= Solution
The <sigma-algebras> $\mathcal G_n=\sigma(S_n,S_{n+1},\ldots)$ decrease with $n$, and (b) gives $S_n/n=\mathbb E[X_1\mid\mathcal G_n]$. The permitted <reverse martingale convergence theorem> therefore gives an <integrable random variable> $L$ such that $S_n/n\to L$ <almost surely> and with <convergence in L1>. In particular,
$$
\mathbb EL=\lim_n\mathbb E[S_n/n]=m.
$$
It remains to prove that $L$ is constant. We must not assume that $\bigcap_n\mathcal G_n$ is the <tail sigma-algebra> of the $X_j$; instead use <tail measurability of limits of sample averages> directly. Define the <random variable> valued in the <extended real numbers> $L^*=\limsup_n S_n/n$. For every fixed $N$,
$$
L^*=\limsup_{n\to\infty}\frac{X_{N+1}+\cdots+X_n}{n},
$$
because $S_N/n\to0$ pointwise. Thus $L^*$ is measurable with respect to $\sigma(X_{N+1},X_{N+2},\ldots)$ for every $N$, and hence to the <tail sigma-algebra>. Also $L^*=L$ <almost surely>.
By the <Kolmogorov zero-one law>, each <tail event> $\{L^*\leq q\}$, for a <rational number> $q$, has probability zero or one. Since $L^*$ is finite <almost surely>, its <distribution function> can only be that of a constant $c$: taking $c=\inf\{q\in\mathbb Q:\mathbb P(L^*\leq q)=1\}$ and using the <density of the rational numbers> gives $L^*=c$ <almost surely>. Its <expected value> identifies $c=m$. This proves the <strong law of large numbers>:
$$
\boxed{\frac{S_n}{n}\longrightarrow m\quad\text{almost surely}.}
$$
Back to article page
