[proofplan] We prove the SLLN using a backwards martingale argument. The sequence $S_n/n$ forms a backwards martingale with respect to the $\sigma$-algebras $\mathcal{G}_n = \sigma(S_n, X_{n+1}, X_{n+2}, \ldots)$. By the backwards martingale convergence theorem, $S_n/n$ converges a.s. and in $L^1$ to a tail-measurable random variable. The [Kolmogorov Zero-One Law](/theorems/512) forces this limit to be a.s. constant, and $L^1$ convergence identifies the constant as $\nu$. [/proofplan] [step:Show $S_n/n$ is a backwards martingale] Define $\mathcal{G}_n = \sigma(S_n, X_{n+1}, X_{n+2}, \ldots)$ for $n \geq 1$. We need $\mathbb{E}[S_{n-1}/(n-1) \mid \mathcal{G}_n] = S_n / n$. Since $S_{n-1} = S_n - X_n$ and $X_n$ is independent of $(X_{n+1}, X_{n+2}, \ldots)$: \begin{align*} \mathbb{E}\left[\frac{S_{n-1}}{n-1} \;\middle|\; \mathcal{G}_n\right] = \frac{S_n}{n-1} - \frac{\mathbb{E}[X_n \mid S_n]}{n-1}. \end{align*} By symmetry of the i.i.d. sequence, $\mathbb{E}[X_k \mid S_n]$ does not depend on $k$ for $1 \leq k \leq n$. Since $\sum_{k=1}^n \mathbb{E}[X_k \mid S_n] = S_n$, we obtain $\mathbb{E}[X_n \mid S_n] = S_n / n$ a.s. Substituting: \begin{align*} \mathbb{E}\left[\frac{S_{n-1}}{n-1} \;\middle|\; \mathcal{G}_n\right] = \frac{S_n}{n-1} - \frac{S_n}{n(n-1)} = \frac{S_n}{n}. \end{align*} [guided] Why is $\mathbb{E}[X_k \mid S_n] = S_n/n$ for each $k$? The key is the exchangeability of the i.i.d. sequence: the joint distribution of $(X_1, \ldots, X_n)$ is invariant under permutations. Therefore $\mathbb{E}[X_k \mid S_n]$ is the same function of $S_n$ for every $k \in \{1, \ldots, n\}$. Since these $n$ identical quantities must sum to $S_n$, each equals $S_n/n$. This symmetry argument replaces the need to compute the conditional expectation directly. [/guided] [/step] [step:Identify the a.s. limit as a tail-measurable constant] By backwards martingale convergence, $S_n/n \to Y$ a.s. and in $L^1$, where $Y = \mathbb{E}[X_1 \mid \mathcal{G}_{-\infty}]$. For each $k$: \begin{align*} Y = \lim_{n \to \infty} \frac{S_n}{n} = \lim_{n \to \infty} \frac{X_{k+1} + \cdots + X_{k+n}}{n}, \end{align*} so $Y$ is $\sigma(X_{k+1}, X_{k+2}, \ldots)$-measurable for every $k$. Therefore $Y$ is measurable with respect to the tail $\sigma$-algebra $\bigcap_k \sigma(X_{k+1}, \ldots)$. By [Kolmogorov's Zero-One Law](/theorems/512), $Y$ is a.s. constant: $Y = c$ a.s. for some $c \in \mathbb{R}$. The $L^1$ convergence gives $c = \mathbb{E}[Y] = \lim_n \mathbb{E}[S_n/n] = \nu$. [/step]