On Independence and Determination of Probability Measures

Abstract

We show that two probability measures defined on the same measure space having the same pairs of independent events are either purely atomic or equal. In the former case, either measures are trivial (taking values in $\{0,1\}$) and singular, or equivalent. We also characterize nonatomic measures having the same events independent of a fixed (nontrivial) event and provide sufficient condition on pairs on independent intervals to determine a probability measure on the real line.

Appendix

 Proof of Lemma 1

First, we observe that any event $A$ with $P(A)>0$ has an event with arbitrarily small but strictly positive measure. Indeed, let such $A$ be given. Then, since $P$ is nonatomic, there exists $A_1\subset A$ with $P(A_1) , P(A-A_1) >0$. Without loss of generality, $P(A_1) \le P(A) /2$. Continue inductively. Given $A_n \subset A$ with $0< P(A_n) \le P(A)/2^n$, there exists $A_{n+1}\subset A_n$, with $0<P(A_{n+1}) < P(A_n)/2$.

We turn to the main claim. Fix $\alpha \in (0,1)$, and let $A_0$ be such that $P(A_0) \in (0, \alpha )$. Continue inductively, assuming $P(A_n) \in (0,\alpha )$. Let $I_n = \{ B \subset A_n^c: P(B) \le \alpha - P(A_n)\}$. Choose $B_{n+1} \in I_n$ that satisfies $P(B_n) \ge (1- \frac{1}{n} ) \sup _{B \in I_n} P(B)$, and let $A_{n+1} = A_n \cup B_{n+1}$. Note that the events $(B_n:n\in \mathbb{N })$ are disjoint. Therefore, $P(B_n) \rightarrow 0$. Consequently, $\sup _{B \in I_n} P(B) \rightarrow 0$. Let $A_\infty = \cup _{n\in \mathbb{N }} A_n$. We continue the proof arguing by contradiction, assuming that $P(A_\infty ) < \alpha$. Then, by the first paragraph, there exists $B_\infty \subset A_\infty ^c$ satisfying $0<P(B_\infty )<\alpha - P(A_\infty )$. Since $A_\infty ^c \subset A_n^c$ for all $n\in \mathbb{N }$, it follows that $B_\infty \in I_n$ for all $n$. But this contradicts the fact that $\sup _{B\in I_n} P(B) \rightarrow 0$. $\square$

Proof of Corollary B

Let $\epsilon = \min (\alpha ,1-\alpha )$. It is enough to show that if $P(A)=P(B) \le \epsilon$, then $Q(A)=Q(B)$. Without loss of generality, $A \cap B= \emptyset$; otherwise, we will consider the pair $A-(A\cap B)$ and $B-(A\cap B)$ instead of $A$ and $B$. Since $A \subset B^c$ and $P(B^c) \ge \alpha$, by Lemma 1, there exists an event $L\subset B^c$, with $P(L)=\alpha$ and $A\subset L$. But since $P((L-A) \cup B)=P(L) - P(A) + P(B) = P(L)= \alpha$, it follows that $Q((L-A) \cup B) =\alpha$. However, $Q((L-A) \cup B) = Q(L) - Q(A) + Q(B) = \alpha - Q(A) +Q(B)$. Thus, $Q(A)=Q(B)$. $\square$