Question about Kolmogorov extension theorem I need some help understanding the relationship between the following two theorems

Theorem 1: Let $\{\mu_n\}$ be a sequence of probability measures on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$, where $\mathcal{B}(\mathbb{R})$ denotes the Borel sets. Then, there exists a probability space $(\Omega,\mathcal{F},\mathbb{P})$ and a sequence of independent random variables $\{X_n\}$ such that $\mathbb{P}(X_n \in B) = \mu_n(B), n\geq 1$, where $B$ is Borel set of $\mathbb{R}$

And a version of Kolmogorov Extension Theorem is given as

Theorem 2: For every $n$, let $\mu^n$ be a probability measure on $(\mathbb{R}^n,\mathcal{B}(\mathbb{R}^n)$. For $1\leq m \leq n$, let $\Pi_{m,n}$ be the projetion map given as:
  $$ \Pi_{m,n}: B \in \mathcal{B}(\mathbb{R}^m) \to \Pi_{m,n}(B) \in \mathcal{B}(\mathbb{R}^n) $$
  $$\Pi_{m,n}(B) = \{(x_1,x_2,\dots,x_n) \in \mathbb{R}^n : (x_1,x_2,\dots,x_m) \in B \}$$ Suppose $\mu^n$ satisfies that $\forall n \geq 1, \forall 1 \leq m \leq n,$ $\mu^n \circ \Pi_{m,n} = \mu^m$. Then, there exists a probability space $(\Omega,\mathcal{F},\mathbb{P})$ such that $\mathbb{P}((X_1,X_2,\dots,X_n) \in B) = \mu^n(B)$, where $B$ is a Borel set in $R^n$.
    *Remark:  Theorem 1 is a special case of Theorem 2 when $\mu^n = \prod_{i=1}^n \mu_i$

I do not understand the remark in the Theorem 2. Why is Theorem 1 a special case of Theorem 2? I am confused since the Theorem 1 shows the existence of an independent sequence of random variables whereas the Theorem 2 does not say whether a sequence is independent. Is there something that I am missing?
 A: First of all, you can check that the sequence of probability measures $\mu^n = \prod_{i=1}^n \mu_i$ satisfies the criteria of Theorem 2. Thus the existence of a probability space $(\Omega,\mathcal{F},\mathbb{P})$ with $\mathbb{P}((X_1,X_2,\dots,X_n) \in B) = \mu^n(B)$ for borel sets $B$ follows from that theorem.
This in turn gives you the existence of random variables $X_i$ which essentially are projections of $(\Omega,\mathcal{F})$ into $(\mathbb{R},\mathcal{B}(\mathbb{R}))$. Now you just need to show their independence. Recall that it is sufficient to check that all finite subsets of these random variables are independent.
Thus pick $k$ different random variables $X_{i_1},\dotsc, X_{i_k}$. And also pick $n \geq i_j \;\forall j \; \in \{1,\dotsc,k\}$. Finally denote by $\widetilde{X}$ the random vector in $\mathbb R^{n-k}$  consisting of the $X_k$ with indices in $\{1,\dotsc,n\}\setminus\{j_1,\dotsc,j_k\}$. For the same indices let $B_k = \mathbb R$.  Finally pick real borel sets $B_{i_1}, \dotsc, B_{i_k}$.
We get for their joint probability distribution:
$$
\begin{aligned}
&\mathbb{P}\left[(X_{i_1},\dotsc,X_{i_k}) \in B_{i_1}\times \dotsc \times B_{i_k}\right]  \\
=\; & \mathbb{P}\left[(X_{i_1},\dotsc,X_{i_k}) \in B_{i_1}\times \dotsc \times B_{i_k}, \widetilde{X} \in \mathbb R^{n-k}\right] \\
= \; &\mathbb{P}\left[(X_1,\dotsc,X_n) \in \prod_{i=1}^n B_i\right]\\
= \;& \mu^n\left(\prod_{i=1}^n B_i\right)\\
= \;& \prod_{i=1}^n \mu_i(B_i)\\
= \;& \prod_{j=1}^k \mu_{i_j}(B_{i_j})\\
= \;& \prod_{j=1}^k \mathbb{P}[X_{i_j} \in B_{i_j}]
\end{aligned}
$$
The second from last equality follows because $\mu_i(\mathbb R)=1 \; \forall \; i$. Thus we have shown the independence of the sequence of random variables $(X_i)_{i\in\mathbb N}$.
A: If in Theorem 2, we choose $\mu^n$ to be the product measure $\prod_{i=1}^n\mu_i$, where the $\mu_i$'s are the measures on $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$ from Theorem 1, then Theorem 2 tells you that you find a space $(\Omega, \mathcal{F}, \mathbb{P})$ and random variables $X_1, X_2, \dots, X_n$ which are independent. Independence follows from the definition of the product measure. If the joint distribution of some random variables is a product measure, then they are independent. They also have the right distribution, since the i-th marginal of $\mu^n$ is $\mu_i$.
However in this form, Theorem 2 doesn't give you a sequence just a finite number of random variables. But there is a more general version of Theorem 2, such that it exactly implies Theorem 1, see Kolmogorov extension theorem.
