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 $P(A_n-C_n)<\color{red}{\displaystyle{\frac{\varepsilon}{2}}}$

1 Answer 1


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 $\sum_{n=1}^{\infty} \frac {\epsilon } {2^{n+1}} =\frac{\epsilon}{2}$.

$$P(A_n\setminus C_n)=P(\cup_{k \leq n} (A_n\setminus C_k))$$ $$\leq P(\cup_{k \leq n} (A_k\setminus B_k))$$ $$ \leq \sum_{k \leq n} P(A_k\setminus B_k)$$ $$\leq \sum_{k=1}^{\infty} P(A_k\setminus B_k) <\frac{\epsilon}{2}$$

  • $\begingroup$ 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 $\endgroup$ Oct 1, 2020 at 13:32
  • 1
    $\begingroup$ Any metric space is Hausdorff. $\endgroup$ Oct 1, 2020 at 14:15

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