Proof*. (Optional material.) We use the monotone class lemma from 3.2, whose notations we also take over. Fix \(C_2 \in \mathcal C_2, \dots, C_n \in \mathcal C_n,\) and define \[\mathcal M_1 :=\bigl\{B_1 \in \mathcal B_1 \,\colon\mathbb{P}(B_1 \cap C_2 \cap \cdots \cap C_n) = \mathbb{P}(B_1)\, \mathbb{P}(C_2) \cdots \mathbb{P}(C_n)\bigr\}\,.\] By assumption, \(\mathcal C_1 \subset \mathcal M_1.\) Moreover, it is easy to verify that \(\mathcal M_1\) is a monotone class. Hence, \[\mathcal M_1 \supset \mathcal M(\mathcal C_1) = \sigma(\mathcal C_1) = \mathcal B_1\,,\] where the second step follows from the monotone class lemma (Proposition 3.8). We conclude: for all \(B_1 \in \mathcal B_1, C_2\in \mathcal C_2, \dots, C_n \in \mathcal C_n,\) we have \[\mathbb{P}(B_1 \cap C_2 \cap \cdots \cap C_n) = \mathbb{P}(B_1)\, \mathbb{P}(C_2) \cdots \mathbb{P}(C_n)\,.\]

We now continue in this fashion, moving on to the second argument. More precisely, fix \(B_1 \in \mathcal B_1, C_3 \in \mathcal C_3, \dots, C_n \in \mathcal C_n\) and define \[\mathcal M_2 :=\bigl\{B_2 \in \mathcal B_2 \,\colon\mathbb{P}(B_1 \cap B_2 \cap C_3 \cap \cdots \cap C_n) = \mathbb{P}(B_1) \, \mathbb{P}(B_2) \, \mathbb{P}(C_3) \cdots \mathbb{P}(C_n)\bigr\}\,.\] As above, it is easy to see that \(\mathcal M_2\) is a monotone class, and by the previous step we know that \(\mathcal C_2 \subset \mathcal M_2.\) By the monotone class lemma, we find that \(\mathcal M_2 \supset \mathcal B_2.\) By repeating this procedure \(n\) times we arrive at the claim.

Proof of Proposition 4.7. Define \(\mathcal D_n :=\sigma(X_1, \dots, X_n)\) (the \(\sigma\)-algebra containing the information up to time \(n\)). Then we claim that \(\mathcal D_n\) and \(\mathcal B_{n+1}\) are independent. This sounds intuitively obvious, as \(\mathcal D_n\) contains information up to time \(n,\) and \(\mathcal B_{n+1}\) information starting from time \(n+1.\) For a rigorous proof, we proceed in two steps.

• For any \(k \geqslant n+1\) we define \(\mathcal B_{n+1,k} :=\sigma(X_{n+1}, \dots, X_k).\) Define the collections \[\begin{aligned} \mathcal C_1 &:=\bigl\{B_1 \cap \cdots \cap B_n \,\colon B_i \in \sigma(X_i) \, \forall i\bigr\}\,, \\ \mathcal C_2 &:=\bigl\{B_{n+1} \cap \cdots \cap B_k \,\colon B_i \in \sigma(X_i) \, \forall i\bigr\}\,. \end{aligned}\] Clearly, these collections are stable under finite intersections and \(\mathcal D_n = \sigma(\mathcal C_n)\) and \(\mathcal B_{n+1,k} = \sigma(\mathcal C_2).\) By Lemma 4.6 and independence of the random variables \((X_n),\) we therefore conclude that \(\mathcal D_n\) and \(\mathcal B_{n+1,k}\) are independent.

• Define the collections \(\mathcal C_1 :=\mathcal D_n\) and \(\mathcal C_2 :=\bigcup_{k \geqslant n+1} \mathcal B_{n+1,k},\) which are clearly stable under finite intersections. Thus, \(\sigma(\mathcal C_1) = \mathcal D_n\) and \(\sigma(\mathcal C_2) = \mathcal B_{n+1}.\) By the previous step and Lemma 4.6, we conclude that \(\mathcal D_n\) and \(\mathcal B_{n+1}\) are independent, as desired.

Next, choose \(\mathcal C_1 :=\bigcup_{n \geqslant 1} \mathcal D_n\) and \(\mathcal C_2 :=\mathcal B_\infty.\) Since \(\mathcal D_n\) and \(\mathcal B_{n+1}\) are independent for all \(n,\) we conclude that \(\mathbb{P}(C_1 \cap C_2) = \mathbb{P}(C_1) \, \mathbb{P}(C_2)\) for all \(C_1 \in \mathcal C_1\) and \(C_2 \in \mathcal C_2.\) By Lemma 4.6, we deduce that \(\sigma(\mathcal C_1) = \sigma(X_1, X_2, \dots) = \mathcal B_1\) and \(\mathcal B_\infty\) are independent. Since \(\mathcal B_\infty \subset \mathcal B_1,\) we conclude that \(\mathcal B_\infty\) is independent of itself! This means that any tail event \(B \in \mathcal B_\infty\) satisfies \(\mathbb{P}(B) = \mathbb{P}(B \cap B) = \mathbb{P}(B)^2,\) from which the zero-one law follows.

Proof. Let \(S_n :=X_1 + \cdots + X_n\) and \(S_0 :=0.\) Let \(a > \mathbb{E}[X_1]\) and define \(M :=\sup_{n \in \mathbb{N}} (S_n - na).\) Thus, \(M\) is a random variable with values in \([0,\infty].\) The core of the proof is to show that \[\tag{4.1} M < \infty \quad \text{a.s.}\]

Let us suppose first that (4.1) has been proved and use it to conclude the proof of the law of large numbers. By definition of \(M\) we have \(S_n \leqslant na + M\) for all \(n\) and hence (4.1) implies, for all \(a > \mathbb{E}[X_1],\) \[\limsup_{n} \frac{S_n}{n} \leqslant a \quad \text{a.s.}\] This implies that \[\tag{4.2} \limsup_{n} \frac{S_n}{n} \leqslant\mathbb{E}[X_1] \quad \text{a.s.}\,,\] since \[\mathbb{P}\biggl(\limsup_{n} \frac{S_n}{n} \leqslant\mathbb{E}[X_1]\biggr) = \mathbb{P}\biggl(\bigcap_{k \in \mathbb{N}^*} \biggl\{\limsup_{n} \frac{S_n}{n} \leqslant\mathbb{E}[X_1] + \frac{1}{k}\biggr\}\biggr) = 1\,,\] where we used that a countable intersection of events of probability one has probability one.

Replacing \(X_n\) with \(-X_n\) we obtain \[\tag{4.3} \liminf_{n} \frac{S_n}{n} \geqslant\mathbb{E}[X_1] \quad \text{a.s.}\] From (4.2) and (4.3) we conclude the law of large numbers.

What remains, therefore, is to prove (4.1). First, we claim that \(\{M < \infty\} \in \mathcal B_\infty.\) Indeed, for all \(k \geqslant 0\) we have \[\{M < \infty\} = \biggl\{\sup_{n \geqslant 0} (S_n - na) < \infty\biggr\} = \biggl\{\sup_{n \geqslant k} \bigl((S_n - S_k) - na\bigr) < \infty\biggr\} \in \sigma(X_{k+1}, X_{k+2}, \dots)\,,\] since \(S_n - S_k = X_{k+1} + X_{k+2} + \cdots + X_n.\) By the zero-one law, Proposition 4.7, to prove (4.1), it therefore suffices to prove that \(\mathbb{P}(M = \infty) < 1.\)

We proceed by contradiction and suppose that \(\mathbb{P}(M = \infty) = 1.\) For all \(k \in \mathbb{N}\) we define \[M_k :=\sup_{0 \leqslant n \leqslant k} (S_n - na)\,, \qquad M_k' :=\sup_{0 \leqslant n \leqslant k} (S_{n+1} - S_1 - na)\,.\] Since \(S_n = X_1 + \cdots + X_n\) and \(S_{n+1} - S_1 = X_2 + \cdots + X_{n+1},\) we conclude that \(M_k \overset{\mathrm d}{=}M_k'\) (recall (2.5)). Moreover, \(M_k\) and \(M'_k\) are increasing sequences that converge from below to their limits \(M\) and \(M'.\) By \(M_k \overset{\mathrm d}{=}M_k',\) we conclude that \(M \overset{\mathrm d}{=}M',\) since \[\mathbb{P}(M' \leqslant x) = \lim_{k \to \infty} \mathbb{P}(M'_k \leqslant x) = \lim_{k \to \infty} \mathbb{P}(M_k \leqslant x) = \mathbb{P}(M \leqslant x)\,.\] Moreover, \[\begin{aligned} M_{k+1} &= \sup \biggl\{0, \sup_{1 \leqslant n \leqslant k+1} (S_n - na)\biggr\} \\ &= \sup \biggl\{0, \sup_{0 \leqslant n \leqslant k} (S_{n+1} - (n+1)a)\biggr\} \\ &= \sup\{0, M'_k + X_1 - a\} \\ &= M'_k - \inf\{a - X_1, M'_k\}\,. \end{aligned}\] Hence, \[\tag{4.4} \mathbb{E}[\inf\{a - X_1, M'_k\}] = \mathbb{E}[M'_k] - \mathbb{E}[M_{k+1}] = \mathbb{E}[M_k] - \mathbb{E}[M_{k+1}] \leqslant 0\,,\] since the sequence \((M_k)\) is nondecreasing. Moreover, since \(M'_k \geqslant 0\) we have \[\lvert \inf\{a - X_1, M'_k\} \rvert \leqslant\lvert a - X_1 \rvert\] for all \(k,\) so that we may apply dominated convergence to (4.4) to get \[\mathbb{E}[\inf\{a - X_1, M'\}] \leqslant 0\,.\]

Now if \(\mathbb{P}(M = \infty) = 1\) then also \(\mathbb{P}(M' = \infty) = 1\) and hence \(\inf\{a - X_1, M'\} = a - X_1.\) But \[\mathbb{E}[a - X_1] > 0\] by assumption. This is the desired contradiction.