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 \(\limsup_n A_n = \liminf_n A_n\).

Part A:

\begin{align*} \limsup_n (A_n \cap B_n) &= \bigcap_{n=1}^\infty \bigcup_{k = n}^\infty (A_k \cap B_k) \\ &\subset \bigcap_{n=1}^\infty \Bigl(\bigcup_{k=n}^\infty A_k \cap \bigcup_{k=n}^\infty B_k\Bigl) \\ &= \Bigl(\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k\Bigl) \cap \Bigl(\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty B_k\Bigl) \\ &= (\limsup_n A_n) \cap (\limsup_n B_n) \end{align*} \begin{align*} (\limsup_n A_n) \cup (\limsup_n B_n) &= \Bigl(\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k\Bigl) \cup \Bigl(\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty B_k\Bigl) \\ &= \bigcap_{n=1}^\infty \Bigl(\bigcup_{k=n}^\infty A_k \cup \bigcup_{k=n}^\infty B_k \Bigl) \\ &= \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty \Bigl(A_k \cup B_k) \\ &= \limsup_n (A_n \cup B_n) \end{align*} \begin{align*} (\liminf_n A_n) \cap (\liminf_n B_n) &= \Bigl(\bigcup_{n=1}^\infty \bigcap_{k=n}^\infty A_k\Bigl) \cap \Bigl(\bigcup_{n=1}^\infty \bigcup_{k=n}^\infty B_k\Bigl) \\ &= \bigcup_{n=1}^\infty \Bigl(\bigcap_{k=n}^\infty A_k \cap \bigcap_{k=n}^\infty B_k\Bigl) \\ &= \bigcup_{n=1}^\infty \bigcap_{k=n}^\infty \Bigl(A_k \cap B_k \Bigl) \\ &= \liminf_n (A_n \cap B_n) \end{align*} \begin{align*} (\liminf_n A_n) \cup (\liminf_n B_n) &= \Bigl(\bigcup_{n=1}^\infty \bigcap_{k=n}^\infty A_k\Bigl) \cup \Bigl(\bigcup_{n=1}^\infty \bigcap_{k=n}^\infty B_k\Bigl) \\ &\subset \bigcup_{n=1}^\infty \Bigl(\bigcap_{k=n}^\infty A_k \cup \bigcap_{k=n}^\infty B_k\Bigl) \\ &\subset \bigcup_{n=1}^\infty \bigcap_{k=n}^\infty \Bigl(A_k \cup B_k\Bigl) \\ &= \liminf_n (A_n \cup B_n) \end{align*}

For the examples: Let \(A_n = \{x\}\) if \(n\) is even and \(A = \emptyset\) when \(n\) is odd. Let \(B_n = \emptyset\) when \(n\) is even and \(B_n = \{x\}\) when \(n\) is odd. Then, for each \(n \in \mathbb{N}\), \(A_n \cap B_n = \emptyset\). So, \(\limsup_n (A_n \cap B_n) = \emptyset\). However, \(\limsup_n A_n = \limsup_n B_n = \{x\}\). So, \((\limsup_n A_n) \cap (\limsup_n B_n) = \{x\}\), which properly contains \(\emptyset = \limsup_n (A_n \cap B_n)\). Also, \(\liminf_n (A_n \cup B_n) = \{x\}\). However, \(\liminf_n A_n = \liminf_n B_n = \emptyset\). So, \((\liminf_n A_n) \cup (\liminf_n B_n) = \emptyset \cup \emptyset = \emptyset \subset \liminf_n (A_n \cup B_n) = \{x\}\).

Part B: TODO

Part C:

\begin{align*} (\liminf_n A_n)^c = \Bigl(\bigcup_{n=1}^\infty \bigcap_{k=n}^\infty A_k\Bigl)^c = \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k^c = \limsup_n A_n^c \\ (\limsup_n A_n)^c = \Bigl(\bigcap_{n=1}^\infty \bigcup_{n=1}^\infty A_k\Bigl)^c = \bigcup_{n=1}^\infty \bigcap_{k=n}^\infty A_k^c = \liminf_n A_n^c \end{align*} \begin{align*} \limsup_n (A_n \cap A_{n+1}^c) &= \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty (A_k \cap A_{k+1}^c) \\ &= \bigcap_{n=1}^\infty \Bigl[\Bigl(\bigcup_{k=n}^\infty A_k\Bigl) \cap \Bigl(\bigcup_{k=n}^\infty A_{k+1}^c\Bigl)\Bigl] \\ &= \bigcap_{n=1}^\infty \Bigl[\Bigl(\bigcup_{k=n}^\infty A_k\Bigl) \cap \Bigl(\bigcup_{k=n+1}^\infty A_{k}^c\Bigl)\Bigl] \\ &= \Bigl[\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k\Bigl] \cap \Bigl[\bigcap_{n=1}^\infty \bigcup_{k=n+1}^\infty A_k^c\Bigl] \\ &= \Bigl[\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k\Bigl] \cap \Bigl[\bigcap_{n=1}^\infty \Bigl(\Bigl\{\bigcup_{k=n}^\infty A_k^c\Bigl\} \setminus A_n^c\Bigl) \Bigl] \\ &= \Bigl[\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k\Bigl] \cap \Bigl[\bigcap_{n=1}^\infty \Bigl(\Bigl(\bigcup_{k=n}^\infty A_k^c\Bigl) \cap A_n\Bigl) \Bigl] \\ &= \Bigl[\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k\Bigl] \cap \Bigl[\bigcap_{n=1}^\infty \Bigl(\bigcup_{k=n}^\infty A_k^c\Bigl) \Bigl] \cap \bigcap_{n=1}^\infty A_n \\ &= \Bigl(\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k\Bigl) \cap \Bigl(\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k^c\Bigl) \\ &= \Bigl(\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k\Bigl) \setminus \Bigl(\bigcup_{n=1}^\infty \bigcap_{k=n}^\infty A_k\Bigl) \\ &= \limsup_n A_n - \liminf_n A_n \end{align*} \begin{align*} \limsup_n (A_n^c \cap A_{n+1}) &= \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty (A_k^c \cap A_{k+1}) \\ &= \bigcap_{n=1}^\infty \Bigl[\Bigl(\bigcup_{k=n}^\infty A_k^c\Bigl) \cup \Bigl(\bigcup_{k=n}^\infty A_{k+1}\Bigl)\Bigl] \\ &= \bigcap_{n=1}^\infty \Bigl[\Bigl(\bigcup_{k=n}^\infty A_k^c\Bigl) \cup \Bigl(\bigcup_{k=n+1}^\infty A_{k}\Bigl)\Bigl] \\ &= \Bigl[\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k^c\Bigl] \cap \Bigl[\bigcap_{n=1}^\infty \bigcup_{k=n+1}^\infty A_k\Bigl] \\ &= \Bigl[\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k^c\Bigl] \cap \Bigl[\bigcap_{n=1}^\infty \Bigl\{\Bigl(\bigcup_{k=n}^\infty A_k\Bigl) \setminus A_n\Bigl\}\Bigl] \\ &= \Bigl[\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k^c\Bigl] \cap \Bigl[\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k\Bigl] \cap \bigcap_{n=1}^\infty A_n^c \\ &= \Bigl[\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k^c\Bigl] \cap \Bigl[\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k\Bigl] \\ &= \Bigl(\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k\Bigl) \setminus \Bigl(\bigcup_{n=1}^\infty \bigcap_{k=n}^\infty A_k \Bigl) \\ &= \limsup_n A_n - \liminf_n A_n \end{align*}

Part D: If \(A_n \rightarrow A\), then \(\limsup_n A_n = \liminf_n A_n = A\). If \(B_n \rightarrow B\), then \(\limsup_n B_n = \liminf_n B_n = B\). We know from part (a) that \(\limsup_n (A_n \cup B_n) = (\limsup_n A_n) \cup (\limsup_n B_n) = A \cup B\). However, this also means that \(\limsup_n (A_n \cup B_n) = (\limsup_n A_n) \cup (\limsup_n B_n) = (\liminf_n A_n) \cup (\liminf_n B_n) \subset \liminf_n (A_n \cup B_n)\). Since we have shown that \(\limsup_n (A_n \cup B_n) \subset \liminf_n (A_n \cup B_n)\), it follows that \(A \cup B = \limsup_n (A_n \cup B_n) = \liminf_n (A_n \cup B_n)\). Hence, if \(A_n \rightarrow A\) and \(B_n \rightarrow B\), then \(A_n \cup B_n \rightarrow A \cup B\). To prove the case for intersections, we show that: if \(D_n \rightarrow D\), then \(D_n^c \rightarrow D^c\) for some sequence of sets \(\{D_n\}_{n \rightarrow \mathbb{N}}\). Since \(\limsup_n D_n = \liminf_n D_n = D\), it follows that \((\limsup_n D_n)^c = (\liminf_n D_n)^c = D^c\). By part (a), we know this means \(\liminf_n D_n^c = \limsup_n D_n^c = D^c\). Therefore, \(D_n^c \rightarrow D^c\). Applying this fact, \(A_n \rightarrow A\) implies \(A_n^c \rightarrow A^c\) and \(B_n \rightarrow B\) implies \(B_n^c \rightarrow B^c\). Using what we just prove about unions, \(A_n^c \cup B_n^c \rightarrow A^c \cup B^c\). Again, applying the fact for complements, we get that \(A_n \cap B_n \rightarrow A \cap B\).