Conditions for Two Probability Measures to Coincide
Theorem
Let space $S$ be a metric space $( S , \rho)$ and a measurable space $(S,\mathcal{B}(S))$.
$\mathcal{O}$ is the set of all open sets, $\mathcal{C}$ is the set of all closed sets, and $P$ and $Q$ are probability measures defined in $(S,\mathcal{B}(S))$.
- [1]: For every open set $O \in \mathcal{O} \subset S$, if $P(O) = Q(O)$ then $P=Q$. In other words, $\mathcal{O}$ is a separating class.
- [2]: For every closed set $C \in \mathcal{C} \subset S$, if $P(C) = Q(C)$ then $P=Q$. In other words, $\mathcal{C}$ is a separating class.
- [3]: If for every $f \in C_{b}(S)$, $Pf = Qf$ implies $P(A)=Q(A)$ for every $A \in \mathcal{B}(S)$, then $P=Q$.
- If for every $A \in \mathcal{B}(S)$ $P(A)=Q(A)$ then it is said to be $P=Q$.
- $C_{b}(S)$ denotes the set of bounded continuous functions defined on $S$. $$ C_{b}(S) := \left\{ f:S \to \mathbb{R} \mid f\text{ is bounded and continuous} \right\} $$
- $\displaystyle \int_{S} f dP$ is simply denoted as $\displaystyle Pf := \int_{S} f dP$.
Notation
Additionally, for the argument, the following notations are introduced:
- For an element $x \in S$ and subsets $A \subset S$, and $\delta >0$: $$ \rho (x, A) := \inf \left\{ \rho (x,a) : a \in A \right\} \\ A^{\delta} := \left\{ x \in S : \rho (x, A) < \delta \right\} $$
- For some fixed $F \subset S$: $$ \begin{align*} f_{\delta}(x) :=& \left( 1 - \rho (x, F) / \delta \right)^{+} \\ =& \begin{cases} 1 &, x \in F \\ 1 - \rho (x,F)/\delta &, x \notin F \land x \in F^{\delta} \\ 0 &, x \notin F^{\delta} \end{cases} \end{align*} $$
Although the formula looks complicated, it is not difficult at all when you see it as a diagram. $\rho (x,A)$ just represents the shortest distance between the point $x \in S$ and the set $A \subset S$.
$A^{\delta}$ is just an open set that has increased by $\delta$ from $A$.
The most challenging one is $f_{\delta}(x)$, but as shown in the figure below, it’s a function whose value is only $1$ at $F$ and continuous values between $0$ and $1$ at places close by $\delta$, and elsewhere, it is $0$. You can guess why this function was created by looking at its shape. It has to be $f_{\delta} \in C_{b}(S)$ because it is bounded and uniformly continuous. Also, it is a good thing that it converges to an easy-to-handle simple function as $\delta \to 0$ when $f_{\delta} (x) \to 1_{F}$.
Proof
Strategy[1], [2]: Given that $\mathcal{O}$ and $\mathcal{C}$ are sets that collect all open and closed sets respectively, showing that they are a pi-system is very easy. Then, using the following property to prove that it becomes a separating class is enough.
Conditions for becoming a separating class and Pi system: If $\mathcal{C}$ holds for $\sigma (\mathcal{C}) = \mathcal{B}(S)$ and for every $A \in \mathcal{C}$ if $P(A) = Q(A)$, then $\mathcal{C}$ is a separating class.
[1]
Since the intersection of open sets is an open set, if $A, B \in \mathcal{O}$ then $A \cap B \in \mathcal{O}$, meaning $\mathcal{O}$ is a pi-system. The Borel sigma field $\mathcal{B}(S)$ being the smallest sigma field that includes all open sets of the metric space $(S, \rho)$, i.e., all elements of $\mathcal{O}$, makes $\sigma (\mathcal{O}) = \mathcal{B} (S)$. Hence, if for every open set $O \in \mathcal{O}$ $P(O) = Q(O)$ is satisfied, then $\mathcal{O}$ becomes a separating class.
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[2]
Since the intersection of closed sets is a closed set, if $A, B \in \mathcal{C}$ then $A \cap B \in \mathcal{C}$, meaning $\mathcal{C}$ is a pi-system. By the definition of closed sets and if $C \in \mathcal{C}$ then $S \setminus C \in \mathcal{O}$, and according to the definition of sigma field, it must be that $C \in \mathcal{B}(S)$. That is $\sigma (\mathcal{O}) = \sigma (\mathcal{C})= \mathcal{B}(S)$. Hence, if for every closed set $C \in \mathcal{C}$ $P(C) = Q(C)$ is satisfied, then $\mathcal{C}$ becomes a separating class.
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[3]
$$ f_{\varepsilon}(x) := \left( 1 - \rho (x,F) / \varepsilon \right)^{+} $$ Let us define $f_{\varepsilon} : S \to \mathbb{R}$ for closed sets $F \in \mathcal{C}$ and $\varepsilon > 0$ as above. Then $f_{\varepsilon}$ satisfies the following two conditions. $$ f_{\varepsilon} \in C_{b}(S) \\ 1_{F}(x) \le f_{\varepsilon}(x) \le 1_{F^{\varepsilon}} (x) $$ If for every $f \in C_{b}(S)$ $P f = Q f$ is assumed, then $\displaystyle \int_{\Omega} f_{\varepsilon} dP = \int_{\Omega} f_{\varepsilon} dQ$ is also met. $$ \begin{align*} P(F) =& \int_{\Omega} 1_{F} dP \\ \le & \int_{\Omega} f_{\varepsilon} dP \\ =& \int_{\Omega} f_{\varepsilon} dQ \\ \le & \int_{\Omega} 1_{F^{\varepsilon}} dQ \\ =& Q(F^{\varepsilon}) \end{align*} $$ Summarizing, $P(F) \le Q ( F^{\varepsilon})$ is obtained. Since $F$ is a closed set, $\displaystyle \lim_{\varepsilon \to 0} F^{\varepsilon} = F$, and if we apply $\displaystyle \lim_{\varepsilon \to 0}$ to both sides of the recently derived formula, according to the continuity of measure, $$ \begin{align*} P(F) =& \lim_{\varepsilon \to 0} P(F) \\ \le & \lim_{\varepsilon \to 0} Q (F^{\varepsilon}) \\ =& Q \left( \lim_{\varepsilon \to 0} F^{\varepsilon} \right) \\ =& Q(F) \end{align*} $$ we get $P(F) \le Q(F)$, and in the same way, it can be shown that $Q(F) \le P(F)$ holds. Hence, for the closed set $F$, if $P(F) = Q(F)$ is met, according to [2], for every $A \in \mathcal{B}(S)$ $P(A) = Q(A)$ holds.
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