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Every bounded sequence of real numbers contains a subsequence with a finite limit of a sequence. If $|x_n|\le M$ for every positive integer $n$, there are integers $n_1<n_2<\cdots$ and a real number $\ell$ such thatThe increasing indices matter: a subsequence retains the original order and selects terms arbitrarily far along the original sequence. The theorem guarantees the existence of at least one such selection; it does not claim that the entire sequence converges or that the subsequential limit is unique.^[ref]

For example, the bounded sequence $x_n=(-1{)}^n+1/(n+1)$ has two distinct subsequential limits:The even terms form a convergent sequence, and so do the odd terms. The full sequence cannot have a limit of a sequence because its two subsequences have different limits. Its limit superior is $1$ and its limit inferior is $-1$.

This is useful when a construction produces many candidate points but does not immediately produce a convergent sequence. For a continuous function on a nonempty closed bounded feasible region in a finite-dimensional Euclidean space, take a sequence whose objective values approach the infimum. A convergent subsequence supplies a candidate point; a closed subset retains that point, and continuity passes the objective values to the limit of a sequence. This is the mechanism behind attainment in the extreme value theorem. The diagonal subsequence argument extends the real-sequence theorem to finite-dimensional Euclidean space.

The theorem is named for Bernard Bolzano and Karl Weierstrass.^[ref]

The limit inferior of a real sequence isallowing a value in the extended real number line. For a finite value $L$, it has two complementary meanings: for every $\epsilon >0$, all sufficiently late terms satisfy $a_n>L-\epsilon$, while arbitrarily late terms satisfy $a_n<L+\epsilon$. The first prevents persistent dips below $L$; the second prevents any larger number from being the true lower edge.

For $a_n=(-1{)}^n(1+1/n)$, the lower bounds in limit inferior and limit superior approach $-1$. Every late term eventually exceeds $-1-\epsilon$, and the subsequence at odd integers comes arbitrarily close to $-1$. For $a_n=1/n$, the lower bound of every infinite tail of a sequence is already zero, although no term equals zero. Thus a limit inferior need not be the minimum of any tail.

The limit inferior and limit superior retain the lower and upper long-run behaviour of a real sequence, even when it has no ordinary limit of a sequence. For each tail of a sequence beginning at $N$, defineDiscarding more initial terms raises or preserves the infimum and lowers or preserves the supremum. Hence $(l_N)$ is nondecreasing and $(u_N)$ is nonincreasing. Their limits exist in the extended real number line:For a bounded sequence, both values are finite and $\mathrm{liminf}{a}_n\le \mathrm{limsup}{a}_n$. These bounds use the entire infinite tail of a sequence, rather than the last few observed terms. ^[ref].

Take the original example $a_n=(-1{)}^n(1+1/n)$, beginning $-2,3/2,-4/3,5/4,\dots$. The subsequence at even integers decreases toward one, while the subsequence at odd integers increases toward minus one. If $e_N$ is the first even integer at least $N$ and $o_N$ is the first odd integer at least $N$, thenThus $\mathrm{liminf}{a}_n=-1$ and $\mathrm{limsup}{a}_n=1$. The finite initial extreme $a_1=-2$ disappears from later lower bounds; neither long-run value is attained by any term of this example.

For a bounded sequence, the limit inferior is the smallest subsequential limit and the limit superior is the largest subsequential limit; both are reached by suitable convergent subsequences. Every other subsequential limit lies between them. In the example there are only two subsequential limits, $-1$ and $1$: a convergent subsequence cannot retain infinitely many terms of both parities, because those two branches stay separated. The whole interval between the two bounds need not consist of subsequential limits. ^[ref].

A finite ordinary limit of a sequence exists exactly when the limit inferior and limit superior agree at the same finite value $L$. The bounds $l_N\le a_N\le u_N$ and the squeeze theorem prove the sufficient direction. For a bounded sequence, its tail diameter of a sequence is $D_N=u_N-l_N$, so the limiting gap is zero exactly when the Cauchy sequence condition holds. In the example the gap approaches two, not zero. The same framework allows infinite bounds: $a_n=n$ has both values $+\infty$, whereas $a_n=(-1{)}^{n}n$ has limit inferior $-\infty$ and limit superior $+\infty$. Equality at infinity describes escape rather than convergence to a real value. ^[ref].

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 $\Omega =\{1,2,3\}$ takeTheir 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,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 byFix a tail event $A\in \mathcal{T}$. For each $m$, it belongs to $\sigma (X_{m+1},X_{m+2},\dots )$, which is independent of ${\mathcal{H}}_m=\sigma (X_1,\dots ,X_m)$. 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 $\mathcal{H}=\sigma (X_1,X_2,\dots )$. 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,This is the Kolmogorov zero-one law, proved here directly.

For each fixed $m$, write $S_n=S_{m-1}+{\sum }_{j=m}^n{X}_j$ for $n\ge m$, with $S_0=0$. The random variable $S_{m-1}$ is finite, so $S_{m-1}/\sqrt{n}\to 0$. Adding a real sequence tending to zero changes neither the extended-real limit inferior nor the extended-real limit superior. HenceThe quantities on the right are measurable with respect to $\sigma (X_m,X_{m+1},\dots )$: write the limit inferior as ${\sup }_N{\inf }_{n\ge N}$ and the limit superior as ${\inf }_N{\sup }_{n\ge N}$, 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.

The final recurrence claim is false under the printed hypotheses. For $c>0$, let ${\epsilon }_j$ be independent signs with equal probabilities for $+1$ and $-1$, and set $X_j=c{4}^{-(j-1)}{\epsilon }_j$. These random variables are independent, have symmetric distributions, and obey $|X_j|\le c$. Nevertheless, the geometric series gives, on every sample path,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 $T_a=\inf \{n:S_n\ge a\}$. 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 thatThere 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 with the added identical-distribution hypothesis, the intended answer is

Assume ${\sup }_n\mathbb{E}|M_n|<\infty$. For every pair of rationals $a<b$, Doob upcrossing inequality givesby monotone convergence theorem. Thus every rational interval is upcrossed only finitely often almost surely. If a real sequence has distinct limit inferior and limit superior, it completes infinitely many upcrossings of some rational interval between them. Hence $M_n$ converges in the extended real line almost surely.

Fatou lemma givesso the limit is finite almost surely and integrable. This is the $L^1$-bounded form of the martingale convergence theorem.

Include the positive integers in $[1,2),[4,8),[16,32),[64,128),\dots$, and exclude the intervening integer intervals and zero. Equivalently, include $[2^{2k},2^{2k+1})\cap \mathbb{Z}$ for $k\ge 0$. Each included block is large enough to raise the running proportion; the following excluded block lowers it again.

With $C_A(N)=|A\cap \{0,\dots ,N-1\}|$, the exact boundary counts follow from a geometric series:The first sum is empty at $k=0$. Dividing by the corresponding boundary, rather than using one denominator for both, givesFor example, $C_A(16)=1+4=5$, whereas $C_A(32)=1+4+16=21$, yielding proportions $5/16$ and $21/32$. For $4^k\le N\le 2\cdot 4^k$, the exact count is $(4^k-1)/3+N-4^k$; for $2\cdot 4^k\le N\le 4^{k+1}$ it stays $(4^{k+1}-1)/3$. Hence the proportion increases through each included block and decreases through each excluded block. These boundary subsequences give the actual limit inferior and limit superior:Since these differ, $A$ has no natural density. Its upper asymptotic density still records its largest limiting proportion.

A subsequential limit of a sequence $(a_n)$ is a value $L$ approached by some subsequence $(a_{n_k})$ with strictly increasing indices $n_k$. For $a_n=(-1{)}^n(1+1/n)$, the subsequences at even integers and odd integers give subsequential limits one and minus one. A finite value appearing among only finitely many initial terms does not become a subsequential limit merely by being a member of the image of a function represented by the sequence.

For a bounded sequence of real numbers, its subsequential limits form a nonempty bounded set that is also a closed set. The Bolzano-Weierstrass theorem guarantees nonemptiness, and its extreme values are the limit inferior and limit superior. The sequence in sparse peaks can determine a limit superior has the two subsequential limits zero and one, despite almost all indices in the natural density sense carrying zero.

Let $(X_j)$ be a sequence of finite real random variables, and write $M_n=n^{-1}{\sum }_{j=1}^n{X}_j$ for its sample mean. For fixed $m$ and $n>m$, put $M_n^{(m)}=n^{-1}{\sum }_{j=m+1}^n{X}_j$. Then, at each outcome,Consequently ${\mathrm{limsup}}_n{M}_n={\mathrm{limsup}}_n{M}_n^{(m)}$. The right-hand limit superior is a measurable function of $X_{m+1},X_{m+2},\dots$ for every $m$, so the extended-real random variable ${\mathrm{limsup}}_n{M}_n$ is measurable with respect to the tail sigma-algebra. The same holds for the limit inferior. Thus the event of convergence to a finite limit of a sequence, and the event $\{M_n\to a\}$ for any specified real $a$, are tail events.

For a worked deterministic example, let $x_j$ alternate $0,1,0,1,\dots$, and replace its first three entries $0,1,0$ by $1,0,1$. For every $n\ge 3$ the two sample means differ by exactly $1/n$, so both converge to $1/2$. Finite edits can change small-sample answers while preserving the infinite limit of a sequence.

Under independence, Kolmogorov's zero–one law makes each specified convergence event have probability zero or one. It does not select which value occurs. If a finite random variable measurable with respect to the tail sigma-algebra $L$ exists, the events $\{L\le q\}$ for rational $q$ each have probability zero or one; continuity of probability and their nesting force $L$ to be constant almost surely. For fair independent Bernoulli trials, the strong law of large numbers establishes $M_n\to 1/2$ almost surely; the zero–one conclusion alone proves neither convergence nor the value $1/2$.