I know that given a space $(E, \epsilon)$ (with some properties, but whatever) and a consistent family of finite dimensional distributions, I have Kolmogorov extension theorem that tells there is a probability $\mu$ on $(E^T, \epsilon ^T)$ with those distributions.
I have been told that if I stopped at distributions of order $m$ I could always find a counterexample where I have a consistent family of distributions up to order $m$ that doesn’t come from any distribution $\mu$ on $(E^T, \epsilon ^T)$.
I don’t see how to construct such counterexample, though.
Let $m=3$, with $T=\{1,2,3\}$ and suppose $(X_1,X_2)$ was uniform on $\{(0,1),(1,0)\}$ and so was $(X_1,X_3)$ and so was $(X_2,X_3)$. And that the $X_i$ are individually uniform on $\{0,1\}$. This specifies all $1$ and $2$-dimensional margins for $\{X_t:t\in S\}$ for all $S\subset T$ with $| S|<3$. There is no joint probability for $(X_1,X_2,X_3)$, because it would put all the mass on elements of $\{0,1\}^3$ whose components were all distinct. There are no such elements, as any combo of 3 zeros and ones must contain a duplicate.
Put another way, $P(X_1=1-X_2)=P(X_3=1-X_2)=1$ from which one might guess $P(X_1=X_3)=1$, but the posited margin for $(X_1,X_3)$ says no.