The Subsequence Criterion
Convergence in measure does not generally mean that the whole sequence converges pointwise almost everywhere. It does, however, ensure that every subsequence contains a further subsequence with almost-everywhere convergence to the same limit. On a space of finite measure, this property also characterizes convergence in measure. The finite-measure hypothesis matters: on an infinite-measure space, almost-everywhere convergence need not imply convergence in measure.
Throughout the theorem and its proof, \((X,\mathcal{M},\mu)\) is a measure space with \(\mu(X)<\infty\). The functions \(f_n,f:X\to\mathbb{R}\) are measurable and finite-valued. All claims of almost-everywhere convergence refer to \(\mu\)-almost every point in \(X\).
A subsequence has indices \(n_1<n_2<\cdots\). A further subsequence of it has the form \(f_{n_{j_k}}\), where \(j_1<j_2<\cdots\). The limit in the criterion below is the original function \(f\), not a limit that may vary with the chosen subsequence.
- \(f_n\to f\) in measure;
- every subsequence of \((f_n)\) has a further subsequence converging to \(f\) \(\mu\)-almost everywhere.
From Convergence in Measure to Almost-Everywhere Convergence
The key construction chooses a new index for each accuracy level. The indices are chosen far enough out that the corresponding error sets have summable measures. Countable subadditivity then shows that almost every point belongs to only finitely many of those error sets.
Proof of the forward implication. Assume \(f_n\to f\) in measure, and start with any subsequence \((f_{n_j})\). For each \(k\ge1\), convergence in measure at threshold \(1/k\) says that \(\mu(\{|f_{n_j}-f|>1/k\})\to0\) as \(j\to\infty\). We can therefore choose strictly increasing indices \(j_k\) such that \[ \mu(E_k)<2^{-k},\qquad E_k=\{x:|f_{n_{j_k}}(x)-f(x)|>1/k\}. \] The sets \(E_k\) are measurable. Their measures satisfy \(\sum_{k=1}^{\infty}\mu(E_k)\le\sum_{k=1}^{\infty}2^{-k}=1\).
Let \(\limsup_k E_k=\bigcap_{N=1}^{\infty}\bigcup_{k\ge N}E_k\); it consists of points that lie in infinitely many \(E_k\). For every \(N\), countable subadditivity gives \[ \mu\!\left(\bigcup_{k\ge N}E_k\right)\le\sum_{k\ge N}\mu(E_k) <\sum_{k\ge N}2^{-k}=2^{1-N}. \] Since \(\limsup_k E_k\) is contained in each such union, its measure is at most \(2^{1-N}\) for every \(N\), and hence is zero. If \(x\notin\limsup_k E_k\), there is a \(K\) such that \(x\notin E_k\) for all \(k\ge K\). Thus \(|f_{n_{j_k}}(x)-f(x)|\le1/k\) for all \(k\ge K\), which tends to zero. The selected further subsequence converges to \(f\) almost everywhere.
This argument requires no assumption that \(\mu(X)\) be finite: the summable error-set construction works on any measure space. Finiteness enters in the converse.
Worked Example: Choosing a Further Subsequence
On \([0,1]\) with Lebesgue measure \(m\), let \(f_n(x)=n\chi_{(0,1/n^2)}(x)\), with \(f(x)=0\). First verify convergence in measure, then check the extracted subsequence \(f_{2^k}\).
Fix \(\varepsilon>0\). If \(n>\varepsilon\), then \(f_n(x)>\varepsilon\) precisely when \(x\in(0,1/n^2)\): inside this interval \(f_n=n>\varepsilon\), and outside it \(f_n=0\). Consequently, \[ m(\{|f_n-f|>\varepsilon\})=\frac{1}{n^2}\longrightarrow0. \] So \(f_n\to0\) in measure.
For the subsequence \(f_{2^k}\), use threshold \(1/k\). Since \(2^k>1/k\) for \(k\ge1\), its error set is \(E_k=(0,1/4^k)\), and \(m(E_k)=4^{-k}\). The sum of these measures is \[ \sum_{k=1}^{\infty}4^{-k}=\frac{1/4}{1-1/4}=\frac13. \] In fact, for every \(x\in[0,1]\), \(x\) belongs to only finitely many \(E_k\): if \(x>0\), then \(4^{-k}\le x\) for all sufficiently large \(k\), and \(x=0\) belongs to none because the intervals are open. Thus \(f_{2^k}(x)\le1/k\) eventually, and \(f_{2^k}(x)\to0\) at every point. This pointwise check agrees with the summable-error-set argument.
From the Subsequence Property to Convergence in Measure
For this direction, use the following elementary consequence of countable additivity. If \(A_1\supseteq A_2\supseteq\cdots\) are measurable and \(\mu(A_1)<\infty\), then \(\mu(A_j)\to\mu(\bigcap_j A_j)\). Indeed, the disjoint sets \(A_j\setminus A_{j+1}\), together with \(\bigcap_j A_j\), partition \(A_1\). Countable additivity expresses \(\mu(A_1)\) as the measure of the intersection plus the sum of the measures of those differences. The measures of the tails of this finite sum tend to zero, giving the stated limit.
Proof of the reverse implication. Assume every subsequence of \((f_n)\) has a further subsequence converging to \(f\) \(\mu\)-almost everywhere. Suppose, to obtain a contradiction, that \(f_n\) does not converge to \(f\) in measure. Negating the definition, there are \(\varepsilon_0>0\), \(\delta>0\), and a subsequence \((f_{n_j})\) such that \[ \mu(\{|f_{n_j}-f|>\varepsilon_0\})\ge\delta \quad\text{for every }j. \] By the assumed property, this subsequence has a further subsequence \((f_{n_{j_k}})\) converging to \(f\) almost everywhere.
Set \(B_k=\{|f_{n_{j_k}}-f|>\varepsilon_0\}\) and \(A_N=\bigcup_{k\ge N}B_k\). At every point where \(f_{n_{j_k}}\to f\), the strict inequality \(|f_{n_{j_k}}-f|>\varepsilon_0\) can hold only finitely often. Therefore \(\bigcap_N A_N=\limsup_k B_k\) is contained in the null set where convergence fails, so \(\mu(\bigcap_N A_N)=0\). The sets \(A_N\) decrease and \(\mu(A_1)\le\mu(X)<\infty\). Continuity from above, justified above, gives \(\mu(A_N)\to0\). Since \(B_N\subseteq A_N\), we obtain \[ \mu(B_N)\le\mu(A_N)\longrightarrow0. \] This contradicts \(\mu(B_N)\ge\delta\) for every \(N\). Hence \(f_n\to f\) in measure. \(\square\)
Worked Example: Applying the Criterion on a Finite Space
On \([0,1]\) with Lebesgue measure, let \(f_n(x)=1/n\) for every \(x\), and let \(f(x)=0\). Check both convergence in measure and the subsequence property.
For fixed \(\varepsilon>0\), the strict error set \(\{|f_n-f|>\varepsilon\}\) is all of \([0,1]\) when \(1/n>\varepsilon\), and is empty when \(1/n\le\varepsilon\). For every integer \(n\ge1/\varepsilon\), it is empty, so its measure is eventually zero. Thus \(f_n\to0\) in measure.
Now choose any subsequence \(f_{n_j}\). Its indices satisfy \(n_j\to\infty\), so \(f_{n_j}(x)=1/n_j\to0\) for every \(x\in[0,1]\). The subsequence itself is already the required further subsequence. This verifies the criterion directly, including the fact that the limit is the same \(f=0\).
Why Finite Measure Cannot Be Dropped
Worked Example: Almost-Everywhere Convergence Without Convergence in Measure
On \(\mathbb{R}\) with Lebesgue measure \(m\), let \(g_n=\chi_{[n,n+1]}\) and \(g=0\). Every subsequence of \((g_n)\) converges to zero at every fixed \(x\): once \(n>x\), \(x\notin[n,n+1]\), so all sufficiently late terms vanish at \(x\). Hence every subsequence already converges almost everywhere to the same limit.
But at the fixed strict threshold \(\varepsilon=1/2\), the error set is \(\{|g_n-g|>1/2\}=[n,n+1]\), whose Lebesgue measure is \(1\) for every \(n\). Therefore the sequence does not converge to zero in measure. The failure is consistent with the theorem: \(m(\mathbb{R})=\infty\), so the finite-measure hypothesis of the converse is absent.
Common pitfall. It is incorrect to reason that almost-everywhere convergence alone forces convergence in measure. For the sequence in the example, pointwise eventual vanishing holds at every fixed \(x\), but at the same threshold \(1/2\) the error-set measures remain \(1\). On a finite measure space, the contradiction proof works because decreasing unions of the remaining error sets have measures tending to zero. On an infinite measure space, that finite-measure step is unavailable. Conversely, convergence in measure alone does not assert almost-everywhere convergence of the entire sequence; the theorem guarantees an almost-everywhere convergent further subsequence of each subsequence.
Check Your Understanding
Keep the strict error-set threshold and the finite-measure hypothesis explicit in your answers.
- In the forward proof, why can the indices \(j_k\) be chosen strictly increasing while also ensuring \(\mu(\{|f_{n_{j_k}}-f|>1/k\})<2^{-k}\)?
- Explain why \(\sum_k\mu(E_k)<\infty\) implies \(\mu(\limsup_k E_k)=0\), using countable subadditivity.
- In the reverse proof, where exactly is \(\mu(X)<\infty\) used?
- For \(g_n=\chi_{[n,n+1]}\) on \(\mathbb{R}\), compute the error-set measure at threshold \(1/2\) and explain why it rules out convergence in measure.
- Does the forward extraction require \(\mu(X)<\infty\)? Identify the step that answers this question.