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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.

Choose $M$ with $|b_n|\le M$. Every tail of a sequence is nonempty and bounded, so its supremum $c_n$ exists and lies in $[-M,M]$. Removing one term cannot increase a supremum, hence $c_{n+1}\le c_n$. The bounded monotone sequence theorem, applied to $-c_n$, provesThus $L$ is the limit superior of $(b_n)$.

For $\epsilon >0$, eventually $c_n<L+\epsilon$, and then $b_n\le c_n<L+\epsilon$. Only finitely many indices can satisfy $b_n\ge L+\epsilon$. If only finitely many satisfied $b_n>L-\epsilon$, some entire tail of a sequence would have $b_n\le L-\epsilon$. Its supremum would also be at most $L-\epsilon$, contradicting $c_n\ge L$.

Inductively choose $n_k>n_{k-1}$ so late that $c_{n_k}<L+1/k$ and $b_{n_k}>L-1/k$. There are infinitely many choices for the latter inequality, so both requirements can be met. Then $|b_{n_k}-L|<1/k$, proving $\boxed{b_{n_k}\to L}$. This constructs a subsequence attaining the limit superior.

At the nonzero indices $n=m!$ with $m\ge 2$,since $m/(m!)=1/(m-1)!\to 0$. At all other sufficiently large indices the coefficient is zero. The initial coefficients do not affect the limit superior, which equals one. The Cauchy-Hadamard theorem therefore gives $\boxed{R=1}$.

The Cauchy-Hadamard theorem states $R^{-1}=\mathrm{limsup}|a_n{|}^{1/n}$. At the prime numbers $n=p$, this root equals $p^{2/p}\to 1$. There are infinitely many prime numbers, so these indices form an unbounded subsequence. For every other positive index the root is zero. The limit superior is therefore one, and $\boxed{R=1}$.

The functions $g_n$ are nonnegative. The change of variables $u=nx$ and the given Gaussian integral implyFor every fixed $\delta >0$,Thus the mass concentrates at the endpoint zero. This is a one-sided approximate identity.

Because $f$ is continuous on the compact interval $[0,1]$, it is a bounded function; put $B={\sup }_{[0,1]}|f|$. Given $\epsilon >0$, choose $\delta \in (0,1)$ with $|f(x)-f(0)|<\epsilon$ for $0\le x<\delta$. Splitting the integral near and away from zero yieldsTaking the limit superior gives a bound of $\epsilon$, and $\epsilon$ was arbitrary. ConsequentlyOnly continuity at zero and boundedness on $[0,1]$ were needed for this final estimate; the assumed continuity everywhere also guarantees the ordinary integrals exist.

For $m\ge 1$, fix signs $x_1,\dots ,x_m$. The constraints $X_j=x_j$ leave exactly two choices for the $(m+1)$ signs $Y_0,\dots ,Y_m$: choose $Y_0$ freely, then set $Y_1=Y_0{x}_1$ and $Y_j=x_{j-1}{x}_j$ for $2\le j\le m$. Each input tuple has probability $2^{-(m+1)}$ by independence. ConsequentlyEvery finite subfamily is a marginal distribution of one such initial segment, so the entire sequence is independent, with fair sign probability distributions.

Using the usual everywhere sign-valued representatives, for every $n\ge 1$ we haveThe right side is measurable with respect to $\sigma (\mathcal{Y},{\mathcal{T}}_n)$. Therefore $Y_0$ is measurable with respect to their intersection $\mathcal{G}$.

Write $\mathcal{T}={\bigcap }_n{\mathcal{T}}_n$. The Kolmogorov zero-one law applies to the independent random variables $(X_n)$, so every event $C\in \mathcal{T}$ has probability zero or one. Also $Y_0$ is independent of $\mathcal{Y}$: this follows first for finite intersections of sign events from the given independence, then for their generated sigma-algebra by part (a).

The sets $D\cap C$ with $D\in \mathcal{Y}$ and $C\in \mathcal{T}$ form a pi-system generating $\mathcal{H}$. If $\mathbb{P}(C)=0$, the event $D\cap C$ has probability zero and is independent of every event in $\sigma (Y_0)$. If $\mathbb{P}(C)=1$, it differs from $D$ by a null set, so the same independence follows from that of $Y_0$ and $\mathcal{Y}$. Another application of part (a) provesIn particular $\mathcal{H}$ cannot contain the nontrivial event $\{Y_0=1\}$, whereas $\mathcal{G}$ does. Thus adjoining a sigma-algebra need not commute with a decreasing intersection.

There is a small version qualification if the assumptions specify only the probability distributions, allowing exceptional values other than signs. Put $q_n=X_n/(Y_1\cdots Y_n)$ when the denominator is nonzero, and $q_n=0$ otherwise. For $k\ge n$, $q_k$ is measurable with respect to $\sigma (\mathcal{Y},{\mathcal{T}}_n)$. The limit superior of $(q_k)$, after scalar clipping to $[-1,1]$, is therefore $\mathcal{G}$-measurable and equals $Y_0$ almost surely. Literal measurability of an arbitrary version of $Y_0$ need not follow without choosing sign-valued representatives or adjoining the ambient null sets. All the independence conclusions are unaffected.

Apply part (i) for every positive integer $K$ and discard the countable union of its exceptional null sets. A sample path bounded below never reaches $-K$ for some such $K$, and must therefore converge to a finite limit of a sequence.

Apply the same argument to the martingale $(-M_n)$. Outside one more null set, every sample path bounded above also converges finitely. Thus a sample path which does not converge finitely must be unbounded in both directions. Each finite initial segment has finite values, so removing any such segment preserves unboundedness. It follows that the limit superior is $+\infty$ and the limit inferior is $-\infty$. ConsequentlyThis proves the bounded-increment martingale convergence-or-oscillation dichotomy; in particular one-sided divergence to infinity is impossible with positive probability.

We prove the Doob-Dynkin lemma here by simple function approximation. Choose $C<\infty$ such that $|Z|\le C$ everywhere, and letEach $Z_m$ is a finite-valued $\sigma (X)$-measurable function and $Z_m\to Z$ pointwise. Write its level sets as $A_{m,j}$, with associated values $z_{m,j}$. By part (i), choose Borel sets $D_{m,j}$ with $A_{m,j}=X^{-1}(D_{m,j})$. Define the Borel measurable functionsAlthough the chosen $D_{m,j}$ need not partition all of $\mathbb{R}$, their inverse images do partition $\Omega$, so $g_m(X)=Z_m$ exactly.

Apply scalar clipping to each $g_m$ with bounds $[-C-1,C+1]$, preserving this equality, and define $g(x)={\mathrm{limsup}}_m{g}_m(x)$. A bounded limit superior of Borel measurable functions is a finite Borel measurable function. At every $\omega \in \Omega$ the limit of a sequence exists and givesThus the factorization is exact for a bounded, exactly $\sigma (X)$-measurable function. If hypotheses are instead given only up to null sets, the corresponding conclusion is a factorization holding almost surely after choosing a representative.

For a fixed finite number of nonnegative terms and $a_L\to \infty$, use ${\max }_j{p}_{j,L}\le {\sum }_j{p}_{j,L}\le r{\max }_j{p}_{j,L}$. The logarithm of $r$, divided by the large-deviation speed, tends to zero, and a finite maximum commutes with the limit superior. If all individual logarithmic limits exist, the limit of the sum is their maximum.

In independent bond percolation on a graph, let $c_n$ be the number of $n$-step self-avoiding walks from a fixed graph vertex and ${\kappa }_n$ the number that are open. Then $\mathbb{E}{\kappa }_n=c_n{p}^n$. The Jensen inequality for the concave function $t^{1/n}$ gives $\mathbb{E}({\kappa }_n^{1/n})\le p{c}_n^{1/n}$. No independence between different self-avoiding walk indicators is needed. When $c_n^{1/n}\to \mu$, the limit superior is at most $p\mu$.

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.

Suppose $q$ is a sublinear functional on the bounded real sequences, dominated by the limit superior and nonpositive on every shift difference $x-Sx$. The Hahn-Banach theorem then yields a Banach limit. Indeed $0\le q(x)+q(-x)$ implies $q(x)\ge -q(-x)\ge \mathrm{liminf}{x}_n$, so $q$ agrees with the ordinary limit on convergent sequences. Extend this linear limit functional by Hahn-Banach to a linear $\psi \le q$. Domination on $x-Sx$ and on its negative forces $\psi (x)=\psi (Sx)$. The two-sided limit bounds give positivity, norm one and agreement with ordinary limits. In finite-shift averaging constructions, the needed bound on shift differences follows by telescoping consecutive averages to $(x_{n+1}-x_{n+k+1})/k$.

Define a sequence by $a_n=1$ when $n$ is a power of two, including $1=2^0$, and $a_n=0$ otherwise. Every tail of a sequence contains both a later power of two and a later nonpower. Therefore $\mathrm{liminf}{a}_n=0$ and $\mathrm{limsup}{a}_n=1$, and the sequence does not converge. Its peak indices become increasingly sparse, yet never stop appearing.

Through index $N\ge 1$, the number of peaks is $1+\lfloor {\log }_{2}N\rfloor$, so their fraction isThe set of powers of two has natural density zero. The same fraction is the Cesàro mean of the first $N$ terms, which tends to zero even though the limit superior stays one. At $N=1024$, eleven peaks give mean $11/1024\approx 0.0107422$. An average and a long-run upper edge answer different questions about the same sequence.

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 $1/2$.

For a subset $A$ of the positive integers, its upper asymptotic density isThe limit superior keeps the largest limiting proportion even when the running proportions oscillate. Because they lie in $[0,1]$, this limit superior always exists in $[0,1]$. If natural density exists, upper asymptotic density equals it.

The convention here uses $1,\dots ,N$, while natural density above uses $0,\dots ,N-1$. Replacing one interval by the other changes a count by at most one, so the proportions differ by at most $1/N$ and have the same limit superior. For the set without natural density above, the count in $\{1,\dots ,16\}$ is six rather than five, because $16$ is included and zero is absent. Nevertheless its upper asymptotic density is exactly $2/3$. The even positive integers have upper asymptotic density $1/2$ and the prime numbers have upper asymptotic density zero.

The useful feature is that a large proportion along arbitrarily large initial integer intervals already supplies structure. The Szemerédi theorem says that positive upper asymptotic density forces arithmetic progressions of every finite length, with positive common difference, even without an existing natural density. Szemerédi's 1975 paper resolved this general finite-length problem; it makes density a bridge from counting to additive structure. ^[ref]