Almost-Sure Convergence Without Summable Error Probabilities

When a Divergent Sum Still Allows Almost-Sure Convergence

A summable sequence of error probabilities is a useful sufficient condition for almost-sure convergence: if the probabilities of errors add to a finite total, only finitely many errors occur almost surely. But the reverse implication does not hold. An error probability sum can diverge even when the errors eventually stop along almost every sample path.

The key is dependence. In this tutorial, the error events are nested: once a sample path leaves the sequence of events, it cannot return. Their probabilities can decrease slowly enough to have a divergent sum, while the probability of belonging to every event is zero. We develop that mechanism using indicator random variables.

Definition: A sequence of events \(A_1,A_2,\ldots\) is decreasing or nested if \(A_{n+1}\subseteq A_n\) for every \(n\). Its indicators are \(X_n=\mathbf{1}_{A_n}\), where \(\mathbf{1}_{A_n}=1\) on \(A_n\) and \(0\) outside \(A_n\). The events occur infinitely often on the event \(\limsup_{n\to\infty} A_n=\bigcap_{m=1}^{\infty}\bigcup_{n\ge m} A_n\).

For nested events, occurring infinitely often has a particularly simple meaning. If a sample point is absent from one event \(A_m\), it is absent from every later event because \(A_n\subseteq A_m\) whenever \(n\ge m\). Thus it can belong to infinitely many of the events only if it belongs to every event.

\limsup_{n\to\infty} A_n=\bigcap_{n=1}^{\infty}A_n \qquad\text{when } A_{n+1}\subseteq A_n.

The Nested-Event Criterion

For a decreasing sequence of events, the probability of the event that they all occur is determined by the limit of their individual probabilities. This is continuity of probability for decreasing sequences: if \(P(A_1)<\infty\), then \(P(\bigcap_n A_n)=\lim_n P(A_n)\). The condition holds automatically on a probability space.

Theorem: Suppose \(A_{n+1}\subseteq A_n\) for all \(n\), and \(P(A_n)\to 0\). Then \(X_n=\mathbf{1}_{A_n}\xrightarrow{a.s.}0\). If also \(\sum_{n=1}^{\infty}P(A_n)=\infty\), the indicators still converge almost surely to zero despite the divergent sum.

Proof. Since the events decrease and \(P(A_1)\le 1\), continuity of probability from above gives

P\left(\bigcap_{n=1}^{\infty}A_n\right) =\lim_{n\to\infty}P(A_n)=0.

For any sample point \(\omega\) outside \(\bigcap_n A_n\), there is some \(m\) such that \(\omega\notin A_m\). For every \(n\ge m\), nesting gives \(A_n\subseteq A_m\), so \(\omega\notin A_n\) and \(X_n(\omega)=0\). Hence \(X_n(\omega)\) is eventually zero for almost every \(\omega\), which is exactly \(X_n\xrightarrow{a.s.}0\). No step of this argument requires the sum of the probabilities to be finite. \(\square\)

For indicators, the same construction also makes convergence in probability transparent. If \(0<\epsilon<1\), then \(X_n\) takes only the values zero and one, so an error larger than \(\epsilon\) occurs exactly on \(A_n\):

P(|X_n-0|>\epsilon)=P(A_n),\qquad 0<\epsilon<1.

Consequently, \(P(A_n)\to0\) yields \(X_n\xrightarrow{p}0\) as well. The theorem’s almost-sure conclusion is stronger: it says that outside one probability-zero set, there is a finite index after which every indicator is zero.

Worked Example: Uniform Thresholds With Harmonic Error Probabilities

Let \(U\sim\operatorname{Uniform}(0,1)\), and define \(A_n=\{U\le 1/n\}\) and \(X_n=\mathbf{1}_{A_n}\), for \(n\ge1\). Find the error probabilities, check whether their sum diverges, and determine the almost-sure limit.

Because \(1/(n+1)<1/n\), the thresholds decrease, so \(A_{n+1}\subseteq A_n\). The uniform distribution assigns an interval its length, giving

P(A_n)=P(U\le 1/n)=\frac{1}{n}.

The sum diverges. For each integer \(j\ge0\), the block from \(2^j\) to \(2^{j+1}-1\) has \(2^j\) terms, each at least \(1/2^{j+1}\). Thus every block contributes at least \(1/2\), and adding infinitely many such blocks makes the harmonic sum infinite.

Now fix any \(u\in(0,1)\). Choose an integer \(N>1/u\). For every \(n\ge N\), \(1/n\le1/N<u\), so \(u\not\le1/n\) and \(X_n=0\). This holds for every possible value of \(U\) in its sample space, so \(X_n\to0\) almost surely. For any fixed \(0<\epsilon<1\), the fixed-tolerance error probability is \(P(|X_n|>\epsilon)=1/n\); its sum diverges even though the errors eventually stop on every sample path.

Different Rates of Rare Events

The mechanism does not depend on the harmonic rate. Any decreasing sequence of thresholds that tends to zero can produce nested events; if its probabilities have a divergent sum, the indicators illustrate the same distinction. The next construction uses an exponential random variable and a slower rate of decay.

Worked Example: Exponential Thresholds With Square-Root Probabilities

Let \(Y\sim\operatorname{Exp}(1)\), so \(P(Y>y)=e^{-y}\) for \(y\ge0\). Define \(A_n=\{Y>\tfrac12\log n\}\) and \(X_n=\mathbf{1}_{A_n}\). Calculate \(P(A_n)\) and verify the convergence claims.

The thresholds increase with \(n\), making the events nested decreasing. Using the exponential survival probability,

P(A_n)=\exp\left(-\frac12\log n\right)=\frac{1}{\sqrt n}.

These probabilities tend to zero, but their sum diverges because \(1/\sqrt n\ge1/n\) for every \(n\ge1\). For each realized value \(y\), which is finite with probability one, choose \(N\) large enough that \(\tfrac12\log N\ge y\). For \(n\ge N\), the event \(y>\tfrac12\log n\) is false, so the indicator is eventually zero. Therefore \(X_n\xrightarrow{a.s.}0\), while for \(0<\epsilon<1\), \(P(|X_n|>\epsilon)=1/\sqrt n\) and the sum of these fixed-tolerance error probabilities diverges.

A discrete random cutoff gives another way to encode nested events. Define the chance that the cutoff reaches at least \(n\), rather than defining the events first. The probabilities of the possible cutoff values are then successive differences of those tail probabilities.

Worked Example: A Discrete Cutoff With Logarithmic Tails

For \(n\ge1\), set \(p_n=1/(1+\log n)\). Verify that there is a positive-integer-valued random variable \(T\) with \(P(T\ge n)=p_n\), and use it to construct almost-surely vanishing indicators with a divergent error-probability sum.

The sequence starts at \(p_1=1\), decreases, and tends to zero. Assign the probability

P(T=k)=p_k-p_{k+1},\qquad k=1,2,\ldots.

Each assigned probability is nonnegative. The probabilities sum to one by telescoping:

\sum_{k=1}^{m}(p_k-p_{k+1})=p_1-p_{m+1} \longrightarrow 1-0=1.

The same telescoping calculation gives the tail probabilities:

P(T\ge n)=\sum_{k=n}^{\infty}(p_k-p_{k+1})=p_n.

Now take \(A_n=\{T\ge n\}\) and \(X_n=\mathbf{1}_{A_n}\). These events decrease. Since \(T\) takes a finite positive-integer value with probability one, \(X_n=0\) for every \(n>T\); hence the indicators converge almost surely to zero. Yet

P(A_n)=\frac{1}{1+\log n}\ge\frac{1}{n}.

For the inequality, \(\log n\le n-1\) for \(n\ge1\), so \(1+\log n\le n\). The harmonic lower bound shows \(\sum_n P(A_n)\) diverges. As before, each fixed tolerance \(0<\epsilon<1\) has error probability \(P(|X_n|>\epsilon)=p_n\).

Why Divergence Alone Does Not Decide the Limit

The examples show that the sum \(\sum_n P(A_n)\) by itself does not determine whether events occur infinitely often. In the nested case, once a path leaves the events, it stays out. The total sum can count a long sequence of overlapping chances along paths that eventually leave, rather than repeated returns on the same path.

The first Borel–Cantelli lemma says that if \(\sum_n P(A_n)<\infty\), then \(P(A_n\text{ infinitely often})=0\). It gives a sufficient condition, not a necessary one. In the examples above, the events occur infinitely often exactly on \(\bigcap_n A_n\), which has probability zero, although the sum diverges.

Common pitfall: “The error probabilities add to infinity, so errors must occur infinitely often.” This conclusion is not justified without additional assumptions. For decreasing events, infinitely many occurrences mean membership in every event, and that intersection can have probability zero even when the probability sum diverges. The correction is to examine how events overlap and, for nested events, calculate \(P(\bigcap_n A_n)=\lim_n P(A_n)\).

A practical way to build such a counterexample is to choose any decreasing sequence \(p_n\) with \(p_n\to0\) but \(\sum_n p_n=\infty\), then use a uniform random variable \(U\) and events \(A_n=\{U\le p_n\}\). These events are nested, \(P(A_n)=p_n\), and \(U>0\) almost surely ensures that each path eventually falls below all the thresholds. This construction separates two questions that should not be conflated: how large each error probability is, and how errors are arranged across sample paths.

Check Your Understanding

Use the nested-event criterion and the fixed-tolerance identity for indicators to answer these questions.

If \(A_{n+1}\subseteq A_n\) and \(P(A_n)\to0\), what is \(P(\bigcap_n A_n)\), and why does this imply that \(\mathbf{1}_{A_n}\to0\) almost surely?
For \(X_n=\mathbf{1}_{A_n}\), what is \(P(|X_n|>\epsilon)\) when \(0<\epsilon<1\)?
Why does the sequence \(P(A_n)=1/\sqrt n\) have a divergent sum even though it tends to zero?
In the discrete-cutoff example, why do the differences \(p_k-p_{k+1}\) define probabilities that sum to one?
What is the error in reasoning that a divergent sum of event probabilities must imply infinitely many errors almost surely?

Almost-Sure Convergence Without Summable Error Probabilities

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