Tight Probability Measures
Definition
Let the space $S$ be a metric space $( S , \rho)$ and a measurable space $(S,\mathcal{B}(S))$.
Let $P$ be a probability measure defined on $S$. It is said to be tight if for all $\varepsilon > 0$, there exists a compact set $K$ such that $P(K) > 1 - \varepsilon$ is satisfied.
Explanation
Generally, in undergraduate level probability, one rarely encounters a probability measure that is not tight. For instance, if there is a probability measure $P_{X}$ induced by a random variable $X$ following a normal distribution, then regardless of what $\varepsilon>0$ is, there must exist a bounded closed set $K$ that satisfies $P(K) > 1 - \varepsilon$, and according to the Heine-Borel theorem, $K$ is compact, showing that $P_{X}$ is tight. In fact, all probability measures induced by random variables defined on $\mathbb{R}$ are tight.
The reason behind considering the concept of tightness naturally arises from the convenience of dealing with compact sets. Saying that $K$ is compact implies that it can be thought of as being partitioned into a finite open cover.
The following theorem guarantees that for whatever $A$ may be, there exists a sequence of compact sets $\left\{ K_n \right\}_{n \in \mathbb{N}}$ such that $P\left( K_{n} \right) \to P(A)$. Considering how compact sets can conveniently be partitioned finitely, one cannot help but appreciate the condition of being tight.
Theorem
If $P$ is tight $\iff$ for all $A \in \mathcal{B}(S)$ $\displaystyle P(A) = \sup_{ K : \text{compact set}} \left\{ P(K) : K \subset A \right\}$
Proof
Let’s suppose $P$ is a probability defined on $(S,\mathcal{B}(S))$. Then, for all $A \in \mathcal{B}(S)$ and $\varepsilon>0$, there exist a closed set $F_{\varepsilon}$ and an open set $G_{\varepsilon}$ satisfying the following. $$ F_{\varepsilon}\subset A \subset G_{\varepsilon} \\ P ( G_{\varepsilon} \setminus F_{\varepsilon}) < \varepsilon $$ According to the above-mentioned property, there exists a closed set $F_{\varepsilon} \subset A$ satisfying $P(A) - P (F_{\varepsilon}) < \varepsilon$. Moreover, since $P$ is tight, there exists a compact set $K$ satisfying $P(K) > 1 - \varepsilon \iff P(K^{c}) < \varepsilon$. $$ P(A) \le P(F_{\varepsilon}) + \varepsilon $$ Now, if we partition $F_{\varepsilon}$ as follows: $$ P(A) \le P \left( F_{\varepsilon} \cap K \right) + P \left( F \setminus K \right) + \varepsilon $$ Then, $$ \begin{align*} P \left( F \setminus K \right) =& P \left( \Omega \setminus K \right) \\ =& P(K^{c}) \\ <& \varepsilon \end{align*} $$ Therefore, $$ P(A) \le P \left( F_{\varepsilon} \cap K \right) + 2 \varepsilon $$
Properties of Compact Sets: If a subset $F$ of a compact set $K$ is a closed set, then $F$ is a compact set.
Thus, since $F_{\varepsilon} \cap K \subset K$ is a closed set, it is a compact set, and hence for all $\varepsilon>0$, there exists a compact set $F_{\varepsilon} \cap K$ satisfying the following: $$ P(A) \le P \left( F_{\varepsilon} \cap K \right) + 2 \varepsilon \\ F_{\varepsilon} \cap K \subset A $$ Therefore, for all $A \in \mathcal{B}(S)$ regarding $\displaystyle P(A) = \sup_{ K : \text{compact set}} \left\{ P(K) : K \subset A \right\} $$ (\Leftarrow) $$ \Omega \in \mathcal{B}(S)$ as well $$ 1 = P ( \Omega ) = \sup_{ K : \text{compact set}} \left\{ P(K) : K \subset \Omega \right\} $$ hence, regardless of what $\varepsilon > 0$ is, there exists a compact set $K$ satisfying $P(K) > 1 - \varepsilon$.
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