= Solution
A <sigma-algebra> on $\Omega$ is a collection $\mathcal G\subseteq\mathcal P(\Omega)$ containing $\Omega$, closed under <complements> relative to $\Omega$, and closed under <countable unions>. These properties also give closure under <countable intersections> by <De Morgan laws>.
Let $\mathcal G=\bigcap_{r\in R}\mathcal F_r$. Every constituent <sigma-algebra> contains $\Omega$, so $\Omega\in\mathcal G$. If $A\in\mathcal G$, its <complement> belongs to every $\mathcal F_r$, hence to $\mathcal G$. Likewise, if $A_j\in\mathcal G$ for all $j$, then $\bigcup_j A_j$ belongs to every $\mathcal F_r$. Thus $\mathcal G$ is a <sigma-algebra>. For an empty indexing set, the usual <intersection> convention gives $\mathcal G=\mathcal P(\Omega)$, which is also a <sigma-algebra>.
A <union> need not work. On $\Omega=\{1,2,3\}$ take
$$
\mathcal F_1=\{\varnothing,\Omega,\{1\},\{2,3\}\},\qquad
\mathcal F_2=\{\varnothing,\Omega,\{2\},\{1,3\}\}.
$$
Their <union> contains $\{1\}$ and $\{2\}$ but not $\{1,2\}$, so it is not a <sigma-algebra>. Put $\mathcal A=\bigcup_r\mathcal F_r$ and let $\mathfrak C$ be the collection of all <sigma-algebras> on $\Omega$ containing $\mathcal A$. This collection is nonempty because it contains the <power set> $\mathcal P(\Omega)$. By the preceding <intersection> argument,
$$
\boxed{\sigma(\mathcal A)=\bigcap_{\mathcal H\in\mathfrak C}\mathcal H}
$$
is a <sigma-algebra> containing every $\mathcal F_r$, and it is contained in every other such <sigma-algebra>. This proves existence and minimality.
For the <independent random variables>, define the <tail sigma-algebra> by
$$
\mathcal T=\bigcap_{m\geq1}\sigma(X_m,X_{m+1},\ldots).
$$
Fix a <tail event> $A\in\mathcal T$. For each $m$, it belongs to $\sigma(X_{m+1},X_{m+2},\ldots)$, which is <independent> of $\mathcal H_m=\sigma(X_1,\ldots,X_m)$. To justify independence of the generated <sigma-algebras>, first use independence of finite collections on cylinder <events>, then extend twice by the <pi-lambda theorem>. In particular, $\mathbb P(A\cap C)=\mathbb P(A)\mathbb P(C)$ for every $C\in\mathcal H_m$ and every $m$.
The increasing <union> $\bigcup_m\mathcal H_m$ is an <algebra of sets> generating $\mathcal H=\sigma(X_1,X_2,\ldots)$. The collection of $C\in\mathcal H$ satisfying this factorization is a <Dynkin system>: it contains $\Omega$, is closed under <complements>, and under countable disjoint <unions>. The <pi-lambda theorem> therefore extends the factorization to all $C\in\mathcal H$. Since $A\in\mathcal H$, taking $C=A$ gives $\mathbb P(A)=\mathbb P(A)^2$. Consequently,
$$
\boxed{\mathbb P(A)\in\{0,1\}\qquad(A\in\mathcal T).}
$$
This is the <Kolmogorov zero-one law>, proved here directly.
For each fixed $m$, write $S_n=S_{m-1}+\sum_{j=m}^nX_j$ for $n\geq m$, with $S_0=0$. The <random variable> $S_{m-1}$ is finite, so $S_{m-1}/\sqrt n\to0$. Adding a real sequence tending to zero changes neither the extended-real <limit inferior> nor the extended-real <limit superior>. Hence
$$
\liminf_n\frac{S_n}{\sqrt n}=\liminf_n\frac{\sum_{j=m}^nX_j}{\sqrt n},\qquad
\limsup_n\frac{S_n}{\sqrt n}=\limsup_n\frac{\sum_{j=m}^nX_j}{\sqrt n}.
$$
The quantities on the right are measurable with respect to $\sigma(X_m,X_{m+1},\ldots)$: write the <limit inferior> as $\sup_N\inf_{n\geq N}$ and the <limit superior> as $\inf_N\sup_{n\geq N}$, using countable operations on <measurable functions>. Therefore both threshold <events> belong to every such <sigma-algebra>, and hence are <tail events>, for every real threshold, including when either limit is infinite.
\b[The final recurrence claim is false under the printed hypotheses.] For $c>0$, let $\varepsilon_j$ be independent signs with equal probabilities for $+1$ and $-1$, and set $X_j=c4^{-(j-1)}\varepsilon_j$. These <random variables> are independent, have <symmetric distributions>, and obey $|X_j|\leq c$. Nevertheless, the <geometric series> gives, on every sample path,
$$
|S_n|\geq c-\sum_{j=2}^n c4^{-(j-1)}\geq c-\frac c3=\frac{2c}{3}>\frac c2.
$$
Thus the proposed repeated-return <event> has <probability> zero, even though none of the increments is degenerate. The <Kolmogorov zero-one law> does not determine which of zero and one occurs.
An additional identical-distribution hypothesis makes the intended conclusion true. Here is a full proof of that qualified version, using the <oscillation of a bounded centered iid random walk>. Symmetry and integrability give $\mathbb E X_j=0$, so $S_n$ is a <martingale> for its <natural filtration>. If the common <probability distribution> is concentrated at zero, every $S_n$ is zero and the conclusion is immediate. Otherwise choose $\delta>0$ with $\mathbb P(|X_1|>\delta)>0$. The <Borel-Cantelli lemmas> then give $|X_n|>\delta$ infinitely often <almost surely>; thus $S_n$ cannot converge to a finite limit.
For $a>0$, let $T_a=\inf\{n:S_n\geq a\}$. Before $T_a$ the sum is below $a$, and at $T_a$ it is at most $a+c$. The <stopped martingale> $a+c-S_{n\wedge T_a}$ is therefore nonnegative. A stopped integrable <martingale> remains a <martingale> at these bounded times, and a nonnegative <martingale> converges finitely <almost surely> by the <martingale convergence theorem>. On $\{T_a=\infty\}$ this would force $S_n$ to converge finitely, which has just been ruled out. Hence $T_a<\infty$ <almost surely>. Apply the same argument to $-S_n$ and take a countable intersection over positive integer $a$. It follows that
$$
\limsup_n S_n=+\infty,\qquad \liminf_n S_n=-\infty\quad\text{almost surely}.
$$
There are therefore arbitrarily late passages from above $c/2$ to below $-c/2$. A path with <bounded increments> of magnitude at most $c$ cannot make such a passage without visiting $[-c/2,c/2]$: jumping directly between its two open complementary half-lines would require an increment larger than $c$. There are infinitely many such visits. Thus \b[with the added identical-distribution hypothesis], the intended answer is
$$
\boxed{\mathbb P\bigl(|S_n|\leq c/2\text{ infinitely often}\bigr)=1.}
$$
Back to article page
