📂Probability Theory

Probability Measures Defined on Polish Spaces are Tight

Theorem

Let’s say that a metric space $(S,\rho)$ is a Polish space. Then, all probability measures defined in $S$ are tight.

Explanation

A Polish space refers to a separable complete metric space. The reason we discuss the tightness of probability measures is precisely that, under these conditions, most probabilities are tight. This, conversely, implies the need to study probabilities defined in non-Polish spaces.

Proof

Strategy: We need to bring in several theorems from topology to follow the definition of tight probability measures.

Given that the space $S$ is both a metric space $( S , \rho)$ and a measurable space $(S,\mathcal{B}(S))$ according to the definition of tight probability measures.

Let $P$ be a probability measure defined in $S$. If there exists a compact set for all $\varepsilon > 0$ such that $P(K) > 1 - \varepsilon$, then $P$ is considered tight.

Furthermore, when distance $\rho$ is given, an open ball centered at $x_{0}$ with radius $\varepsilon$ is represented as $B_{\rho} ( x_{0} ; \varepsilon )$, and the closed ball as $B_{\rho} [ x_{0} ; \varepsilon]$.

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Since $S$ is a separable space, a dense countable set $D:= \left\{ a_{1}, a_{2}, \cdots \right\}$ of $S$ can be chosen. For all $\delta > 0$, $\displaystyle \bigcup_{k=1}^{\infty} B_{\rho} \left( a_{k} ; \delta \right) = S$ holds, and by the continuity of measure, $\displaystyle P(S) = \lim_{n \to \infty} P \left( \bigcup_{k=1}^{n} B_{\rho} \left( a_{k} ; \delta \right) \right)$ follows. Now, if we set $\varepsilon>0$, for all $m \in \mathbb{N}$ there exists $n_{m}$ satisfying the following: $$P \left( \bigcup_{k=1}^{n_{m}} B_{\rho} \left( a_{k} ; {{ 1 } \over { m }} \right) \right) > P(S) - 2^{-m}\varepsilon$$ Let’s define $K \subset S$ as follows: $$K := \bigcap_{m=1}^{\infty} \bigcup_{k=1}^{n_{m}} B_{\rho} \left[ a_{k} ; {{ 1 } \over { m }} \right]$$ $K$ is a set obtained by taking an infinite intersection of a finite union of closed balls, thus $K$ is a closed set in $S$, and by selecting $m$ such that for all $\delta$, $m > 1 / \delta$ holds, $$K \subset \bigcup_{k=1}^{n_{m}} B_{\rho} \left[ a_{k} ; {{ 1 } \over { m }} \right] \subset \bigcup_{k=1}^{n_{m}} B_{\rho} \left( a_{k} ; \delta \right)$$ Hence, $K$ is a totally bounded space.

Properties of Complete Metric Spaces: Since $K$ is a totally bounded space $\iff$ $X$, the closed set $K$ is compact

Therefore, the closed set $K$ is compact, and according to the property of the probability measure $P$, $$\begin{align*} P(S \setminus K) =& P \left( \bigcup_{m=1}^{\infty} \left( S \setminus \bigcup_{k=1}^{n_{m}} B_{\rho} \left[ a_{k} ; {{ 1 } \over { m }} \right] \right) \right) \\ =& \sum_{m=1}^{\infty} P \left( S \setminus \bigcup_{k=1}^{n_{m}} B_{\rho} \left[ a_{k} ; {{ 1 } \over { m }} \right] \right) \\ =& \sum_{m=1}^{\infty} \left[ P \left( S \right) - P \left( \bigcup_{k=1}^{n_{m}} B_{\rho} \left[ a_{k} ; {{ 1 } \over { m }} \right] \right) \right] \\ <& \sum_{m=1}^{\infty} 2^{-m} \varepsilon \\ =& \varepsilon \end{align*}$$

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