densities $p$. (In general, the measures may not have densities, so Kolmogorov's version covers some situations that Van Kampen's doesn't.)
If we assume the densities $p$ exist, then Van Kampen's (1) implies Kolmogorov's (1) by taking the integrals
$$
\nu_{t_1,\dots,t_n}(F_1 \times \dots \times F_n) = \int_{F_1} \dots \int_{F_n} p(x_1, t_1; \dots; x_n, t_n)\ dx_n \dots dx_1,$$
and the converse holds by taking the derivatives
$$
p(x_1,t_1;\dots;x_n,t_n) = \frac{\partial^n}{\partial x_1 \dots \partial x_n} \nu_{t_1,\dots,t_n}((-\infty, x_1] \times \dots \times (-\infty, x_n]).$$
Given Kolmogorov's (1) and (2), we have
$$
\nu_{t_n,\dots,t_1}(\mathbb R \times F_{n-1} \times \dots \times F_1)
= \nu_{t_{n-1},\dots,t_1}(F_{n-1} \times \dots \times F_1),
$$
which then proves Van Kampen's (2) by differentiation, as above.
Similarly, given Van Kampen's (1) and (2), we can use integration to deduce Kolmogorov's (2) in the case $m=1$; applying this result repeatedly gives the general form of Kolmogorov's (2).
2.
Yes, the conditions (1) and (2) are trivial, if the measures $\nu$ or densities $p$ represent the finite-dimensional distributions of a stochastic process. The point is that these conditions are necessary to check whether or not such a process could exist.
For example, suppose my measures $\nu_{t_1,t_2}$ place all their mass on $(0, 1)$, for any values of $t_1$ and $t_2$. Then this would imply a "stochastic process" $X$ whose finite-dimensional distributions satisfy:
$$\begin{align*}
P(X_0 = 0, X_1 = 1) &= 1\\
P(X_1 = 1, X_0 = 0) &= 0\end{align*}$$
Obviously this is nonsense, and no such process $X$ exists, but to exclude these nonsense measures $\nu$ from the theorem, I need condition (1). Likewise, condition (2) prevents the measures from contradicting themselves when you add a new time $t_n$.
The interesting part of the theorem is that (1) and (2) are the only conditions required. If you check these two things, the measures must make sense as the finite-dimensional distributions of a stochastic process; there's nothing else that can go wrong.
