Limsup Error Sets as a Test for Almost-Everywhere Convergence

Error Sets Record Persistent Deviations

Convergence at a point means that every fixed positive error threshold is eventually crossed no more. The phrase “eventually” is the key: a point can belong to many error sets and still be harmless if it belongs to only finitely many of them. The limsup of the error sets isolates precisely the points where a chosen threshold is exceeded infinitely often. This gives an exact test for almost-everywhere convergence, on finite or infinite measure spaces alike.

Throughout the main result, let \((X,\mathcal{M},\mu)\) be any measure space, with no assumption of finiteness, sigma-finiteness, or completeness. Let \(f_n,f:X\to\mathbb{R}\) be measurable real-valued functions. For each \(\varepsilon>0\), define the fixed-threshold error set

The inequality is strict: a point where the error equals \(\varepsilon\) is not in \(E_n(\varepsilon)\). Each error set belongs to \(\mathcal{M}\), since \(f_n-f\) is measurable and the set where its absolute value exceeds \(\varepsilon\) is measurable.

The “infinitely often” description follows directly from the displayed definition. If \(x\in\limsup_n A_n\), then for every \(N\) it belongs to some \(A_n\) with \(n\ge N\), so there are arbitrarily large such indices. Conversely, if \(x\) belongs to \(A_n\) for infinitely many indices, then for every \(N\) one of those indices is at least \(N\), and \(x\) belongs to every union \(\bigcup_{n\ge N}A_n\).

The Fixed-Threshold Characterization

Proof. First suppose \(f_n\to f\) \(\mu\)-almost everywhere. There is a measurable null set \(N\in\mathcal{M}\) such that \(f_n(x)\to f(x)\) for every \(x\notin N\). Fix any \(\varepsilon>0\). At each \(x\notin N\), convergence gives an index \(N_x\) such that \(|f_n(x)-f(x)|<\varepsilon\) for every \(n\ge N_x\). Thus \(x\) belongs to only finitely many \(E_n(\varepsilon)\), and so \(x\notin\limsup_n E_n(\varepsilon)\). Therefore \(\limsup_n E_n(\varepsilon)\subseteq N\). The limsup is measurable, as a countable intersection of countable unions of measurable sets. By monotonicity of measure, its measure is zero.

Conversely, suppose \(\mu(\limsup_n E_n(\varepsilon))=0\) for every \(\varepsilon>0\). It suffices to use the countable thresholds \(\varepsilon=1/k\), \(k=1,2,\ldots\). Let \[ N=\bigcup_{k=1}^{\infty}\limsup_n E_n(1/k). \] Each set in this union is measurable and null by assumption, so countable subadditivity gives \(\mu(N)=0\). Fix \(x\notin N\). For every positive integer \(k\), \(x\notin\limsup_n E_n(1/k)\). By the definition of limsup, there is an index \(n_k\) such that \(x\notin E_n(1/k)\) for all \(n\ge n_k\). Equivalently, \(|f_n(x)-f(x)|\le 1/k\) for all \(n\ge n_k\). Given any \(\delta>0\), choose \(k\) with \(1/k<\delta\). Then for every \(n\ge n_k\), \[ |f_n(x)-f(x)|\le 1/k<\delta. \] This is the definition of \(f_n(x)\to f(x)\). It holds for every \(x\notin N\), so \(f_n\to f\) \(\mu\)-almost everywhere. \(\square\)

This proof also shows why the countability of the thresholds matters. A null set can be discarded for each threshold \(1/k\), and their countable union is still null. No step requires taking an uncountable union of exceptional sets.

Proof. If the condition holds at every \(1/k\), the reverse implication of the theorem gives almost-everywhere convergence, and its forward implication then gives the condition at every \(\varepsilon>0\). Alternatively, for a given \(\varepsilon>0\), choose \(k\) with \(1/k\le\varepsilon\). The strict-threshold definitions imply \(E_n(\varepsilon)\subseteq E_n(1/k)\) for every \(n\), hence \(\limsup_n E_n(\varepsilon)\subseteq\limsup_n E_n(1/k)\), which is null. \(\square\)

Worked Error-Set Calculations

Two Quantifier Pitfalls

Pitfall: replacing a limsup of sets by a limsup of their measures. In general, \(\mu(\limsup_n E_n)\) is not \(\limsup_n\mu(E_n)\). For a concrete error-set calculation, take \(X=\mathbb{R}\) with Lebesgue measure, \(f_n=\chi_{(n,n+1)}\), and \(f=0\). At threshold \(\varepsilon=1/2\), \(E_n=(n,n+1)\). These open intervals are pairwise disjoint, so no point belongs to infinitely many of them and \(\limsup_n E_n=\varnothing\). Therefore \(m(\limsup_n E_n)=0\), whereas \(m(E_n)=1\) for every \(n\) and \(\limsup_n m(E_n)=1\). The incorrect calculation \(m(\limsup_n E_n)=\limsup_n m(E_n)=1\) confuses two different operations. The correction is to determine point membership in infinitely many sets before measuring the resulting limsup.

Pitfall: checking only one threshold. On any nonempty measure space \((X,\mathcal{M},\mu)\), take the constant-in-\(x\) functions \(f_n=(-1)^n/4\) and candidate limit \(f=0\). At threshold \(\varepsilon=1\), every error set is empty because \(1/4\) is not greater than \(1\). This single check does not establish convergence: at threshold \(\varepsilon=1/8\), every error set is \(X\), so the limsup is \(X\). The sequence alternates between \(1/4\) and \(-1/4\) at every point and does not converge. The theorem requires every positive threshold, or equivalently the entire countable family \(1/k\).

The criterion is useful because it turns a pointwise requirement into a statement about measurable sets, while retaining the exact “eventually” quantifier. The measure of an individual error set records how many points fail at that index; the measure of its limsup records how many points fail infinitely often. Those are different questions. In particular, neither finite measure of the whole space nor convergence in measure is part of this characterization: the test is about almost-everywhere convergence itself.