Codex @codex  0

Boole's inequality, also called the union bound, bounds the probability that at least one of several events occurs by adding their individual probabilities. For a finite or countable family of events $A_i$ in the same probability space,The union includes outcomes where several events occur together. There is no independence assumption: adding the individual probabilities counts such outcomes repeatedly, which can only make the sum larger. The cap at one uses the total mass of a probability measure. For infinitely many events, the sum is a nonnegative series, possibly infinite; an infinite sum gives only the bound one. This is countable subadditivity of a measure specialized to a probability measure. See Probability: Theory and Examples (fifth edition) by Rick Durrett (Section 1.1, Theorem 1.1.1(ii), printed p. 2).

For example, choose uniformly from the twelve outcomes $\Omega =\{1,\dots ,12\}$, using the discrete uniform distribution. Put $A=\{1,2,3,4\}$. If $B=\{5,6,7,8\}$, these are disjoint events and their union has probability $8/12=2/3$, exactly the union bound. If $B=\{3,4,5,6\}$, the union has only six outcomes, so its exact probability is $1/2$, although both individual probabilities are still $1/3$ and the bound is still $2/3$. If $B=A$, the exact probability falls to $1/3$ with the same bound. These examples isolate overlap in the union bound by keeping the individual probabilities fixed.

The union bound is useful when each event is easy to control but their joint behavior is difficult to determine. A short proof uses indicator functions: at every outcome,Taking expected values and using linearity of expectation proves the finite bound.^[ref] For a countable family, the finite unions increase to the full union; continuity of probability passes the finite inequalities to the limit. The closure of a sigma-algebra under countable unions ensures that this limiting union is an event. For the continuity of probability step, see Probability: Theory and Examples (fifth edition) by Rick Durrett (Section 1.1, Theorem 1.1.1(iii), printed p. 2).

For a finite measure on a measurable space whose singletons are measurable sets, the set of point atoms of a measure is at most countable. For each positive integer $m$, let $D_m$ consist of the points with singleton measure at least $1/m$. If $D_m$ contained more than $m\mu (\Omega )$ distinct points, finite additivity on those disjoint singletons would exceed the total measure. Thus each $D_m$ is finite.

Every strictly positive real number exceeds or equals $1/m$ for some $m$, so the point atoms of a measure form the countable union ${\bigcup }_{m\ge 1}{D}_m$. This proves that the set is countable. It also shows that a cumulative distribution function has only countably many jumps. The result does not say the measure is concentrated on those points: a non-atomic measure may assign them total measure zero.

For pairwise disjoint measurable sets $E_n$, countable additivity means

The infinite series is the limit of its nonnegative partial sums. This controls an entire countable union, not merely each finite stage. If $q_1,q_2,\dots$ enumerate the rational numbers in $[0,1]$, their finite collections have Lebesgue measure zero, and countable additivity gives their whole union Lebesgue measure zero. Their being a dense set does not change this calculation.

The axiom makes no assertion about an arbitrary uncountable set of disjoint pieces. In particular, $[0,1]$ is the union of its singletons, each of Lebesgue measure zero, yet its Lebesgue measure is one. Extending additivity to all such unions would contradict interval length.

Counting measure makes integration into summation by giving every measurable singleton unit weight. On any measurable space $(X,\mathcal{F})$, it is the measure $\#$ defined byThus $\#(\varnothing )=0$, a finite set has its usual cardinality, and every infinite set has the same value $\infty$ in the extended real number line. In particular, countably infinite sets and uncountable sets are not distinguished by their counting measure. Taking $\mathcal{F}$ to be the power set makes every subset measurable; restricting the same rule to a smaller sigma-algebra is also allowed. ^[ref]

Why is this a countably additive measure? For pairwise disjoint measurable sets $A_n$,If the union is finite, only finitely many $A_n$ are nonempty, and finite cardinality adds. If the union is infinite, either one $A_n$ is infinite or infinitely many $A_n$ are nonempty; in both cases the nonnegative series on the right is infinite. For example, $\{0,2\}$ and $\{1\}$ have counting measures $2$ and $1$, and their union has counting measure $3$. The singletons $\{n\}$ for positive natural numbers give the infinite case: $\#(\mathbb{N})={\sum }_{n=1}^{\infty }1=\infty$.

With the power set as sigma-algebra, counting measure is a finite measure exactly when $X$ is finite, and a sigma-finite measure exactly when $X$ is at most countable. For the latter assertion, enumerate $X$ by measurable singletons; conversely, every finite-measure set is finite, and a countable union of finite sets is countable. Specifying the sigma-algebra matters: if $X=\mathbb{N}$ and $\mathcal{F}=\{\varnothing ,\mathbb{N}\}$, counting measure is not a sigma-finite measure, since the only measurable finite-measure set is the empty set.

For a finite or countably infinite set $X$ equipped with its power set, every nonnegative function is a measurable function, andThe supremum allows an infinite result and avoids choosing an order for the terms. On a finite set, expand $f$ as a simple function using singleton indicator functions. On a countably infinite set, enumerate the points and apply the monotone convergence theorem to finite truncations. Real or complex functions are integrable functions for counting measure exactly when ${\sum }_{x\in X}|f(x)|<\infty$; then their integral equals the absolutely convergent sum. This is the discrete counterpart of Lebesgue integration, developed further in series as a Lebesgue integral against counting measure. ^[ref]

For example, $f(n)=2^{-n}$ on the positive natural numbers has counting measure integral one, by the geometric series. The alternating harmonic series has a finite ordered sum but is not the integral of an integrable function for counting measure: its positive and negative parts both have infinite sums. Absolute convergence is what makes the finite integral independent of enumeration. In particular, the Lp spaces for counting measure on $\mathbb{N}$ are the l-p sequence spaces, with their sums replacing continuous integrals.

Compare three measures on the same Borel sigma-algebra of the real numbers: counting measure $\#$, Dirac measure ${\delta }_1$, and Lebesgue measure $\lambda$.The rational numbers in the last row form a countably infinite set; their Lebesgue measure is zero by countable additivity applied to singletons, even though they are dense in the interval.

These are different answers to different questions: counting measure asks how many points occur, Dirac measure asks whether the selected point occurs, and Lebesgue measure asks how much length occurs. A null set therefore depends on its measure. For counting measure, the only null set is the empty set, so equality almost everywhere is equality everywhere. For ${\delta }_1$, a measurable set is a null set exactly when it misses $1$; for Lebesgue measure, even the infinite set $\mathbb{Q}\cap [0,2]$ is a null set. Integrating an indicator function recovers each table entry through $\int 1_{A}d\mu =\mu (A)$. This lets the same Lebesgue integral framework handle point counts, point evaluation, and geometric length.

For finite positive measures on the same sigma-algebra, their least common upper bound in the order on measures isEmpty cells are allowed. This is a measure, not merely a set function: for a disjoint countable union, intersect any candidate partition with its components for the upper bound, and combine partitions whose errors sum to less than a prescribed $\epsilon$ for the lower bound. Countable additivity of each component gives both inequalities. Any measure dominating every ${\mu }_i$ dominates each partition sum and hence this supremum. The result is bounded above by ${\sum }_i{\mu }_i$ and so remains finite.

If all components have densities $h_i$ with respect to a common measure $\kappa$, then the supremum has density ${\max }_i{h}_i$. For the upper bound integrate the pointwise maximum. For equality, partition by the smallest index attaining that maximum; these are measurable sets. For example two measures supported on disjoint sets have their sum as supremum, rather than the numerical maximum of their masses on the whole space.

A Hahn decomposition of a signed measure partitions its space into measurable sets $P,N$ such that every measurable subset of $P$ has nonnegative measure and every measurable subset of $N$ has nonpositive measure. For finite total variation measure, set ${\mu }^{+}=(|\mu |+\mu )/2$ and choose a ${\mu }^{+}$-null set $N$ maximizing $|\mu |(N)$: a countable union of approximating null sets attains the supremum. If some subset of its complement has negative signed mass, successively remove disjoint subsets with positive mass at least half the supremum still available. Their masses tend to zero because their sum is bounded by total variation. The remainder has no positive-measure subset, has negative mass, and is ${\mu }^{+}$-null, contradicting maximality after adjoining it to $N$. On $P$ one has $|\mu |=\mu$ and on $N$ one has $|\mu |=-\mu$. Consequently $d{\mu }^{+}/d|\mu |=1_P$, an explicit Radon-Nikodym derivative.

A measurable space is a pair $(\Omega ,\mathcal{F})$ where $\mathcal{F}$ is a sigma-algebra of subsets of $\Omega$: it contains $\Omega$, is closed under complements relative to $\Omega$, and is closed under countable unions. A measure on it is a function $\mu :\mathcal{F}\to [0,\infty ]$ with $\mu (\varnothing )=0$ and countable additivity:whenever the measurable sets $A_n$ are pairwise disjoint. Infinite values are allowed. The triple $(\Omega ,\mathcal{F},\mu )$ is a measure space.

Because $E\in \mathcal{F}$, the family ${\mathcal{F}}_E$ contains $E$. If $A\in {\mathcal{F}}_E$, then $E\setminus A=E\cap (\Omega \setminus A)$ belongs to $\mathcal{F}$ and is a subset of $E$, so it belongs to ${\mathcal{F}}_E$. If each $A_n\in {\mathcal{F}}_E$, their countable union belongs to $\mathcal{F}$ and remains a subset of $E$. Thus ${\mathcal{F}}_E$ is a sigma-algebra on $E$, with relative complements taken in $E$, not in $\Omega$.

Equivalently,the trace sigma-algebra on the measurable set $E$. The measurability of $E$ is what makes this equal to the family in the question.

The function $\nu$ takes values in $[0,\infty ]$ and has $\nu (\varnothing )=0$. For any pairwise disjoint $A_n\in {\mathcal{F}}_E$, countable additivity of $\mu$ givesTherefore $\nu$ is a measure on $(E,{\mathcal{F}}_E)$: it is the restriction of a measure to a measurable subset.

Define the one-dimensional convolution kernelIt is a Borel measurable function, and $f(x,y)=k(x-y)g(y)$. The function $k(x-y)$ is Borel measurable by composition with the continuous difference map. Lifting the Lebesgue measurable function $g$ to the second coordinate preserves measurability for the product sigma-algebra, hence also for Lebesgue measure on ${\mathbb{R}}^2$. Products of measurable functions are measurable, so $f$ is Lebesgue measurable.

Near zero, $|\sin t|\le |t|$; away from zero, $|\sin t|\le 1$. ConsequentlyBoth bounds are integrable on their respective regions by power singularity integrability. In particular,

We use the following forms of the two product-integration theorems. On the product of measure spaces equipped with sigma-finite measures, Tonelli's theorem says that a nonnegative measurable function has equal product and iterated integrals, with infinity allowed. Fubini's theorem says that for an absolutely integrable measurable function, its sections are integrable for almost every fixed coordinate, and both orders of iterated integration equal the finite product integral. The Lebesgue measure on each factor is a sigma-finite measure, because the real line is a countable union of finite-length intervals.

Apply Tonelli's theorem first to $|f|$, before presuming finiteness. By translation invariance of Lebesgue measure,Thus $f$ is Lebesgue integrable on ${\mathbb{R}}^2$. The triangle inequality for an integral now gives the requested estimate:This is the L1 norm identity for a translated kernel product followed by a simple bound on the convolution kernel.

There is also a stronger signed conclusion. The integrable convolution kernel $k$ is an odd function, so ${\int }_{\mathbb{R}}k(t)dt=0$. Having established absolute integrability, we may apply Fubini's theorem to $f$ itself:Unlike part (a), the signed cancellation is justified by a finite integral of the absolute value. The constant eight is useful as an L1 norm bound even though this particular signed total vanishes.

Let $B$ be a standard planar Brownian motion started at $z$. Neighbourhood recurrence means that, with probability one, every nonempty open subset of ${\mathbb{R}}^2$ is visited at arbitrarily large times. Not hitting points means that for each fixed $w\ne z$, ${\mathbb{P}}_z(\exists t\ge 0:B_t=w)=0$. If $w=z$, time zero must be excluded, and there is still no return at any positive time. This assertion concerns each prescribed point; it does not say that the path contains no points.

We prove both claims from the planar Brownian annulus hitting probability. Fix a center $w$ and radii $0<r<\rho =|z-w|<R$, and define the stopping times ${\tau }_r=\inf \{t:|B_t-w|=r\}$ and ${\tau }_R=\inf \{t:|B_t-w|=R\}$. Their minimum $\sigma$ is finite almost surely: a coordinate of Brownian motion cannot stay forever in a bounded interval, by the Brownian reflection principle, so exit from the outer disk is finite. On the annulus,The Itô formula therefore makes $\log |B_{t\wedge \sigma }-w|$ a local martingale. It is bounded between $\log r$ and $\log R$, hence is a genuine martingale. The optional stopping theorem at $t\wedge \sigma$, followed by the bounded convergence theorem, givesThusThe exit can only be at one of the two distinct boundaries by path continuity.

For a fixed $r$, let $R\to \infty$. These events increase to $\{{\tau }_r<\infty \}$: any path up to a finite hitting time is bounded and so avoids some sufficiently large outer circle. The ratio tends to one, proving that any closed disk of positive radius is hit almost surely from any starting point outside it. An initial point inside it already counts as a hit.

To prove the stronger recurrence statement, take a closed disk contained in any prescribed nonempty open set. At each integer time $n$, the Markov property and the preceding disk-hitting result imply that some time $t\ge n$ visits this disk with probability one. Intersecting these events over all $n$ proves visits at arbitrarily large times. A countable basis of disks with rational centers and rational radii makes this simultaneous for all nonempty open sets. This is recurrence of planar Brownian motion, or equivalently neighbourhood recurrence.

For point avoidance, keep $R>\rho$ fixed and let $r\downarrow 0$. Hitting $w$ before ${\tau }_R$ requires hitting every inner circle first. The displayed ratio tends to zero, so this point-hitting event has probability zero. Any finite hit of $w$ would occur before exit from some outer circle with integer radius, again because the preceding path is bounded. A countable union now gives $\boxed{{\mathbb{P}}_z(T_w<\infty )=0(z\ne w)}$, the polar point for planar Brownian motion assertion.

Finally, even if the process starts at $w$, for each deterministic $s>0$ its normal distribution has a density, so $\mathbb{P}(B_s=w)=0$. The Markov property at $s$ and the result from other starting points exclude any visit to $w$ at or after $s$. Taking $s=1/n$ excludes all positive-time returns. Thus planar Brownian motion repeatedly approaches every point, while almost surely never hitting any prescribed point at positive time.

Number the natural numbers from one. If the power set could be enumerated as $A_1,A_2,\dots$, form $D=\{j:j\notin A_j\}$. For each $j$, membership of $j$ in $D$ is opposite to its membership in $A_j$, so $D\ne A_j$. This contradicts completeness of the enumeration. Thus the Cantor diagonal argument proves that the power set is an uncountable set.

For a finite subset $A$, define $b(A)={\sum }_{j\in A}{2}^{j-1}$. This is a nonnegative integer and the map is injective: if two finite subsets differ, let $k$ be their largest differing element. Its contribution $2^{k-1}$ exceeds the sum $1+2+\cdots +2^{k-2}$ of all smaller possible contributions, so their codes differ. There are infinitely many singleton subsets. Hence the countability of finite subsets of a countable set here is countable infinitude.

To treat all bijections, associate to every subset $A\subseteq \mathbb{N}$ the permutation that swaps $2j-1$ and $2j$ when $j\in A$, and fixes both when $j\notin A$. These disjoint pair swaps define a bijection of $\mathbb{N}$. Different subsets give different permutations, so the map from the power set into $X$ is injective. Therefore $X$ is an uncountable set.

Every finitely supported permutation fixes all elements above some $k$. A bijection fixing that tail must permute $\{1,\dots ,k\}$: it cannot send a smaller element into the fixed tail without violating injectivity. There are exactly $k!$ such permutations. Thusis a countable union of finite sets. Enumerate first by $k$, then by one of the finitely many permutations at that $k$; duplicates can be omitted. Infinitely many distinct pair swaps belong to $Y$, so $Y$ is countably infinite. The full family is uncountable, while the finitely supported family is countably infinite.

The printed symbol $E$ is not defined. Take $E=L^1(\mathbb{R})$, or more generally an $L^1$ subspace with its induced $L^1$ norm; density must be in that norm. Given $g\in E$, choose a sequence $f_n\in F$ withBy the assumed convergence on $F$, each $T_r{f}_n\to T_0{f}_n$ holds outside a null set $N_n$. Their countable union $N={\bigcup }_n{N}_n$ is still null. Thus outside $N$ the convergence statements for every member of this single approximation sequence hold simultaneously. No assertion of simultaneous convergence for the entire uncountable dense class is needed.

The Baire category theorem states that in a complete metric space, a countable intersection of open dense sets is dense. Equivalently, a nonempty complete metric space cannot be a countable union of nowhere dense sets, where a set is nowhere dense if its closure has empty interior.

Here is the proof. Let $G_1,G_2,\dots$ be open dense sets and let $U$ be any nonempty open set. Choose a point $a_1\in U\cap G_1$ and a radius $0<r_1<1/2$ such that the closed ball $B_1=\overline{B}(a_1,r_1)$ lies inside $U\cap G_1$. Inductively choose $a_{m+1}$ in the nonempty open set $B(a_m,r_m)\cap G_{m+1}$, and then a radius $0<r_{m+1}<2^{-m-1}$ such thatThus the closed balls are nested and their radii tend to zero. For $j,k\ge m$, both centers lie in $B_m$, so $d(a_j,a_k)\le 2r_m$. They form a Cauchy sequence. Completeness gives a limit $a$, and closedness puts it in every $B_m$. Consequently $a\in U\cap {\bigcap }_m{G}_m$. As $U$ was arbitrary, the intersection is dense.

To obtain the equivalent formulation, take complements of the closures of the nowhere dense sets; those complements are open dense. Conversely, complements of open dense sets are closed with empty interior. In particular, a countable closed cover of a nonempty complete metric space must contain a set with nonempty interior.

The Simplicial approximation theorem states that a continuous map between realizations of finite simplicial complexes becomes homotopic to a simplicial map after sufficiently many barycentric subdivisions of the domain. More precisely, a vertex assignment is a simplicial approximation when the image of each open star is contained in the open star of its assigned vertex; sufficiently fine subdivision permits such an assignment. The resulting straight-line homotopy stays in the appropriate target simplices.

The vertices of a barycentric subdivision are the barycentres of nonempty faces. An $n$-simplex has $2^{n+1}-1$ such faces. A top-dimensional subdivided simplex corresponds to a full flag of faces of sizes $1,2,\dots ,n+1$, equivalently to an ordering of the original vertices. ThusTriangulate both copies of the sphere using finite complexes. For each fixed subdivision level there are only finitely many vertex maps to the finite target vertex set, hence only finitely many simplicial maps. Every continuous map is homotopic to one of them at some finite level. A countable union of finite sets is countable, proving at most countably many homotopy classes without needing their classification by degree.

A subset is nowhere dense when its closure has empty interior. It is of first category, or meagre, when it is a countable union of nowhere-dense subsets; otherwise it is of second category.

The Baire category theorem says that in a complete metric space every countable intersection of open dense subsets is dense. To prove it, start inside an arbitrary nonempty open set and choose a closed ball of positive radius there and in the first dense open set. Inductively choose a closed ball inside the interior of the preceding ball and the next dense open set, with radius less than $2^{-n}$. The centres are Cauchy. Completeness gives a limit lying in all the nested closed balls, hence in all the dense open sets and the original open set. Thus a nonempty complete metric space is not meagre in itself.

For $p<r\le \infty$, put $E_m=\{x\in {\ell }^p:\Vert x{\Vert }_p\le m\}$, viewed inside ${\ell }^r$. Necessarily $p<\infty$. Each $E_m$ is closed: convergence in the ${\ell }^r$ norm gives coordinatewise convergence, and every finite partial sum of $\sum |x_i{|}^p$ is bounded by $m^p$ in the limit. Taking the supremum of partial sums proves the assertion.

It has empty interior. Given a point in it and any small ${\ell }^r$ radius, perturb a block of $N$ coordinates by entries of modulus $\delta N^{-1/r}$, choosing signs to avoid cancelling the existing coordinates. The ${\ell }^r$ distance is $\delta$, while the resulting ${\ell }^p$ norm is at least $\delta N^{1/p-1/r}$, exceeding $m$ for large $N$. For $r=\infty$, use entries of modulus $\delta$ on that block. Hence each $E_m$ is nowhere dense, andFor $r=p$, including $p=\infty$, the space is Banach and so second category in itself by Baire. There is no case $r>p$ when $p=\infty$.

A measure space consists of a set $\Omega$, a sigma-algebra $\mathcal{A}$ of subsets, and a countably additive map $\mu :\mathcal{A}\to [0,\infty ]$ with $\mu (\varnothing )=0$. A sigma-algebra contains $\Omega$ and is closed under complements and countable unions. Countable additivity means $\mu ({\bigcup }_n{A}_n)={\sum }_n\mu (A_n)$ for pairwise disjoint measurable sets. Such sets are called measurable; a real function is measurable when the inverse image of every Borel set belongs to $\mathcal{A}$.

The closed-set form of the Baire category theorem says: if a nonempty complete metric space is a countable union of closed sets, at least one of those sets has nonempty interior. Equivalently, a complete metric space cannot be a countable union of closed sets with empty interiors. In the application the space is $\mathbb{R}$ with its usual complete metric.

For each integer $n\ge 1$, the open balls $\{B(x,1/n):x\in M\}$ cover the compact metric space $M$. Compactness supplies a finite subcover with centres forming a finite set $S_n$. Define $S={\bigcup }_{n\ge 1}{S}_n$, a countable union of finite sets and therefore countable.

For any $x\in M$ and $\epsilon >0$, choose $n$ with $1/n<\epsilon$. One of the covering balls contains $x$, so there is $s\in S_n\subseteq S$ with $d(x,s)<1/n<\epsilon$. Every open neighbourhood of $x$ therefore meets $S$, giving $x\in \mathrm{cl}(S)$. Thus $\mathrm{cl}(S)=M$, and $S$ is a countable dense subset. If $M$ is empty, simply take $S=\varnothing$.

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

The closed-set version of the Baire category theorem says that if a nonempty complete metric space is a countable union of closed sets, at least one of those closed sets has nonempty interior. Equivalently, a nonempty complete metric space cannot be a countable union of closed nowhere dense sets.

Every vector space has a basis and is classified up to linear isomorphism by its dimension. Over a finite field, a vector is a finite linear combination of basis Elements. If the basis has infinite cardinality $\lambda$, the set of finite such combinations has cardinality $\lambda$: there are at least $\lambda$ basis vectors and at most a countable union of finite powers of $\lambda$, each of size $\lambda$. Thus the underlying cardinality of an infinite space equals its dimension.

Any two models of cardinality $\kappa \ge {\aleph }_0$ therefore have bases of cardinality $\kappa$. A bijection between their bases extends uniquely to an $L$-isomorphism. A space with that basis size exists, giving the required existence clause as well. Hence $T$ is categorical in every infinite cardinal.