I’m reading Tao’s article: https://terrytao.wordpress.com/tag/kolmogorov-extension-theorem/
I am stuck in the proof for Kolmogorov extension theorem (Theorem 62).
Let $\{((X_\alpha,\mathscr{B}_\alpha),\mathscr{F}_\alpha)\}_{\alpha\in A}$ be a family of measurable spaces $(X_\alpha,\mathscr{B}_\alpha)$ equipped with a topology $\mathscr{F}_\alpha$
For each finite $B\subset A$, let $\mu_B$ be an inner regular probability measure on $\mathscr{B}_B:=\prod_{\alpha\in B} \mathscr{B}_\alpha$ with the product topology $\mathscr{F}_B:=\prod_{\alpha\in B} \mathscr{F}_\alpha$. Assume that $(\pi_{C\leftarrow B})_*\mu_B=\mu_C$ whenever $C\subset B\subset A$ are two nested finite susbsets of $A$.
Let $\mathscr{B}_0$ be the set of all subsets of $X_A$ that are of the form $\pi_B^{-1}(E_B)$ for some finite $B\subset A$ and some $E_B\in \mathscr{B}_A$.
Define $\mu_0(E):=\mu_B(E_B)$ whenever $E$ takes the form $\pi^{-1}_B(E_B)$ for some finite $B\subset A$ and $E_B\in \mathscr{B}_B$. By the compatibility condition, this $\mu_0$ is well-defined.
Let $\{F_n\}$ be a decreasing sequence in $\mathscr{B}_0$ such that $\bigcap_n F_n=\emptyset$.
Suppose $\lim_{n\to\infty} \mu_0(F_n)>0$.
Then, there exists $\epsilon>0$ such that $\mu_0(F_n)>\epsilon$ for all $n$. As each $F_n$ lies in $\mathscr{B}_0$, we have $F_n=\pi^{-1}_{B_n}(G_n)$ for some finite sets $B_n\subset A$ and some $\mathscr{B}_{B_n}$-measurable sets $G_n$. By enlarging each $B_n$ as necessary we may assume that the $B_n$ are increasing in $n$. The decreasing nature of the $F_n$ then gives the inclusion $G_{n+1}\subset \pi^{-1}_{B_n\leftarrow B_{n+1}}(G_n)$.
By inner regularity, one can find a compact subset $K_n$ of each $G_n$ such that $\mu_{B_n}(K_n)\geq \mu_{B_n}(G_n)-\epsilon/2^{n+1}$.
Define $K_n’:= \bigcap_{m=1}^n \pi_{B_m\leftarrow B_n}^{-1}(K_m)$.
Then we see that each $K_n’$ is compact and $\mu_{B_n}(K_n’)\geq \mu_{B_n}(G_n)-\epsilon/2^n\geq \epsilon - \epsilon/2^n$
I do not get where this first inequality comes from. I also checked Tao’s book “An introduction to measure theory” and here he defines $K_n’$ as the union instead of intersection. However, this again does not make sense because $K_n’$ need not be compact in this case..
Note that $$\bigcap_{m=1}^n \pi_{B_m\leftarrow B_n}^{-1}(G_m) \setminus \bigcap_{m=1}^n \pi_{B_m\leftarrow B_n}^{-1}(K_m) \subset \bigcup_{m=1}^n \pi_{B_m\leftarrow B_n}^{-1}(G_m) \setminus \pi_{B_m\leftarrow B_n}^{-1}(K_m) $$
and that $\bigcap_{m=1}^n \pi_{B_m\leftarrow B_n}^{-1}(G_m)=G_n$.
Hence. $$\mu_{B_n}(G_n)-\mu_{B_n}(K_n)\leq \mu_{B_n} (\bigcup_{m=1}^n \pi_{B_m\leftarrow B_n}^{-1}(G_m) \setminus \pi_{B_m\leftarrow B_n}^{-1}(K_m) ) \leq \sum_{m=1}^n \epsilon/2^{m+1}$$.