Summable Error Sets and Almost-Everywhere Subsequence Limits

From Summable Errors to Almost-Everywhere Convergence

Suppose a sequence of measurable functions approaches a target except on sets where its error exceeds a chosen threshold. If, for each fixed positive threshold, the measures of those error sets have a finite sum, then almost every point can belong to only finitely many of them. Applying this observation at a countable collection of thresholds tending to zero gives convergence almost everywhere.

Let \((X,\mathcal{M},\mu)\) be any measure space: no assumption of finite measure, sigma-finiteness, or completeness is required. Let \(f_n:X\to\mathbb{R}\) and \(f:X\to\mathbb{R}\) be measurable, real-valued functions. For \(\varepsilon>0\), define the strict threshold-error set

E_n(\varepsilon)=\{x\in X:|f_n(x)-f(x)|>\varepsilon\}.

These sets are measurable because the functions are measurable. The strict inequality matters: outside \(E_n(\varepsilon)\), the error is at most \(\varepsilon\), including points where it equals \(\varepsilon\).

Theorem (Summable Error Sets): Suppose that for every positive rational number \(q\), \(\sum_{n=1}^{\infty}\mu(E_n(q))<\infty\). Then \(f_n\to f\) \(\mu\)-almost everywhere on \(X\). The conclusion means that there is a measurable set \(N\) with \(\mu(N)=0\) such that \(f_n(x)\to f(x)\) for every \(x\in X\setminus N\).

The proof has two stages. First, summability at one fixed threshold shows that almost every point has only finitely many errors exceeding that threshold. Then countability lets us impose this conclusion simultaneously at all positive rational thresholds.

A Tail Estimate for Sets

Lemma: Let \((A_n)\) be measurable sets in \((X,\mathcal{M},\mu)\). If \(\sum_{n=1}^{\infty}\mu(A_n)<\infty\), then the set of points belonging to infinitely many \(A_n\) has measure zero.

Proof. Define \(B_N=\bigcup_{n=N}^{\infty}A_n\). Countable subadditivity gives \(\mu(B_N)\le\sum_{n=N}^{\infty}\mu(A_n)\). The series is finite, so its tails tend to zero as \(N\to\infty\). A point belongs to infinitely many \(A_n\) exactly when it belongs to every \(B_N\); thus the set in the claim is \(\bigcap_{N=1}^{\infty}B_N\). It is measurable and is contained in every \(B_N\), so its measure is at most \(\sum_{n=N}^{\infty}\mu(A_n)\) for every \(N\). Letting \(N\to\infty\) bounds its measure by zero. \(\square\)

This proof is sometimes called the first Borel–Cantelli argument. It uses only countable subadditivity and convergence of a nonnegative series. In particular, it does not require independence of the sets or finiteness of the measure space.

Worked Example: Shrinking Error Sets on a Finite Interval

On \(X=[0,1]\) with its Lebesgue sigma-algebra and Lebesgue measure \(m\), take \(f=0\) and \(f_n=\chi_{(0,2^{-n})}\). Verify the summability condition and apply the theorem.

For a rational threshold \(q\) with \(0<q<1\), the function \(f_n\) equals \(1\) on \((0,2^{-n})\) and \(0\) elsewhere. Therefore the strict error set is \(E_n(q)=(0,2^{-n})\), and its measure is \(2^{-n}\). The sum of these measures is

\sum_{n=1}^{\infty}\mu(E_n(q)) =\sum_{n=1}^{\infty}2^{-n} =\frac{1/2}{1-1/2} =1.

For \(q\ge1\), the strict inequality \(|f_n-f|>q\) never holds, because the error is at most \(1\). Thus \(\mu(E_n(q))=0\) for every \(n\), and the sum is \(0\). The required condition holds for every positive rational \(q\), so \(f_n\to0\) almost everywhere. In fact, convergence holds at every point: \(x=0\) lies in none of the open intervals, and for each \(x>0\), \(2^{-n}\le x\) for all sufficiently large \(n\).

Check: the tail bound at threshold \(q=1/2\) is \(\sum_{n=N}^{\infty}2^{-n}=2^{1-N}\). At \(N=1\) this gives \(1\), matching the full sum; as \(N\to\infty\), it tends to \(0\), as required by the lemma.

Why Rational Thresholds Suffice

Proof of the theorem. Fix a positive rational number \(q\). Apply the lemma to \(A_n=E_n(q)\). The summability assumption implies that outside a measurable null set \(N_q\), each point belongs to only finitely many \(E_n(q)\). Equivalently, for each \(x\notin N_q\), there is an index \(N\), possibly depending on \(x\) and \(q\), such that \(|f_n(x)-f(x)|\le q\) for all \(n\ge N\).

Let \(N\) be the union of \(N_q\) over all positive rational \(q\). The positive rationals are countable, so \(N\) is measurable and has measure zero by countable subadditivity. Fix \(x\notin N\), and let \(\varepsilon>0\) be any real number. By density of the rationals, choose a positive rational \(q<\varepsilon\). Since \(x\notin N_q\), there is an \(N_0\) such that \(|f_n(x)-f(x)|\le q\) whenever \(n\ge N_0\). Hence \(|f_n(x)-f(x)|<\varepsilon\) for every \(n\ge N_0\). This is the definition of \(f_n(x)\to f(x)\). The same null set \(N\) works for every \(x\) outside it, proving \(f_n\to f\) \(\mu\)-almost everywhere. \(\square\)

The countability step is essential to this proof: an uncountable union of null sets need not be null. Rational thresholds provide a countable set of tests, while their density ensures that passing every test eventually is enough to pass any positive real threshold.

Worked Example: Summable Errors Need Not Give Uniform Convergence

On \([0,1]\) with Lebesgue measure, set \(I_n=(2^{-n},2^{1-n})\), \(f_n=\chi_{I_n}\), and \(f=0\). Check the summability condition and compare almost-everywhere and uniform convergence.

For \(0<q<1\), the strict error set is \(E_n(q)=I_n\), whose measure is its interval length: \[ m(I_n)=2^{1-n}-2^{-n}=2^{-n}. \] Thus \(\sum_{n=1}^{\infty}m(E_n(q))=\sum_{n=1}^{\infty}2^{-n}=1\). For \(q\ge1\), the error never exceeds \(q\), so the error sets are empty. The theorem gives almost-everywhere convergence to zero.

Here the intervals are pairwise disjoint: \(I_{n+1}=(2^{-(n+1)},2^{-n})\) ends at the left endpoint of \(I_n\), and neither interval contains that endpoint. Thus each point belongs to at most one \(I_n\), which also verifies pointwise convergence at every point. On the other hand, every \(I_n\) is nonempty, so \(\sup_{x\in[0,1]}|f_n(x)-f(x)|=1\) for every \(n\). In particular, the sequence does not converge uniformly. Check: the essential supremum is also \(1\), since the set where the error equals \(1\) has measure \(2^{-n}>0\) for every \(n\).

Summability Is Sufficient, Not Necessary

The theorem is a criterion, not an if-and-only-if characterization. Its hypothesis provides a convenient quantitative guarantee, but a.e. convergence can occur even if the error measures are not summable. Conversely, checking only one threshold—even a large one—does not establish convergence, because convergence requires control at arbitrarily small error levels.

Worked Example: Convergence Despite a Divergent Error Sum

On \([0,1]\) with Lebesgue measure, take \(f=0\) and \(f_n=\chi_{(0,1/n)}\). For \(0<q<1\), the strict error set is \((0,1/n)\), so \(m(E_n(q))=1/n\). The sum \(\sum_{n=1}^{\infty}1/n\) diverges: for every integer \(k\ge0\), the terms with \(2^k\le n<2^{k+1}\) have sum at least \(2^k\cdot 2^{-(k+1)}=1/2\). Hence the summability hypothesis fails at every rational \(q\) with \(0<q<1\).

Nevertheless, \(f_n(x)\to0\) for every \(x\in[0,1]\). At \(x=0\), all indicators vanish. If \(x>0\), choose \(N>1/x\); then \(1/n<x\) for every \(n\ge N\), so \(x\notin(0,1/n)\) and \(f_n(x)=0\). This example shows that divergent error sums do not imply failure of a.e. convergence; the summability condition is sufficient, not necessary.

A Common Threshold Mistake

It is not enough to show summability at one fixed threshold. For example, on \([0,1]\) let \(f=0\) and \(f_n(x)=(-1)^n/2\). At threshold \(q=1\), every strict error set is empty, since \(|f_n(x)-f(x)|=1/2\) is not greater than \(1\). The resulting sum is \(0\), but the sequence alternates between \(1/2\) and \(-1/2\), so it does not converge at any point.

The correction is to require summability at every positive rational threshold. At \(q=1/4\), for instance, every error set is all of \([0,1]\), so its measure is \(1\) and the sum of the measures is \(\sum_{n=1}^{\infty}1=\infty\). The full hypothesis correctly does not apply. In a construction, it is therefore not enough to control errors above a single fixed level: the estimates must reach arbitrarily small levels.

The criterion is useful whenever a sequence has error estimates that can be summed. A particularly direct application is to a sequence already chosen for approximation: if its threshold-error measures have finite sums at every positive rational level, then its limit agrees with the target almost everywhere. The conclusion concerns the values of the functions outside one null set; it does not claim uniform convergence, convergence at every point, or convergence in any norm.

Check Your Understanding

For each question, keep the strict inequality in the definition of the error set in view.

Why does \(\sum_n\mu(E_n(q))<\infty\) imply that almost every point belongs to only finitely many \(E_n(q)\)?
In the theorem’s proof, where is countability used, and why are rational thresholds sufficient?
For \(f_n=\chi_{(0,2^{-n})}\) on \([0,1]\), compute the sum of the error-set measures when \(q=1\).
For \(f_n=\chi_{(0,1/n)}\) on \([0,1]\), explain why pointwise convergence holds although the error measures are not summable for \(0<q<1\).
Give a reason why summability at the single threshold \(q=1\) does not imply convergence of \(f_n(x)=(-1)^n/2\).