Price of Kolmogorov's extension theorem? From Wikipedia

an alternative way of stating Kolomogorov's extension theorem is that, provided that the above consistency conditions hold, there exists a (unique) measure $\nu$ on $(\mathbb{R}^n)^T$ with marginals $\nu_{t_{1} \dots t_{k}}$ for any finite collection of times $t_{1} \dots t_{k}$. The remarkable feature of Kolmogorov's extension theorem is that it does not require $T$ to be countable, but the price to pay for this level of generality is that the measure $\nu$ is only defined on the product σ-algebra of $(\mathbb{R}^n)^T$, which is not very rich.

I was wondering in what sense "the measure $\nu$ is only defined on the product σ-algebra of $(\mathbb{R}^n)^T$" is a price? $\nu$ not being able to be defined on some other sigma algebras?
What does "which is not very rich" mean precisely?
Thanks and regards!
 A: Here is a typical example: Let $T=[0,1]$, the unit interval and $\lambda$ Lebesgue measure, and take the coin-flipping measure on $\{0,1\}^T$. Let $f_x$ be the $x$-coordinate of $f\in\{0,1\}^T$. You might want to know the probability of the set $$B=\big\{f:\lambda\{x\in T:f_x=1\}=1/3\big\},$$
the probability that the fraction of $1$s is exactly one third. This event is not in the product $\sigma$-algebra. 
This follows from the general result that if $A\in\sigma(\mathcal{F})$, then there exists a countable family $\mathcal{C}\subseteq \mathcal{F}$ such that $A\in\sigma(\mathcal{C})$. One can show this by verifying that the family of sets generated by a countable sub-family of $\mathcal{F}$ forms a $\sigma$-algebra that contains $\mathcal{F}$.
Applied to our case, an event in the product $\sigma$-algebra must be generated by countably many coordinates. So take any $f\in B$ and let $H_f$ be the set of functions $g\in\{0,1\}^T$ such that $\{x:f_x\neq g_x\}$ is countable. Then $H_f$ intersects every nonempty set generated by countably many coordinates and has therefore outer measure $1$. Clearly, $H_f\subseteq B$, so $B$ has measure $1$ too. But by a similar argument, $B^C$ has outer measure $1$ too. It follows that $B$ is not measurable in the product $\sigma$-algebra. 
