# Application of regularity theorem. Some doubts.

I quote Billingsley (1995):

Theorem (Regularity). Suppose that $$\mu$$ is a measure on $$\mathbb{R}^k$$ such that $$\mu(A) < \infty$$ if $$A$$ is bounded.
For $$A\in\mathbb{R}^k$$ and $$\varepsilon>0$$, there exists a closed $$C$$ and an open $$G$$ such that $$C \subset A \subset G$$ and $$\mu(G - C) < \varepsilon$$.

Given that, let us start by considering $$T=\{1,2\ldots\}$$, a probability space $$(\Omega,\mathcal{A}, P)$$ and a collection $$[X_t:t\in T]$$ of random variables - that is, a stochastic process - on it. For each $$k-$$tuple $$(t_1,\ldots,t_k)$$ of distinct elements of $$T$$, the random vector $$(X_{t_1},\ldots X_{t_k})$$ has over $$\mathbb{R}^k$$ some distribution $$\mu_{t_1\cdots t_k}$$: $$\mu_{t_1\cdots t_k}(H)=P[(X_{t_1},\ldots,X_{t_k})\in H],\hspace{2cm}H\in\mathbb{R}^k\tag{1}$$ Consider now the following set (which is a cylinder): $$A_n=[X\in\mathbb{R}^T:(X_{t_1},\ldots,X_{})\in H_n],\hspace{2cm}H_n\in\mathbb{R}^n\tag{2}$$ We have that $$P(A_n)=\mu_{t_1,\ldots,t_n}(H_n)$$. My doubts concern the following part:

By Theorem (Regularity), there exists inside $$H_n$$ a compact set $$K_n$$ such that $$\mu_{t_1,\ldots,t_n}(H_n-K_n)<\displaystyle{\frac{\varepsilon}{2^{n+1}}}$$.
If $$B_n=[X\in\mathbb{R^T}: (X_{t_1},\ldots, X_{t_n})\in K_n]$$, then $$P(A_n-B_n)<\displaystyle{\frac{\varepsilon}{2^{n+1}}}$$.
Put $$C_n=\bigcap_{k=1}^{n}B_k$$. Then, $$C_n\subset B_n\subset A_n$$ and $$P(A_n-C_n)<\displaystyle{\frac{\varepsilon}{2}}$$

I was asking myself how Theorem (Regularity) is applied in the immediately above result. Specifically, my doubts are the following:

1. We know that $$K_n$$ is a compact set inside $$H_n$$. Hence - so as for Theorem (Regularity) to be appliable - is a compact set closed by definition?
2. Is the quantity $$\displaystyle{\frac{\varepsilon}{2^{n+1}}}$$ arbitrary small or is there some reason for it to be exactly that way?
3. Why if we choose $$C_n=\bigcap_{k=1}^{n}B_k$$, it follows that $$P(A_n-C_n)<\color{red}{\displaystyle{\frac{\varepsilon}{2}}}$$?
In general, I understand that, given the definition of $$C_n$$, $$P(A_n-C_n)>P(A_n-B_n)$$, but I cannot understand why it $$\color{red}{\text{exactly}}$$ (see the $$\color{red}{\text{red r.h.s. of the inequaility}}$$) holds that $$P(A_n-C_n)<\color{red}{\displaystyle{\frac{\varepsilon}{2}}}$$

Compact subset of Hausdorff spaces are closed. So compact subsets of $$\mathbb R^{n}$$ are certainly closed.
$$\frac {\epsilon } {2^{n+1}}$$ is chosen because $$\sum_{n=1}^{\infty} \frac {\epsilon } {2^{n+1}} =\frac{\epsilon}{2}$$.
$$P(A_n\setminus C_n)=P(\cup_{k \leq n} (A_n\setminus C_k))$$ $$\leq P(\cup_{k \leq n} (A_k\setminus B_k))$$ $$\leq \sum_{k \leq n} P(A_k\setminus B_k)$$ $$\leq \sum_{k=1}^{\infty} P(A_k\setminus B_k) <\frac{\epsilon}{2}$$
• Thank you a lot! Could you please just give me a hint to understand why compact subsets of $\mathbb{R}^n$ are certainly closed, given that compact subset of Hausdorff spaces are closed? @KaviRamaMurthy Oct 1, 2020 at 13:32