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If $E_n\downarrow E$ and $\mu (E_1)<\infty$, then $\mu (E_n)\downarrow \mu (E)$. Apply continuity from below of a measure to the increasing differences $E_1\setminus E_n$.

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 nonnegative measurable function $f:X\to [0,\infty ]$ on a measure space, defineThe supremum may be infinite. A simple function assigns finitely many heights to measurable sets, so its integral is height times measure, summed over those pieces. This definition groups points by their values rather than requiring a partition into short intervals. See MIT lecture 11, Definition 108.

A real-valued function is Lebesgue integrable when ${\int }_X|f|d\mu <\infty$. Its integral is ${\int }_X{f}_{+}d\mu -{\int }_X{f}_{-}d\mu$, where $f_{+}=\max (f,0)$ is its positive part and $f_{-}=\max (-f,0)$. This finiteness requirement avoids subtracting two infinities. For functions taking values in the complex numbers, integrate the real part and imaginary part separately.

Consider the Dirichlet function $f=1_{\mathbb{Q}\cap [0,1]}$ on $[0,1]$. It is a simple function, andIts Riemann integral does not exist. For every partition of an interval $0=x_0<\cdots <x_n=1$, each nondegenerate $[x_{i-1},x_i]$ contains both rational numbers and irrational numbers, by their respective density of the rational numbers and density of the irrational numbers. Thus its infimum there is $0$ and its supremum is $1$. The lower Darboux sum is ${\sum }_{i}0(x_i-x_{i-1})=0$, whereas the upper Darboux sum is ${\sum }_{i}1(x_i-x_{i-1})=1$. Refinement never closes this gap. A finite scatterplot cannot represent this behavior: the assertion concerns every subinterval and every partition of an interval.

Another view builds the Lebesgue integral from horizontal layers. For any nonnegative measurable function, the layer cake representation statesBoth sides may be infinite. The right side integrates the measure of a superlevel set against Lebesgue measure in the height variable $t$. See Gautam Iyer's measure theory notes, Theorem 9.5.

For $f(x)=x^2$ on $[0,1]$ with Lebesgue measure, the superlevel set at $0\le t<1$ is $(\sqrt{t},1]$, of Lebesgue measure $1-\sqrt{t}$; for $t\ge 1$ it is empty. ConsequentlyAt height $t=1/4$, the layer extends from $x=1/2$ to $x=1$, a width of $1/2$. The picture shows why the same area can be accumulated vertically or horizontally. In a probability space, the identical formula expresses the expected value of a nonnegative random variable through its exceedance probabilities.

A proof outline avoids extra assumptions on the measure space. For a nonnegative simple function $s={\sum }_j{a}_j{1}_{A_j}$ with disjoint measurable sets $A_j$, the superlevel set has measure ${\sum }_{a_j>t}\mu (A_j)$. Integrating this finite sum gives ${\sum }_j{a}_j\mu (A_j)=\int sd\mu$. Next use approximation by nonnegative simple functions $s_n\uparrow f$. For every $t$, the superlevel sets $\{s_n>t\}$ increase to $\{f>t\}$. Continuity from below of a measure and the monotone convergence theorem on the height variable give the formula for $f$.

The uniqueness theorem for measures says that two measures agreeing on a generating pi-system agree on its generated sigma-algebra, provided they are finite with the same total mass, or there is a common increasing exhaustion by pi-system sets of finite measure.

For the finite case let $\mathcal{D}=\{A:\mu (A)=\nu (A)\}$. Equality of total masses gives closure under complements, and countable additivity gives closure under disjoint countable unions. Thus $\mathcal{D}$ is a Dynkin system containing the generating pi-system. The pi-lambda theorem puts the entire generated sigma-algebra inside $\mathcal{D}$.

For the exhaustion version, let $E_n\uparrow E$ with $E_n$ in the pi-system and $\mu (E_n)=\nu (E_n)<\infty$. The finite measures $A\mapsto \mu (A\cap E_n)$ and $A\mapsto \nu (A\cap E_n)$ agree on the generating pi-system, since it is intersection-closed. The finite proof applies, and continuity from below of a measure as $n\to \infty$ gives $\mu (A)=\nu (A)$ for every measurable $A$. A finiteness hypothesis is essential to this standard extension-uniqueness statement.

Assume $\theta (1/2)>0$. The event that some infinite percolation cluster exists is invariant under lattice translations. By translation ergodicity of Bernoulli percolation, its positive probability is therefore one.

Let $D_N$ be the event that an infinite percolation cluster meets the finite graph vertex box $B_N=[-N,N{]}^2$. These increasing events increase with $N$, and their union is the event that some infinite percolation cluster exists. Continuity from below of a measure givesAn infinite percolation cluster meeting $B_N$ contains an infinite simple open ray in a graph from a graph vertex of the box. This follows from local finiteness and the König infinity lemma. That ray visits the finite box only finitely many times; its last box graph vertex lies on the boundary, and the subsequent ray is outside. Thus $D_N$ is also the event that the box boundary intersects an infinite percolation cluster, and is the union of the four exterior-side arm events.

Consequently, for every $ϵ>0$, choose $N$ withFor $ϵ\ge 1$ any sufficiently large $N$ still satisfies the strict inequality. No uniqueness assertion is needed for this preliminary step.

Work on the common probability space from part (c). For a configuration of all the uniforms, the set $S=\{r\in [0,1]:I_r\text{ occurs}\}$ is an upper interval, possibly with or without its lower endpoint. It is nonempty because $I_1$ occurs, and the coupled onset parameter for origin percolation is $M=\inf S$. For $0<p\le 1$,Indeed, occurrence at some $r<p$ implies $M<p$. Conversely, if $M<p$, the definition of infimum yields occurrence at some $s<p$, and a rational $r$ between $s$ and $p$ also satisfies $I_r$. This also proves measurability of $M$ through its strict sublevel sets.

Choose a rational sequence increasing to $p$. By continuity from below of a measure and the monotone coupling of Bernoulli percolation,Moreover, $\{M<p\}\subseteq I_p\subseteq \{M\le p\}$. Hence $I_p$ is the disjoint union of $\{M<p\}$ and $I_p\cap \{M=p\}$, givingIt is important to retain the intersection with $I_p$: occurrence at the infimum is not automatic. At $p=0$ there is no left limit within the parameter domain, so the displayed left-limit assertion is for $p>0$.

The pointwise limit $f$ is a measurable function, since it can be expressed as ${\sup }_n{\inf }_{m\ge n}{f}_m$. Thus each set $E_n^{(k)}$ is a countable intersection of measurable sets within $A$. Deleting the condition at index $m=n$ weakens the requirements, whereas replacing $1/k$ by $1/(k+1)$ strengthens them. HenceThese are inclusions, not necessarily strict inclusions. The displayed sets describe eventual uniform error tolerances at an individual point.

Fix $k\ge 1$ and $x\in A$. By pointwise convergence, there is an $N=N(x,k)$ such that $|f_m(x)-f(x)|\le 1/k$ for every $m\ge N$. This is exactly $x\in E_N^{(k)}$, provingTogether with monotonicity in $n$ and finite measure of $A$, continuity from below of a measure gives $\lambda (E_n^{(k)})\uparrow \lambda (A)$. ConsequentlyThis quantitative form is what allows a single large-measure set to work for every tolerance, rather than choosing a different exceptional set for each point.

Put $I=\{x:|C(x)|=\infty \}$, the union of all infinite percolation clusters. Since $p>p_c$, the percolation probability $\theta (p)$ is positive. The translation ergodicity of Bernoulli percolation makes $\{I\ne \varnothing \}$ a zero-one event; it has positive probability because ${\mathbb{P}}_p(0\in I)=\theta (p)>0$, hence it occurs almost surely.

Let $D_m=\{B(m)\cap I\ne \varnothing \}$. These events increase to $\{I\ne \varnothing \}$, so continuity from below of a measure gives ${\mathbb{P}}_p(D_m)\to 1$. On $D_m$, some vertex of $B(m)$ has an infinite open path in a graph. For every $N\ge m$, the segment up to its first visit to $\partial B(N)$ lies entirely in $B(N)$. Thus even with this restriction on the connecting path in a graph,This proves the requested uniformity and supplies the version needed for crossings later.

The Beppo Levi theorem, or monotone convergence theorem, states that if $(f_n)$ is a sequence of nonnegative measurable functions on a measure space andpointwise almost everywhere, then $f$ is measurable andwhere either side may be infinite.

Measurability of $f={\sup }_n{f}_n$ follows from measurability of countable suprema. Since $f_n\le f$, monotonicity of the Lebesgue integral givesFor the reverse inequality, let $s$ be a nonnegative simple function with $s\le f$, and fix $0<c<1$. DefineThen $E_n\uparrow X$ up to the null set on which convergence fails. Since $f_n\ge c,s{1}_{E_n}$,by continuity from below of a measure, applied to the finitely many level sets of $s$. Hence ${\lim }_n\int f_n\ge c\int s$. Letting $c\uparrow 1$ and taking the supremum over all simple $s\le f$, as in the definition of the Lebesgue integral, gives the reverse inequality and proves the theorem.

First, $\mathcal{B}\subseteq \mathcal{C}$ by taking $B=A$. If $A\in \mathcal{C}$, thenso $\mathcal{C}$ is closed under complements. Now let $A_i\in \mathcal{C}$ and put $A={\bigcup }_i{A}_i$. Because $\mu (X)<\infty$, continuity from below of a measure gives an $N$ such thatChoose $B_i\in \mathcal{B}$ with $\mu (A_i△B_i)<ϵ/(2N)$ and let $B={\bigcup }_{i=1}^N{B}_i\in \mathcal{B}$. The union bound for symmetric differences givesThus $\mathcal{C}$ is a sigma-algebra containing $\mathcal{B}$, and

Finiteness is essential. Take $X=\mathbb{N}$ with counting measure and let $\mathcal{B}$ be the finite-cofinite Boolean algebra of sets. Then $\sigma (\mathcal{B})=\mathcal{P}(\mathbb{N})$. For $ϵ<1$, the condition $\mu (A△B)<ϵ$ forces $A=B$, so $\mathcal{C}=\mathcal{B}$. Any infinite coinfinite subset of $\mathbb{N}$ belongs to $\sigma (\mathcal{B})$ but not to $\mathcal{C}$.

Within the monotone coupling, $I_p$ increases with $p$ andTaking a countable cofinal sequence $p^{\prime }\uparrow p$ and using continuity from below of a measure givesSince $\{M<p\}\subseteq I_p$, subtraction from $\theta (p)=\mathbb{P}(I_p)$ proves

Every finite positive Borel measure on Euclidean space is inner regular by compact sets and outer regular by open sets. For an open set $U$, the compact sets $\{x:|x|\le n,\mathrm{dist}(x,U^c)\ge 1/n\}$ increase to $U$, so continuity from below of a measure gives inner approximation. The empty complement is handled by compact balls. The class of Borel sets admitting both approximations is closed under complements: interchange inner and outer approximations, then truncate the inner closed set by a large compact ball, using finiteness of the measure. For a countable union, approximate its components from outside with summable errors, and approximate finitely many components from inside after making the remaining union's measure small. Thus this class is a sigma-algebra containing the open sets. In particular, a finite measure carried by a Borel null set has compact subsets of that carrier capturing arbitrarily nearly all its mass.