Measure Theoretic Probability Part 3

Note that \(x \in \limsup_n A_n\) if and only if \(x\) lies in infinitely many of the \(A_n\). Also, \(x \in \liminf_n A_n\) is \(x\) lies in all but finitely many of the \(A_n\). Therefore, if \(x \in \liminf_n A_n\), then \(x \in \limsup_n A_n\) since if it lies in all but finitely many of the \(A_n\), it lies in infinitely many of the \(A_n\). Hence, \(\liminf_n A_n \subset \limsup_n A_n\). To check that \(A_n \rightarrow A\), we only have to show \(\limsup_n A_n \subset \liminf_n A_n\).

We first translate the notation in the book to something more familiar first:

\begin{align*} \limsup_n x_n &= \inf_{n \in \mathbb{N}} \sup_{k \ge n} x_k \\ \liminf_n x_n &= \sup_{n \in \mathbb{N}} \inf_{k \ge n} x_k \end{align*}

Suppose that \(\omega \in \limsup_n A_n := A\). Then, \(I_A(\omega) = 1\). Also, \(\omega\) is in infinitely many of the \(A_n\)'s. That is, \(I_{A_n}(\omega) = 1\) for infinitely many \(n \in \mathbb{N}\). In other words, for every \(n \in \mathbb{N}\), there exists \(k \ge n\) so that \(I_{A_k}(\omega) = 1\). That is, \(\sup_{k \ge n} I_{A_k}(\omega) = 1\) since the range of \(I_{A_k}\) is \(\{0, 1\}\). Therefore, \(\limsup_n I_{A_n}(\omega) = 1\). Now, suppose that \(I_A(\omega) = 0\). Then, for some \(n \in \mathbb{N}\), \(x \not\in A_k\) for every \(k \ge n\). That is, \(I_{A_k}(\omega) = 0\). Then, \(\inf_{n \in \mathbb{N}} \sup_{k \ge n} I_{A_k}(\omega) = 0\), implying that \(\limsup_n I_{A_n}(\omega) = 0\). This shows that \(I_A = I_{\limsup_n A_n} = \limsup_n I_{A_n}\). We now similarly prove that \(I_A = \liminf_n I_{A_n}\) where \(A = \liminf_n A_n\). Suppose \(\omega \in A\). Then, \(I_A(\omega) = 1\). Also, \(\omega \in A_n\) for all but finitely many \(n \in \mathbb{N}\). Then, for all but finitely many \(n \in \mathbb{N}\), \(I_{A_n}(\omega) = 1\). That is, for every \(n \in \mathbb{N}\), there exists \(k \in \mathbb{N}\) so that \(I_{A_k}(\omega) = 1\) for all \(k \ge m \ge n\). Consequently, for any particular \(n \in \mathbb{N}\), \(\inf_{k \ge n} I_{A_k}(\omega) = 1\). Then, \(\liminf_n I_{A_n}(\omega) = 1\). Now suppose \(\omega \not\in A\). Then, \(I_A(\omega) = 0\). There exists \(N \in \mathbb{N}\) so that \(\omega \not\in A_n\) for all \(n \ge N\). That is, \(I_{A_n}(\omega) = 0\) for all \(n \ge N\). Consider any fixed \(n \in \mathbb{N}\). If \(n \ge N\), then \(\inf_{k \ge n} I_{A_k}(\omega) = 0\) because \(I_{A_k}(\omega) = 0\) for all \(k \ge n \ge N\). If \(n < N\), then still \(\inf_{k \ge n} I_{A_k}(\omega) = 0\). So, \(I_{A_k}(\omega) = 0\) for all \(k \ge N > n\). Therefore, \(\liminf_n I_{A_n}(\omega) = 0\). So, \(I_A = \liminf_n I_{A_n}\).

Suppose \(\limsup_n A_n = \liminf A_n\). Let \(\omega\) be arbitrary. We show \(\limsup_n I_{A_n}(\omega) = \liminf_n I_{A_n}(\omega)\). Based on what we just prove, this is equivalent to showing that \(I_{\limsup_n A_n}(\omega) = I_{\liminf_n A_n}(\omega)\). Since \(\limsup_n A_n = \liminf_n A_n\), this is true, and \(\limsup_n I_{A_n}(\omega) = \liminf_n I_{A_n}(\omega)\) for every \(\omega\), so \(\lim_n I_{A_n}(\omega)\) exists for every \(omega\). Now, suppose \(\lim_n I_{A_n}(\omega)\) exists for every \(\omega\). Then, \(\limsup_n I_{A_n}(\omega) = \liminf_n I_{A_n}(\omega)\) for every \(\omega\). By what we just prove, this is equivalent to saying \(I_{\limsup_n A_n}(\omega) = I_{\liminf_n A_n}(\omega)\). This would mean that \(\omega \in \limsup_n A_n\) if and only if \(\omega \in \liminf_n A_n\). Then, \(\limsup_n A_n \subset \liminf_n A_n\), so \(\limsup_n A_n = \liminf_n A_n\).