= Continuity of probability
{title2=$A_n\uparrow A\text{ or }A_n\downarrow A\ \Longrightarrow\ \mathbb P(A_n)\to\mathbb P(A)$}
<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 $A_n,B_n\in\mathcal F$. Then
$$
\begin{aligned}
A_1\subseteq A_2\subseteq\cdots,\quad A=\bigcup_{n=1}^{\infty}A_n
&\quad\Longrightarrow\quad \mathbb P(A_n)\uparrow\mathbb P(A),\\
B_1\supseteq B_2\supseteq\cdots,\quad B=\bigcap_{n=1}^{\infty}B_n
&\quad\Longrightarrow\quad \mathbb P(B_n)\downarrow\mathbb P(B).
\end{aligned}
$$
The <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 $\mathbb P(B_1)\leq1$.<Probability 1 lecture notes, Chapter 1: Axioms of probability by Durham Statistics and Probability teaching team>{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,
$$
\mathbb P(A_n)=1-2^{-n},\qquad \mathbb P(B_n)=2^{-n},\qquad
\mathbb P\left(\bigcup_{n=1}^{\infty}A_n\right)=1,\qquad
\mathbb P\left(\bigcap_{n=1}^{\infty}B_n\right)=0.
$$
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 $1/512>1/1000$.
\Image[/probability-axioms-continuity-nested-limits.png]
{title=Exact fair-coin model: success by toss n increases to probability one, while the all-tails prefix probability decreases to zero}
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 $n\geq2$. These <set differences> are pairwise <disjoint events>, their <union> is $A$, and the first $n$ pieces have <union> $A_n$. Consequently
$$
\mathbb P(A)=\sum_{j=1}^{\infty}\mathbb P(D_j)
=\lim_{n\to\infty}\sum_{j=1}^{n}\mathbb P(D_j)
=\lim_{n\to\infty}\mathbb P(A_n).
$$
For decreasing <events>, their <complements> increase to $B^c$; applying the increasing result and the <complement rule of probability> gives $\mathbb P(B_n)=1-\mathbb P(B_n^c)\to1-\mathbb P(B^c)=\mathbb P(B)$. The finite-total-mass condition is essential for this subtraction. For comparison, the <counting measure> of each <set> $\{n,n+1,\ldots\}$ 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 $R_n\downarrow\varnothing\Rightarrow\mathbb P(R_n)\to0$ satisfies <countable additivity>. For pairwise <disjoint events> $D_j$ with $E=\bigcup_jD_j$, the remainders $R_n=E\setminus\bigcup_{j=1}^nD_j$ decrease to the empty <set>, and
$$
\mathbb P(E)=\sum_{j=1}^{n}\mathbb P(D_j)+\mathbb P(R_n).
$$
Taking 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> $C_n=\bigcup_{j\geq n}E_j$ decrease, and the <union bound> gives $\mathbb P(C_n)\leq\sum_{j\geq n}\mathbb P(E_j)\to0$. <Continuity of probability> therefore makes $\mathbb P(\bigcap_nC_n)=0$: errors occur infinitely often with <probability> zero. This is the <Borel-Cantelli first lemma>, and it requires no <independence>. For the illustrative bounds $\mathbb P(E_j)\leq2^{-j}$, 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> $\{X\leq x+1/n\}$ decrease to $\{X\leq x\}$. 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>.
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