Agreement of Almost-Everywhere and In-Measure Limits

Two Modes, One Almost-Everywhere Limit

Almost-everywhere convergence describes the behavior of a sequence at nearly every individual point. Convergence in measure describes how much of the space has an error larger than a chosen threshold. These descriptions use different quantifiers, but they cannot assign genuinely different limits to the same sequence: if both modes hold, the limits agree almost everywhere. The measure space need not have finite measure, and no integrability assumption is required.

Throughout, \((X,\mathcal{M},\mu)\) is an arbitrary measure space; it need not be finite, sigma-finite, or complete. All functions in this tutorial are measurable, finite-valued real functions on \(X\). For such functions \(f_n\) and \(f\), the statement \(f_n\to f\) \(\mu\)-a.e. means there is a measurable null set \(N\) such that \(f_n(x)\to f(x)\) for every \(x\in X\setminus N\). The statement \(f_n\to g\) in measure means that for every \(\varepsilon>0\),

The inequality in the error set is strict. Convergence in measure is defined using the measure of these sets, not by requiring pointwise convergence. In particular, a null set where two candidate limits differ is invisible to convergence in measure.

A Summable-Error Subsequence

The key step is to select a subsequence whose errors relative to the in-measure limit occur on sets with summable measures. We first give the elementary measure argument that makes this useful. For measurable sets \(E_k\), their limsup is the set of points that belong to infinitely many \(E_k\): \(\limsup_k E_k=\bigcap_{r=1}^{\infty}\bigcup_{k\ge r}E_k\).

Proof. Fix a positive integer \(r\). Since \(\limsup_k E_k\subseteq\bigcup_{k\ge r}E_k\), countable subadditivity gives \(\mu(\limsup_k E_k)\le\sum_{k\ge r}\mu(E_k)\). The series has a finite sum, so its tails tend to zero as \(r\to\infty\). Thus \(\mu(\limsup_k E_k)=0\). \(\square\)

This lemma is the first Borel–Cantelli lemma, in the direction needed here. Its conclusion says that outside a null set, a point belongs to only finitely many of the error sets. It makes no claim about independence, and it applies on an arbitrary measure space.

Proof. For each positive integer \(k\), convergence in measure, applied with threshold \(1/k\), gives an integer \(N_k\) such that for every \(n\ge N_k\),

Choose indices \(n_k\) recursively so that \(n_k\ge N_k\) and \(n_k>n_{k-1}\) when \(k\ge2\). This is possible because the displayed bound holds for every sufficiently large \(n\), not just for one index. Define the measurable error sets \(E_k=\{x:|f_{n_k}(x)-g(x)|>1/k\}\). Then \(\mu(E_k)<2^{-k}\), and \(\sum_{k=1}^{\infty}2^{-k}=1\), so \(\sum_k\mu(E_k)<\infty\). The lemma implies that \(\limsup_k E_k\) is null.

If \(x\notin\limsup_k E_k\), then \(x\) belongs to only finitely many \(E_k\). Consequently, there is an index \(K(x)\) such that for every \(k\ge K(x)\), \(x\notin E_k\), and hence \(|f_{n_k}(x)-g(x)|\le1/k\). Given any \(\varepsilon>0\), choose \(k\) sufficiently large that \(k\ge K(x)\) and \(1/k<\varepsilon\). The bound then gives \(|f_{n_k}(x)-g(x)|<\varepsilon\). Therefore \(f_{n_k}(x)\to g(x)\).

By the assumed almost-everywhere convergence, there is a null set \(N\) such that \(f_n(x)\to f(x)\) for every \(x\notin N\). At each such point, the subsequence \(f_{n_k}(x)\) also tends to \(f(x)\): the definition of convergence of the original sequence applies in particular to all sufficiently late indices \(n_k\). Thus, outside \(N\cup\limsup_k E_k\), the same real sequence \(f_{n_k}(x)\) tends both to \(f(x)\) and to \(g(x)\). Limits in \(\mathbb{R}\) are unique: if two candidate limits had distance \(d>0\), eventually each would be within \(d/3\) of the sequence, forcing their distance to be at most \(2d/3<d\), a contradiction. Therefore \(f(x)=g(x)\) outside this union of two null sets. The union is null by subadditivity, proving \(f=g\) \(\mu\)-a.e. \(\square\)

The proof uses convergence in measure only to choose the subsequence. It does not try to show that the whole sequence converges pointwise to \(g\). Once the selected subsequence converges to both limits on a common set of full measure, uniqueness of pointwise limits finishes the argument.

Worked Examples

What the Theorem Does—and Does Not—Say

The agreement theorem is not the claim that almost-everywhere convergence always implies convergence in measure. That implication requires additional conditions, such as finite measure, and is a separate result. The theorem here has a different logical form: it assumes both modes of convergence and concludes that their limits agree almost everywhere. Its proof uses no estimate on \(\mu(X)\), since the subsequence is selected by the error-set measures themselves.

A common mistake is to use the theorem backwards, assuming that almost-everywhere convergence alone guarantees convergence in measure on any space. For example, on \(\mathbb{R}\) with Lebesgue measure, take \(f_n=\chi_{[n,n+1]}\). For each fixed \(x\), \(x\notin[n,n+1]\) for all sufficiently large \(n\), so \(f_n(x)\to0\) everywhere. But at threshold \(\varepsilon=1/2\), the error set \(\{|f_n|>1/2\}\) is \([n,n+1]\), whose measure is \(1\) for every \(n\). The error-set measures do not tend to zero. This example does not contradict the agreement theorem: it lacks convergence in measure.

Another mistake is to conclude that the two limits must agree at every point. The null-modification example shows why the strongest general conclusion is equality almost everywhere. Convergence in measure cannot distinguish functions that differ only on a null set, and almost-everywhere convergence itself disregards a null exceptional set. When applying the theorem, state the conclusion with the correct qualifier: \(f=g\) \(\mu\)-a.e., not necessarily everywhere.