Tail-Measurable Random Variables Are Almost Surely Constant

Why Tail Information Cannot Encode a Nonconstant Real Value

A tail event is determined by what happens arbitrarily far along a sequence: changing or deleting finitely many initial coordinates does not change whether the event occurs. For an independent sequence, Kolmogorov’s zero–one law says that every such event has probability zero or one.

The same principle constrains not only events but also real-valued random variables. If a random variable is measurable using only tail information, then each of its sublevel events is a tail event. In particular, its value cannot have a genuinely nontrivial distribution: it must equal one fixed real number almost surely.

Tail Measurability and the Zero–One Law

Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space, and let $(X_n)_{n\in\mathbb N}$ be an independent sequence of real-valued random variables. The tail sigma-algebra of this sequence is

\mathcal T = \bigcap_{m\in\mathbb N}\sigma(X_n:n\ge m).

Thus an event in $\mathcal T$ belongs to the sigma-algebra generated by every tail of the sequence. Kolmogorov’s zero–one law states that, under the independence assumption, every $A\in\mathcal T$ satisfies $\mathbb P(A)\in\{0,1\}$.

Definition: A real-valued random variable $Y$ is tail-measurable if it is measurable as a map from $(\Omega,\mathcal T)$ to $(\mathbb R,\mathcal B(\mathbb R))$, where $\mathcal B(\mathbb R)$ is the Borel sigma-algebra.

For such a $Y$, every Borel level set belongs to $\mathcal T$. We will use the sublevel sets $\{Y\le q\}$, where $q\in\mathbb Q$. The rational thresholds are enough: they are countable, and they distinguish every pair of different real numbers.

The Rational-Threshold Argument

The key step does not require an independent sequence. It applies to any real-valued random variable whose rational sublevel probabilities are all zero or one. This isolates the distributional reasoning before we apply the zero–one law.

Proposition: Suppose $Y$ is a real-valued random variable and $\mathbb P(Y\le q)\in\{0,1\}$ for every $q\in\mathbb Q$. Then there exists $c\in\mathbb R$ such that $\mathbb P(Y=c)=1$.

Proof. Set $$ S=\{q\in\mathbb Q:\mathbb P(Y\le q)=1\}. $$ This set is nonempty. Indeed, the events $\{Y\le n\}$, for $n\in\mathbb N$, increase to $\Omega$, since $Y$ is real-valued. Their probabilities therefore increase to one, so at least one has positive probability; by the assumed zero–one property, its probability is one. Also, $S$ is bounded below: the events $\{Y\le -n\}$ decrease to $\varnothing$, so their probabilities decrease to zero. For some $n$, the probability is less than one and hence is zero. Monotonicity then gives $\mathbb P(Y\le q)=0$ for every rational $q\le -n$.

Consequently, $c=\inf S$ is a finite real number. If $q\in\mathbb Q$ and $q>c$, the definition of infimum gives some $r\in S$ with $r<q$. Since $\{Y\le r\}\subseteq\{Y\le q\}$, we have $\mathbb P(Y\le q)=1$. If $q<c$, then $q\notin S$, so the zero–one property gives $\mathbb P(Y\le q)=0$.

The rationals greater than $c$ form a countable set, and $$ \{Y\le c\} = \bigcap_{\substack{q\in\mathbb Q\\q>c}}\{Y\le q\}. $$ Every event in this intersection has probability one, so $\mathbb P(Y\le c)=1$. Likewise, $$ \{Y<c\} = \bigcup_{\substack{q\in\mathbb Q\\q<c}}\{Y\le q\}. $$ Every event in this union has probability zero, so $\mathbb P(Y<c)=0$. Together these statements give $\mathbb P(Y=c)=1$.

$\square$

The finite-valued assumption matters in this argument. Since $Y$ is real-valued, the events $\{Y\le n\}$ increase to the whole space and $\{Y\le -n\}$ decrease to the empty set. This is what ensures that the cutoff $c$ is a finite real number rather than an infinite endpoint.

Main Result: Tail-Measurable Variables Are Almost Surely Constant

Theorem: If $(X_n)_{n\in\mathbb N}$ is an independent sequence of real-valued random variables and $Y$ is a real-valued random variable measurable with respect to its tail sigma-algebra $\mathcal T$, then there is a constant $c\in\mathbb R$ such that $\mathbb P(Y=c)=1$.

Proof. For every rational $q$, tail measurability gives $\{Y\le q\}\in\mathcal T$. Kolmogorov’s zero–one law therefore gives $\mathbb P(Y\le q)\in\{0,1\}$ for every $q\in\mathbb Q$. The proposition now supplies a real number $c$ with $\mathbb P(Y=c)=1$.

$\square$

The conclusion is almost sure constancy, not necessarily pointwise constancy. The variable can take other values on a set of probability zero. The zero–one law constrains probabilities of tail events; it does not rule out modifications on null sets.

Worked Example: Infinitely Many Successes in an Independent Bernoulli Sequence

Let $(X_n)_{n\in\mathbb N}$ be independent Bernoulli random variables with common success probability $p$, where $0<p<1$. Define $$ E=\{X_n=1\text{ for infinitely many }n\}, \qquad Y=\mathbf 1_E. $$ For every $m$, the event $E$ can be written as $\bigcap_{k\ge m}\bigcup_{n\ge k}\{X_n=1\}$, so it belongs to $\sigma(X_n:n\ge m)$. Thus $E\in\mathcal T$ and $Y$ is tail-measurable.

For fixed $m$, the probability of no successes from $m$ through $N$ is $(1-p)^{N-m+1}$, which tends to zero as $N$ tends to infinity. Hence the probability of no successes at all after $m$ is zero. The event $E^c$ is the union, over $m$, of the events that there are no successes after $m$; therefore $\mathbb P(E^c)=0$. It follows that $Y=1$ almost surely, in agreement with the theorem.

Worked Example: A Tail Event with Probability Zero

Let $(X_n)_{n\in\mathbb N}$ be independent Bernoulli random variables with $\mathbb P(X_n=1)=1/(n+1)^2$. As above, let $E=\{X_n=1\text{ for infinitely many }n\}$. This is a tail event, so its indicator $Y=\mathbf 1_E$ is tail-measurable.

For each $m$, the event $E$ is contained in the event that at least one success occurs at an index $n\ge m$. The union bound gives $$ \mathbb P(E) \le \mathbb P\!\left(\bigcup_{n\ge m}\{X_n=1\}\right) \le \sum_{n\ge m}\frac{1}{(n+1)^2}. $$ The series converges, so its tail tends to zero as $m$ tends to infinity. Since the bound holds for every $m$, $\mathbb P(E)=0$, and $Y=0$ almost surely. This gives the other possible constant value for a tail indicator.

Worked Example: A Real-Valued Variable Built from a Tail Event

Let $E\in\mathcal T$, and choose two real numbers $a\ne b$. Define $Y=a$ on $E$ and $Y=b$ on $E^c$. For any Borel set $B\subseteq\mathbb R$, the preimage $Y^{-1}(B)$ is one of $\varnothing$, $E$, $E^c$, or $\Omega$, according to whether $a$ and $b$ belong to $B$. Each of these events belongs to $\mathcal T$, so $Y$ is tail-measurable.

Because $E$ is a tail event, the zero–one law gives either $\mathbb P(E)=1$ or $\mathbb P(E)=0$. In the first case, $\mathbb P(Y=a)=1$; in the second, $\mathbb P(Y=b)=1$. Although $Y$ takes two distinct values in its definition, it cannot take both with positive probability.

How to Read the Constant from the Sublevel Events

The proof identifies the constant through the transition between rational thresholds that have probability zero and those that have probability one. Below $c$, the event $\{Y\le q\}$ has probability zero; above $c$, it has probability one. The value at the threshold itself follows from countable intersections and unions, not from assuming that a distribution function has any particular behavior at $c$.

This is why rational thresholds are especially useful. There are only countably many of them, so probability-one statements can be intersected without losing probability one, and probability-zero statements can be united without gaining positive probability. Density of $\mathbb Q$ then pins down the real value.

Scope and a Common Pitfall

The theorem requires measurability with respect to the tail sigma-algebra, not merely measurability with respect to the sigma-algebra generated by the entire sequence. A variable that depends on an initial coordinate need not be tail-measurable and need not be constant. For example, a nondegenerate Bernoulli variable $X_1$ is measurable with respect to $\sigma(X_n:n\in\mathbb N)$, but its sublevel events need not belong to $\mathcal T$.

Independence is used through Kolmogorov’s zero–one law. The variables do not need to be identically distributed. Conversely, being a real-valued random variable on a probability space is not enough: the crucial assumption is that every rational sublevel event is a tail event, which follows from tail measurability.

A useful version of the conclusion also applies when a variable has a tail-measurable version. If $Y$ is tail-measurable and $Z=Y$ almost surely, then the theorem gives a constant $c$ with $\mathbb P(Z=c)=1$ as well. This follows because the events $\{Z=c\}$ and $\{Y=c\}$ can differ only on the null event where $Z\ne Y$.

Takeaway: For an independent sequence, a real-valued random variable that can be determined from every tail has rational sublevel events of probability zero or one. Those threshold probabilities force the variable to equal a single real constant almost surely.

Check Your Understanding

Use the rational-threshold argument and the tail zero–one law to answer these questions.

Why does tail measurability of $Y$ imply that $\{Y\le q\}$ belongs to $\mathcal T$ for every rational $q$?
Where in the proof that the cutoff $c$ is finite is the assumption that $Y$ is real-valued used?
Why is it enough to know that $\mathbb P(Y\le q)$ is zero or one for rational $q$, rather than for every real threshold?
Can a tail-measurable variable take more than one value on the whole sample space? Can it take two values each with positive probability?
Which assumption supplies the zero–one probabilities in the main theorem?