From Small Errors to Almost-Everywhere Convergence
Convergence in measure controls one error threshold at a time: for each fixed \(\varepsilon>0\), the set where the error exceeds \(\varepsilon\) becomes small in measure. Almost-everywhere convergence asks for a stronger pointwise conclusion: outside one null set, the errors must eventually be small at every positive tolerance. The key is to select a subsequence carefully. Its error sets can be made summable, which ensures that almost every point belongs to only finitely many of them.
Throughout, \((X,\mathcal{M},\mu)\) is an arbitrary measure space; it need not be finite, complete, or sigma-finite. All functions in the theorem are measurable and real-valued, and statements of almost-everywhere convergence mean convergence at every point outside a \(\mu\)-null set.
The theorem below converts this convergence into almost-everywhere convergence along a subsequence. The index choices are allowed to depend on the error tolerance: convergence in measure guarantees that sufficiently late terms have small error sets at any one fixed tolerance.
Summable Error Sets
Proof. For every \(N\ge1\), \(\limsup_k A_k\subseteq\bigcup_{k\ge N}A_k\). Countable subadditivity therefore gives \(\mu(\limsup_k A_k)\le\sum_{k\ge N}\mu(A_k)\). Since the series of measures is finite, its tails tend to zero as \(N\to\infty\). Thus \(\mu(\limsup_k A_k)=0\). \(\square\)
This argument needs no finiteness assumption on \(\mu(X)\). It uses only countable subadditivity and the fact that the tails of a convergent series of nonnegative numbers tend to zero.
Riesz's Subsequence Theorem
Proof. For each positive integer \(k\), convergence in measure applied with the fixed threshold \(1/k\) says that \(\mu(\{|f_n-f|>1/k\})\to0\) as \(n\to\infty\). Consequently, there is an integer \(N_k\) such that \(\mu(\{|f_n-f|>1/k\})<2^{-k}\) whenever \(n\ge N_k\). Choose \(n_1\ge N_1\), and recursively choose \(n_k\ge N_k\) with \(n_k>n_{k-1}\). This is possible because every integer at least \(N_k\) satisfies the measure bound, so one can take an integer larger than both \(N_k\) and \(n_{k-1}\).
Define \(A_k=\{|f_{n_k}-f|>1/k\}\). These sets are measurable, and the choices give \(\sum_{k=1}^{\infty}\mu(A_k)\le\sum_{k=1}^{\infty}2^{-k}=1\). By the lemma, \(N=\limsup_k A_k\) is null. If \(x\notin N\), then \(x\) belongs to only finitely many of the sets \(A_k\). Hence there is an index \(K(x)\) such that for every \(k\ge K(x)\), \(x\notin A_k\), so \(|f_{n_k}(x)-f(x)|\le1/k\). Given any \(\varepsilon>0\), choose \(K\ge K(x)\) large enough that \(1/K<\varepsilon\). For all \(k\ge K\), \(|f_{n_k}(x)-f(x)|\le1/k\le1/K<\varepsilon\). Therefore \(f_{n_k}(x)\to f(x)\). This holds for every \(x\notin N\), proving almost-everywhere convergence. \(\square\)
The same null set works for every tolerance: once \(x\) lies outside the limsup of the selected error sets, the bound \(1/k\) applies to all sufficiently large \(k\). That is what turns the separate measure estimates into a pointwise limit.
Worked Example: Shrinking Supports on a Finite Measure Space
Let \(X=[0,1]\) with Lebesgue measure \(m\), and set \(f_n=\chi_{(0,1/n)}\) and \(f=0\). Each function is measurable and real-valued. For \(0<\varepsilon<1\), the strict error set \(\{|f_n-f|>\varepsilon\}\) is exactly \((0,1/n)\), of measure \(1/n\); for \(\varepsilon\ge1\), it is empty because the function takes only the values zero and one. Thus \(f_n\to0\) in measure.
Choose \(n_k=2^k\), which is strictly increasing. At the threshold \(1/k\), the error set \(A_k=\{|f_{2^k}|>1/k\}\) is empty when \(k=1\), since \(1\) is not strictly greater than \(1\). For \(k\ge2\), it is \((0,2^{-k})\), since \(1>1/k\). Therefore \(\sum_{k=1}^{\infty}m(A_k)=\sum_{k=2}^{\infty}2^{-k}=1/2\). The lemma gives almost-everywhere convergence. In fact, for each \(x>0\), \(2^{-k}<x\) for all sufficiently large \(k\), so \(x\notin(0,2^{-k})\) thereafter; at \(x=0\), every indicator is zero. Thus this chosen subsequence converges to zero everywhere. Check: the first three error-set measures at the specified thresholds are \(0\), \(2^{-2}=1/4\), and \(2^{-3}=1/8\); their full sum from \(k=2\) is \(1/4+1/8+\cdots=1/2\), as computed.
Worked Example: Why the Theorem Does Not Require Finite Measure
Let \(X=\mathbb{R}\) with Lebesgue measure \(m\), so \(m(X)=\infty\). Define \(f_n=\chi_{[n,n+1/n]}\) and \(f=0\). If \(0<\varepsilon<1\), then \(\{|f_n-f|>\varepsilon\}=[n,n+1/n]\), whose measure is \(1/n\). If \(\varepsilon\ge1\), the strict error set is empty. It follows that \(f_n\to0\) in measure despite the infinite measure of the underlying space.
Take \(n_k=2^k\). At the fixed threshold \(1/2\), the error set is \([2^k,2^k+2^{-k}]\), of measure \(2^{-k}\), and \(\sum_{k=1}^{\infty}2^{-k}=1\). The intervals are disjoint: for \(k\ge1\), \(2^k+2^{-k}<2^{k+1}\), so the \(k\)-th interval ends before the next begins. In particular, each \(x\in\mathbb{R}\) belongs to at most one of them, and \(f_{2^k}(x)=0\) eventually for every \(x\). The subsequence converges to zero everywhere. Check: the first two interval lengths are \(2^{-1}=1/2\) and \(2^{-2}=1/4\); the total length sum is \(1/2+1/4+\cdots=1\). Neither the theorem nor this verification uses finiteness of \(m(\mathbb{R})\).
Worked Example: Extracting from a Shrinking Spike
On \(X=[0,1]\) with Lebesgue measure \(m\), define \(f_n(x)=x+n\chi_{(0,1/n^2)}(x)\) and \(f(x)=x\). The error is \(n\) on \((0,1/n^2)\) and zero elsewhere. For a fixed \(\varepsilon>0\), the strict error set is \((0,1/n^2)\) when \(n>\varepsilon\), with measure \(1/n^2\), and is empty when \(n\le\varepsilon\). Its measure tends to zero, so \(f_n\to f\) in measure.
Choose \(n_k=2^k\). At threshold \(1/k\), the error height \(2^k\) is strictly larger than \(1/k\), so \(A_k=(0,2^{-2k})\) and \(m(A_k)=4^{-k}\). The sum is \(\sum_{k=1}^{\infty}4^{-k}=(1/4)/(1-1/4)=1/3\), so the summable-error lemma gives convergence almost everywhere. Indeed, at any \(x>0\), \(2^{-2k}<x\) for all sufficiently large \(k\), and then \(x\) lies outside the spike; at \(x=0\) the open interval excludes the point at every \(k\). Thus \(f_{2^k}(x)=x\) eventually at every \(x\in[0,1]\). Check: the first two selected error-set measures are \(4^{-1}=1/4\) and \(4^{-2}=1/16\), whose geometric tail sums to \(1/3\).
The Selection Is the Essential Step
Convergence in measure alone supplies error-set measures tending to zero at each fixed threshold; it does not supply a summable sequence of such measures for an arbitrary choice of indices and thresholds. In the proof, the threshold \(1/k\) is fixed first, and convergence in measure then permits an index far enough out to make the corresponding error-set measure less than \(2^{-k}\). The indices must also be chosen strictly increasingly, but this adds no obstruction: once the measure bound holds for all sufficiently large indices, one can choose an index beyond the previous choice.
Common mistake. A tempting but invalid calculation is to choose indices with \(\mu(A_k)<1/k\) and then assert that the error-set measures are summable. The relevant comparison is \(\sum_{k=1}^{\infty}\mu(A_k)\le\sum_{k=1}^{\infty}1/k\), and the harmonic series on the right diverges. The summable-error lemma therefore does not apply to that bound. This does not assert that a subsequence chosen with the weaker bound must fail to converge almost everywhere; it means only that this argument gives no conclusion. The correction is to use convergence in measure to select indices with \(\mu(A_k)<2^{-k}\), for which the geometric sum is finite. The threshold and the target measure bound have different roles: \(1/k\) tends to zero to force convergence, while \(2^{-k}\) is summable to control the exceptional set.
The result is useful whenever a conclusion is easiest to establish pointwise but the available hypothesis is convergence in measure. It gives a subsequence that retains the same limit and converges outside one null set. The theorem asserts the existence of such a subsequence; it does not claim that the original sequence converges almost everywhere, and the selection may depend on the sequence and on the chosen summable bounds.
Check Your Understanding
Use the strict error-set definition and identify the role of summability in each question.
- In the subsequence theorem, why must the index at stage \(k\) be chosen beyond the previous index as well as beyond the index required by the measure bound?
- For \(f_n=\chi_{[n,n+1/n]}\) on \(\mathbb{R}\), compute the error-set measure for a fixed threshold \(\varepsilon=1/2\).
- Explain why the estimate \(\mu(A_k)<1/k\) does not establish the hypothesis of the summable-error lemma.
- In the shrinking-spike example, write the error set at threshold \(1/k\) for the selected index \(n_k=2^k\), and compute its measure.
- Does Riesz's subsequence theorem require \(\mu(X)<\infty\) or completeness of the measure space? Identify the step in the proof that answers this.