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Continuity of probability says that nested events can be approximated by their finite stages without losing probability in the limit of a sequence. In a probability space $(\Omega ,\mathcal{F},\mathbb{P})$, let $A_n,B_n\in \mathcal{F}$. ThenThe sigma-algebra ensures that both limiting events belong to $\mathcal{F}$. These are continuity from below of a measure and continuity from above of a measure. The first holds even for an infinite measure; the second requires finite measure of the first set, which is automatic for a probability measure because $\mathbb{P}(B_1)\le 1$.^[ref]

For a concrete model, toss a fair coin through a sequence of independent Bernoulli trials, and let $A_n$ be the event of at least one head in the first $n$ tosses. Its complement $B_n=A_n^c$ is the event of $n$ initial tails. On the sample space of infinite head-or-tail sequences,Every finite waiting horizon has positive tail probability, yet the event of waiting forever has probability zero. That intersection is not empty: it contains the all-tails sequence. Thus a zero-probability event need not be impossible as an outcome, and eventual success almost surely does not supply a deterministic deadline. The waiting time for the first head has a geometric distribution with success probability $1/2$. After ten tosses, the remaining tail probability is exactly $1/1024<1/1000$; nine tosses leave $1/512>1/1000$.

A proof outline exposes what makes countable additivity stronger than finite additivity of a set function. For increasing events, put $D_1=A_1$ and $D_n=A_n\setminus A_{n-1}$ for $n\ge 2$. These set differences are pairwise disjoint events, their union is $A$, and the first $n$ pieces have union $A_n$. ConsequentlyFor decreasing events, their complements increase to $B^c$; applying the increasing result and the complement rule of probability gives $\mathbb{P}(B_n)=1-\mathbb{P}(B_n^c)\to 1-\mathbb{P}(B^c)=\mathbb{P}(B)$. The finite-total-mass condition is essential for this subtraction. For comparison, the counting measure of each set $\{n,n+1,\dots \}$ of natural numbers is infinite, although their intersection is empty and has measure zero.

Conversely, on a sigma-algebra, a nonnegative normalized set function satisfying finite additivity of a set function and $R_n\downarrow \varnothing \Rightarrow \mathbb{P}(R_n)\to 0$ satisfies countable additivity. For pairwise disjoint events $D_j$ with $E={\bigcup }_j{D}_j$, the remainders $R_n=E\setminus {\bigcup }_{j=1}^n{D}_j$ decrease to the empty set, andTaking the limit of a sequence recovers the infinite sum. This equivalence explains why continuity of probability is part of the foundations rather than a separate assumption about smooth curves.

One useful application controls infinitely many possible errors. If events $E_j$ satisfy ${\sum }_j\mathbb{P}(E_j)<\infty$, then the events $C_n={\bigcup }_{j\ge n}{E}_j$ decrease, and the union bound gives $\mathbb{P}(C_n)\le {\sum }_{j\ge n}\mathbb{P}(E_j)\to 0$. Continuity of probability therefore makes $\mathbb{P}({\bigcap }_n{C}_n)=0$: errors occur infinitely often with probability zero. This is the Borel-Cantelli first lemma, and it requires no independence. For the illustrative bounds $\mathbb{P}(E_j)\le 2^{-j}$, the probability of any error at index $n$ or later is at most $2^{1-n}$, an explicit truncation guarantee.

The word continuity here concerns nested sets, not the absence of atoms in a probability distribution. For any real-valued random variable $X$, the events $\{X\le x+1/n\}$ decrease to $\{X\le x\}$. Hence its cumulative distribution function is right-continuous even when it has jumps, as for a geometric distribution. The nesting hypothesis matters: an arbitrary sequence of events need not admit either displayed interchange of limit and probability.

The cumulative distribution function of a real random variable $X$ isIt records the probability accumulated up to a threshold. It is a nondecreasing function and is right-continuous; its limits are zero at $-\infty$ and one at $+\infty$. These properties follow from ordered threshold events, continuity from above of a measure, and continuity from below of a measure. A cumulative distribution function determines the entire probability distribution on the real line. ^[ref].

For $a<b$,where $F_X(x^{-})={\lim }_{t\uparrow x}{F}_X(t)$ is its left-hand limit. The second formula measures a jump. Thus $\mathbb{P}(a\le X\le b)=F_X(b)-F_X(a^{-})$; changing an endpoint is harmless only when its point probability is zero.

For the finite example under probability distribution, $F_X(x)$ is $0$ below zero, $0.2$ on $[0,1)$, $0.7$ on $[1,3)$, and $1$ on $[3,\infty )$. In particular $F_X(1)=0.7$, but $F_X(1^{-})=0.2$, so the jump gives $\mathbb{P}(X=1)=0.5$. For the density example, $F_Y(y)=y^2$ on $[0,1]$, and its increase from $1/4$ to $3/4$ is $9/16-1/16=1/2$.

When an absolutely continuous probability distribution has probability density function $f$, its cumulative distribution function is $F(x)={\int }_{-\infty }^{x}f(t)dt$ and $F^{\prime }=f$ almost everywhere. A continuous cumulative distribution function by itself does not guarantee such a density: the Cantor distribution is an important exception.

For a Lebesgue measurable set $E\subseteq \mathbb{R}$ of finite measure, the displayed function satisfies $|H_E(t)-H_E(s)|\le |t-s|$. Indeed, for $s<t$, finite additivity and monotonicity of a measure give $H_E(t)-H_E(s)=m(E\cap (s,t])\le t-s$. Thus $H_E$ is a nondecreasing Lipschitz function and is uniformly continuous even if $E$ is unbounded.

Continuity from above of a measure gives $H_E(t)\to 0$ as $t\to -\infty$, while continuity from below of a measure gives $H_E(t)\to m(E)$ as $t\to \infty$. Every strictly intermediate value is attained by the intermediate value theorem. The corresponding half-line cut of $E$ realizes any prescribed measure between zero and $m(E)$; the endpoints are realized by the empty set and $E$ itself.

For independent identically distributed mean-zero variables with finite positive variance, the central limit theorem gives a positive eventual lower bound for $\mathbb{P}(S_n/\sqrt{n}\ge K)$ for every finite $K$. The decreasing events ${\bigcup }_{n\ge m}\{S_n/\sqrt{n}\ge K\}$ have probabilities bounded below, so continuity from above of a measure gives positive probability that the limsup is at least $K$. The limsup itself is unchanged by removing any finite initial sum, hence is tail measurable. Kolmogorov zero-one law makes each event “limsup at least $K$” certain; intersect over integer $K$. The result is false for zero variance. One should apply the zero-one law to the limsup, rather than assume the exact threshold infinitely-often event is unaffected by a vanishing change.

Choose a strictly decreasing positive sequence $q_0=1,q_1,q_2,\dots$. Begin with $C_0=[0,1]$ and at step $n+1$ keep two closed end intervals of length $q_{n+1}/2^{n+1}$ from every component interval of $C_n$. This constructs a compact set $C={\bigcap }_n{C}_n$ withThe second equality is continuity from above of a measure. Every stage has positive gaps and maximum component length at most $2^{-n}$, so $C$ contains no nondegenerate interval and is a nowhere dense set. It has no isolated points: both endpoints of the stage-$n$ component containing a point remain in $C$, and at least one distinct endpoint is within $2^{-n}$. Thus its topological sparseness is compatible with positive Lebesgue measure.

Taking $q_n=(2/3{)}^n$ gives the usual Cantor set of measure zero. Taking $q_n=1/2+2^{-(n+1)}$ gives a generalized construction of measure $1/2$. This is one useful family of generalized constructions; the removed intervals need not have the same proportions at every stage.

For the increasing family, define the measurable setsThey are pairwise disjoint, with $A_N={\bigcup }_{n=1}^N{D}_n$ and ${\bigcup }_n{A}_n={\bigcup }_n{D}_n$. By countable additivity,The nonnegative series is defined by its increasing partial-sum limit, so this proof also covers an infinite value. This is continuity from below of a measure.

For the decreasing family, put $C_n=B_1\setminus B_n$. These measurable sets increase, andSince $\mu (B_1)<\infty$, finite additivity gives legitimate finite subtractions:Apply the first result to $C_n$ and subtract from $\mu (B_1)$ to obtainThis is continuity from above of a measure. Its finiteness hypothesis matters: for Lebesgue measure and $B_n=[n,\infty )$, every $B_n$ has infinite measure but their intersection is empty. The proof must not subtract $\infty -\infty$.

For $x<y$, finite additivity on the disjoint pieces of the half-line givesBy monotonicity of a measure and the Lebesgue measure of an interval,Exchanging the two points if necessary provesThus the cumulative measure of a finite-measure set is a Lipschitz function with Lipschitz constant at most one, hence is uniformly continuous: given $\epsilon >0$, the choice $\delta =\epsilon$ works simultaneously at every point.

To locate its endpoint limits, the measurable sets $E\cap (-\infty ,N]$ increase to $E$, so continuity from below of a measure gives $h(N)\to m(E)$. The measurable sets $E\cap (-\infty ,-N]$ decrease to the empty set and all have finite measure, so continuity from above of a measure gives $h(-N)\to 0$. Since $h$ is nondecreasing,For $0<c<m(E)$, choose $R>0$ with $h(-R)<c<h(R)$. The intermediate value theorem on $[-R,R]$ gives $t$ with $h(t)=c$. ThenThe endpoint values $c=0,m(E)$ can also be achieved by $A=\varnothing ,E$. This proves divisibility of Lebesgue measure for arbitrary finite-measure Lebesgue measurable sets on the real line, including unbounded ones. The non-atomic measure structure permits prescribed fractions of measure; a measure concentrated at one point would not have this property.

Set $C_0=[0,1]$. Inductively, replace each component interval of $C_n$ by its two closed end intervals, each of length ${\ell }_{n+1}:=q_{n+1}/2^{n+1}$. The current component length is ${\ell }_n=q_n/2^n$. Because $q_{n+1}<q_n$, the removed central gap has positive lengthThis is a generalized Cantor set construction. The three required properties are checked in parts i, ii and iii.

The sets are closed, measurable, decreasing, and contained in $[0,1]$, which has finite Lebesgue measure. By continuity from above of a measure,This limit exists because $(q_n)$ is a bounded decreasing monotone sequence. To exclude a nondegenerate interval $J\subseteq C$, note that its connectedness forces it into a single component interval of every $C_n$: crossing two components would cross a removed gap. Its length would therefore be at most $2^{-n}$ for every $n$, which is impossible for a positive-length interval. Thus $C$ contains no nondegenerate interval, even when its Lebesgue measure is positive.

The exact criterion isNecessity follows already from one approximation with error less than one: a compact set $K\subseteq \mathbb{R}$ is bounded, hence has finite Lebesgue measure, so $m(E)=m(K)+m(E\setminus K)<\infty$.

Conversely, suppose $m(E)<\infty$. By continuity from above of a measure, the decreasing sets $E\setminus [-N,N]$ have measures tending to zero. Choose $N\ge 1$ so that $m(E\setminus [-N,N])<\epsilon /2$. Put $B=[-N,N]\setminus E$. By part (b)(i), there is an open set $V\supseteq B$ with $m(V\setminus B)<\epsilon /2$. DefineIt is closed and bounded, hence a compact set by the Heine-Borel theorem, and $K\subseteq E$. The discarded portion satisfiesThe second term is at most $m(V\setminus B)$. This proves the claimed inner regularity of Lebesgue measure with an arbitrarily small absolute error.

For $\epsilon >0$, define the measurable decreasing tail eventsIf $f_n\to f$ almost everywhere, every point outside its null exceptional set eventually leaves each such tail event, so $\mu ({\bigcap }_n{A}_n(\epsilon ))=0$. Continuity from above of a measure applies because $\mu (E)<\infty$, giving $\mu (A_n(\epsilon ))\to 0$.

Conversely, assume these measures tend to zero. For each positive integer $k$, the intersection ${\bigcap }_n{A}_n(1/k)$ has measure zero. Outside the union of these countably many null sets, every $k$ has an index after which $|f_m-f|\le 1/k$ for all later $m$. Hence $f_m\to f$ there. This proves the desired equivalence. The argument uses all positive thresholds; replacing the intersection of tail events by a single strict limsup inequality at the same threshold would mishandle equality cases.

Let $D_n=B_n\setminus B_{n+1}$. These sets are pairwise disjoint andBy countable additivity, ${\sum }_n\mu (D_n)=\mu (B_1)-\mu (B)<\infty$. The tails of this convergent nonnegative series tend to zero, soThis proves continuity from above of a measure and explains the finite-measure assumption.

Continue with ${\sigma }^2>0$. For each integer $K\ge 1$, let $B_m(K)={\bigcup }_{n\ge m}\{S_n/\sqrt{n}\ge K\}$. These events decrease with $m$, and their probabilities are at least $c_K>0$ by (iii). Continuity from above of a measure givesIt is important not to assume that this exact threshold event is automatically a tail event: a vanishing change of partial sums can affect infinitely many exact comparisons at the threshold. Instead useFor any fixed $m$, subtracting $S_m/\sqrt{n}\to 0$ shows that $L={\mathrm{limsup}}_{n\to \infty }(X_{m+1}+\cdots +X_n)/\sqrt{n}$. Thus $L$ is measurable with respect to every tail sigma-algebra. The event $\{L\ge K\}$ has positive probability by the preceding bound, so Kolmogorov zero-one law makes its probability one. Intersecting over all positive integers givesFor any specified real $K>0$, this conclusion implies infinitely many $n$ with $S_n/\sqrt{n}\ge K$, completing both requests. In the excluded zero-variance case the limsup is zero.

Use consistently wired boundary conditions in increasing finite boxes and let ${\varphi }_{p,q}^{\mathrm{w}}$ be the usual wired infinite-volume random-cluster measure. The finite-volume comparison just proved applies with boundary graph vertices identified, since it is still a comparison on a finite graph. Passing to the infinite-volume limit preserves the inequality for every increasing event determined by finitely many edges.

For fixed $m$, let $A_m$ be the event that the origin connects by an open path in $[-m,m{]}^2$ to its boundary. It is such a finite increasing event. Therefore, for $q_1\le q_2$,The events decrease to the event that the origin has an infinite percolation cluster, by local finiteness. Continuity from above of a measure givesDefine the random-cluster critical probability by $p_c(q)=\inf \{p:{\theta }^{\mathrm{w}}(p,q)>0\}$. The set in which percolation occurs at $q_2$ is a subset of that at $q_1$, whenceThe same proof works if free boundary conditions are used consistently throughout. It compares critical probabilities; it does not presume whether an infinite percolation cluster exists at the critical parameter itself.

First, the Cramér model has infinitely many successes almost surely, so every $P_n$ is finite. Indeed, for each fixed $K\ge 3$, independence of the Bernoulli random variables givesThe sum diverges, for example because $1/\log j\ge 1/j$ for $j\ge 3$. Continuity from above of a measure and a countable union bound exclude a final success.

Fix $ϵ>0$ and put $L_k=\lceil (1+ϵ)(\log k{)}^2\rceil$. Let $A_k$ be the event that the whole interval $k+1,\dots ,k+L_k$ consists of failures. By independence and $1-v\le e^{-v}$,Since $L_k=o(k)$, the exponent divided by $\log k$ tends to $-(1+ϵ)$. Thus $\mathbb{P}(A_k)\le k^{-1-ϵ/2}$ for all sufficiently large $k$. The Borel-Cantelli first lemma shows that almost surely every sufficiently large such interval contains a success. Taking $k=P_n$ then gives $P_{n+1}-P_n\le L_{P_n}$ for all sufficiently large $n$.

For each fixed $ϵ$, the limit superior of the normalized gap is therefore at most $1+ϵ$ almost surely. Intersect the probability-one events for $ϵ=1/r$, $r\in \mathbb{N}$, to obtain the Cramér model prime-gap upper bound:

Let $A_n$ be the event that the origin has an open graph path to the boundary of $[-n,n{]}^d\cap {\mathbb{Z}}^d$. The graph path can be stopped on its first boundary visit, so $A_n$ depends on only finitely many sites. Its probability $f_n(p)={\mathbb{P}}_p(A_n)$ is a finite sum of terms $p^a(1-p{)}^b$, and is therefore a continuous function of $p$. Also $A_{n+1}\subseteq A_n$, and the König infinity lemma gives ${\bigcap }_n{A}_n=\{0\leftrightarrow \infty \}$. By continuity from above of a measure,The monotone coupling of Bernoulli percolation described below shows that $\theta$ is nondecreasing. For $p<1$, its right limit of a function $\theta (p+)$ exists and is at least $\theta (p)$. For every fixed $n$,Taking the infimum in $n$ proves $\theta (p+)=\theta (p)$. This proves right continuity of percolation probability throughout $[0,1)$; at $1$ right continuity is understood relative to the parameter domain. The argument does not interchange two uncontrolled limits.

We use the uniqueness of the infinite percolation cluster on ${\mathbb{Z}}^d$: for $p>p_c$, there is almost surely exactly one infinite percolation cluster, denoted $I$. Fix $m$ and let $N>m$. Two observations account for the possible distinct endpoints of the face connections.

First, the probability that some finite percolation cluster meeting $B(m)$ reaches $\partial B(N)$ tends to zero as $N\to \infty$. There are only finitely many vertices in $B(m)$, and each of their finite percolation clusters has finite radius; apply the union bound and continuity from above of a measure.

Second, all vertices of $I\cap B(m)$ are joined to one another inside $B(N)$ with probability tending to one. On each configuration, this is a finite set of vertices of one connected percolation cluster. Choose a finite open path in a graph from one such vertex to each of the others; the union of these finitely many paths is contained in some finite box. The assertion is vacuous if the set is empty.

By part (b.ii), each of the two face-connection events within $B(N)$ fails with probability at most ${\delta }_m^{1/(2d)}$. Outside the two exceptional events just described, their endpoints in $B(m)$ belong to $I$ and can be joined inside $B(N)$. Concatenating the left-face path, that joining path, and the right-face path produces a crossing of $B(N)$. ConsequentlyNow let $m\to \infty$. Since ${\delta }_m\to 0$,The order of limits matters: $m$ is fixed while the finitely many relevant percolation clusters are connected inside increasingly large boxes.

The events $A_n$ decrease, and an open path from $0$ reaches every $\partial {\Lambda }_n$ exactly when the open cluster of $0$ is infinite. Therefore continuity from above of a measure givesEach $A_n$ depends on finitely many sites, so $p\mapsto {\mathbb{P}}_p(A_n)$ is a polynomial and hence continuous. Given $\epsilon >0$, choose $n$ with ${\mathbb{P}}_p(A_n)<\theta (p)+\epsilon$. For $p^{\prime }\downarrow p$, monotonicity and the finite-event continuity yieldThus $\theta (p^{\prime })\to \theta (p)$ from the right.

A sigma-algebra $\mathcal{F}$ on $\Omega$ contains $\Omega$, is closed under complements relative to $\Omega$, and is closed under countable unions. Consequently it contains the empty set and is closed under countable intersections. A measure space is a triple $(\Omega ,\mathcal{F},\mu )$ with a map $\mu :\mathcal{F}\to [0,\infty ]$ satisfying $\mu (\varnothing )=0$ and countable additivity on pairwise disjoint measurable sets.

For the increasing sequence, set $C_1=A_1$ and $C_n=A_n\setminus A_{n-1}$ for $n\ge 2$. These measurable sets are disjoint, their union is $A$, and $A_N={\bigcup }_{n=1}^N{C}_n$. Thus countable additivity and finite additivity giveThe identity permits the value infinity; it uses no subtraction of infinite measures. This is continuity from below of a measure.

For a decreasing sequence, the analogous assertion is not always true. On $\Omega =\mathbb{N}$ with the counting measure, take $B_n=\{n,n+1,\dots \}$. Then ${\bigcap }_n{B}_n=\varnothing$, so $\mu (B)=0$, whereas $\mu (B_n)=\infty$ for every $n$ and their limit is infinity.

A sufficient extra condition for continuity from above of a measure is $\mu (B_1)<\infty$. Indeed $B_1\setminus B_n$ increases to $B_1\setminus B$, and the increasing-sequence result givesMore generally it suffices that one of the $B_n$ has finite measure. The missing finiteness assumption is exactly what the counterexample violates.

For independent bond percolation on an infinite graph with a specified root $o$, let $C_o$ be its percolation cluster. The percolation probability isIt concerns the root, rather than the existence of an infinite percolation cluster somewhere else. Even above the percolation critical probability of the square lattice, an isolated root has probability $(1-p{)}^4$, because its four incident edges can all be closed. Thus $\theta (p)\le 1-(1-p{)}^4<1$ when $p<1$. Positive percolation probability does not mean every vertex is connected to infinity.

On the square lattice, the events that $o$ is joined by an open path to the boundary of $[-n,n{]}^2$ decrease as $n$ increases. Their probabilities converge to $\theta (p)$ by continuity from above of a measure. This relates an infinite question to finite observations, without equating any one finite boundary connection with survival forever. The monotone coupling of Bernoulli percolation shows that $\theta$ is a monotone function of $p$.

For independent site percolation or bond percolation on a rooted locally finite graph, $\theta (p)$ is right continuous on $[0,1)$. Finite-radius connection events have polynomial probabilities in $p$ and decrease to infinite connection. Thus $\theta$ is their infimum by continuity from above of a measure. This infimum of continuous functions is upper semicontinuous; parameter monotonicity from the monotone coupling of Bernoulli percolation then gives right continuity.