From Convergence in Measure to Convergence in \(L^1\)
Convergence in measure controls how large the set of substantial errors can be, but it does not by itself control the integral of the error. A function can be very large on a very small set: the set may shrink to measure zero while the integral on it remains significant. The Lebesgue Integration results from earlier in this course, especially the Dominated Convergence Theorem, provide one way to control such errors. A different route is to control the large values of the whole sequence uniformly.
This tutorial develops the Vitali convergence argument. Its central estimate separates the space into a set where \(f_n\) and its limit differ by at most a chosen small amount, and an exceptional set whose measure tends to zero. Uniform integrability controls the integrals over that exceptional set. We will also distinguish two uses of finite measure: it is needed to make the error on the nonexceptional set small, but it is not needed to extract an almost-everywhere convergent subsequence from convergence in measure.
Uniform Integrability and Small Sets
Let \((X,\mathcal{F},\mu)\) be a measure space and suppose \(\mu(X)<\infty\). For a sequence of measurable real-valued functions, the tail criterion below says that the integrals contributed by values larger than \(K\) become uniformly small as \(K\) increases. This is the form of uniform integrability we will use.
The tail condition also implies that the integrals of the \(f_n\) are uniformly bounded. Indeed, choose \(K\) so that the supremum of the tail integrals is at most \(1\). Splitting \(X\) into \(\{|f_n|\leq K\}\) and \(\{|f_n|>K\}\) gives $$ \int_X|f_n|\,d\mu \leq K\mu(X)+\int_{\{|f_n|>K\}}|f_n|\,d\mu \leq K\mu(X)+1 $$ for every \(n\). More importantly for the convergence proof, the same split works on any measurable subset.
Proof. Choose \(K>0\) so large that $$ \sup_{n\geq1}\int_{\{|f_n|>K\}}|f_n|\,d\mu<\frac{\varepsilon}{2}. $$ For every measurable \(A\), splitting its integral according to whether \(|f_n|\leq K\) or \(|f_n|>K\) gives $$ \int_A|f_n|\,d\mu \leq K\mu(A)+\int_{\{|f_n|>K\}}|f_n|\,d\mu. $$ Set \(\eta=\varepsilon/(2K)\). If \(\mu(A)<\eta\), then \(K\mu(A)<\varepsilon/2\), and the tail term is also less than \(\varepsilon/2\). Thus \(\int_A|f_n|\,d\mu<\varepsilon\) for every \(n\), proving the lemma.
Worked Example: Moving Indicators on a Finite Interval
On \(X=[0,1]\) with Lebesgue measure, define $$ f_n=\mathbf{1}_{[0,\,1/2+1/(2n)]}, \qquad f=\mathbf{1}_{[0,\,1/2]}. $$ The interval defining \(f_n\) has length \(1/2+1/(2n)\), which is at most \(1\), including when \(n=1\). The functions differ only on the interval between \(1/2\) and \(1/2+1/(2n)\), up to endpoints of measure zero. Consequently, $$ \mu(\{|f_n-f|>0\})=\frac{1}{2n}, \qquad \int_0^1|f_n-f|\,d\mu=\frac{1}{2n}. $$ In particular, \(f_n\to f\) in measure and in \(L^1\). The sequence is uniformly integrable: \(|f_n|\leq1\) everywhere, so for every \(K\geq1\) the set \(\{|f_n|>K\}\) is empty. This simple case illustrates that bounded sequences satisfy the tail criterion immediately.
A Subsequence Fact That Does Not Need Finite Measure
Convergence in measure is often paired with almost-everywhere convergence of a subsequence. The extraction itself does not require \(\mu(X)<\infty\). This distinction matters in proofs: finite measure may be essential for a later estimate without being necessary for the subsequence argument.
Proof. For each positive integer \(k\), convergence in measure allows us to choose the indices successively, with \(n_k>n_{k-1}\), so that $$ \mu\bigl(\{|f_{n_k}-f|>1/k\}\bigr)<2^{-k}. $$ Write \(A_k=\{|f_{n_k}-f|>1/k\}\). For every \(m\), countable subadditivity gives $$ \mu\left(\bigcup_{k\geq m}A_k\right) \leq\sum_{k\geq m}2^{-k} =2^{1-m}. $$ The set of points that belong to infinitely many \(A_k\) is contained in \(\bigcup_{k\geq m}A_k\) for every \(m\). Its measure is therefore at most \(2^{1-m}\) for every \(m\), and hence is zero. At every point outside this null set, there is an index after which \(|f_{n_k}-f|\leq1/k\). Since \(1/k\to0\), the subsequence converges to \(f\) there. No assumption on \(\mu(X)\) was used.
The Vitali Convergence Theorem
We can now state the main result. In addition to convergence in measure, it assumes uniform control of the tails of the sequence. The finite-measure hypothesis will be used to control the error on the part of the space where \(f_n\) is already close to \(f\).
Proof. First suppose \(\mu(X)=0\). Every measurable function has integral of its absolute value zero on \(X\), so the conclusion holds. Now suppose \(0<\mu(X)<\infty\).
Apply the almost-everywhere subsequence lemma to obtain a subsequence \(f_{n_k}\) converging to \(f\) almost everywhere. Fatou’s Lemma applied to \(|f_{n_k}|\) gives $$ \int_X|f|\,d\mu \leq\liminf_{k\to\infty}\int_X|f_{n_k}|\,d\mu. $$ The right-hand side is finite because the tail condition and \(\mu(X)<\infty\) imply a uniform bound on \(\int_X|f_n|\,d\mu\), as shown above. Thus \(f\) is integrable.
We also need control of the integral of \(f\) on small sets. Fix \(\varepsilon>0\). Choose \(K>0\) so that $$ \sup_n\int_{\{|f_n|>K\}}|f_n|\,d\mu<\varepsilon. $$ For any \(A\in\mathcal{F}\), Fatou’s Lemma on \(A\), using the almost-everywhere convergence of \(f_{n_k}\), yields $$ \int_A|f|\,d\mu \leq\liminf_{k\to\infty}\int_A|f_{n_k}|\,d\mu \leq K\mu(A)+\varepsilon. $$ Since \(\varepsilon\) can be made arbitrarily small by choosing \(K\) sufficiently large, this proves that the integral of \(|f|\) is small on sets of sufficiently small measure. The small-set lemma gives the corresponding uniform control for all the \(f_n\).
Now fix a target error \(\gamma>0\). Choose \(\delta>0\) such that $$ \delta\mu(X)<\frac{\gamma}{3}. $$ By the small-set lemma, choose \(\eta>0\) so that \(\mu(A)<\eta\) implies $$ \sup_n\int_A|f_n|\,d\mu<\frac{\gamma}{3} \quad\text{and}\quad \int_A|f|\,d\mu<\frac{\gamma}{3}. $$ The second condition follows from the small-set control for \(f\) just established. Define \(A_n=\{|f_n-f|>\delta\}\). Since \(f_n\to f\) in measure, \(\mu(A_n)\to0\). Therefore, for all sufficiently large \(n\), \(\mu(A_n)<\eta\).
On \(X\setminus A_n\), the difference is at most \(\delta\); on \(A_n\), it is at most \(|f_n|+|f|\). Hence, for all sufficiently large \(n\), $$ \begin{aligned} \int_X|f_n-f|\,d\mu &\leq \int_{X\setminus A_n}\delta\,d\mu +\int_{A_n}|f_n|\,d\mu+\int_{A_n}|f|\,d\mu\\ &\leq \delta\mu(X)+\int_{A_n}|f_n|\,d\mu+\int_{A_n}|f|\,d\mu\\ &<\frac{\gamma}{3}+\frac{\gamma}{3}+\frac{\gamma}{3} =\gamma. \end{aligned} $$ This proves convergence in \(L^1\). The finite-measure hypothesis is used in the estimate \(\delta\mu(X)\) on the complement of the exceptional set. It is not needed for the almost-everywhere subsequence extraction.
Worked Example: An Unbounded Uniformly Integrable Sequence
On \((0,1)\) with Lebesgue measure, let $$ f_n(x)=x^{-1/2}\mathbf{1}_{[1/n,\,1]}(x), \qquad f(x)=x^{-1/2}. $$ For each \(x\in(0,1)\), eventually \(1/n\leq x\), so \(f_n(x)\to f(x)\). In particular, the sequence converges pointwise and therefore in measure on this finite-measure space.
For \(K>0\), the set \(\{|f_n|>K\}\) is contained in \(\{x:x^{-1/2}>K\}\), which is contained in \((0,K^{-2})\). Thus $$ \int_{\{|f_n|>K\}}|f_n(x)|\,dx \leq\int_0^{\min(1,K^{-2})}x^{-1/2}\,dx \leq\frac{2}{K}. $$ The last bound holds also when \(K<1\), since the integral over \((0,1)\) is \(2\leq2/K\). Taking the supremum over \(n\) and then letting \(K\to\infty\) verifies uniform integrability. The exact \(L^1\) error is $$ \int_0^1|f_n-f|\,dx =\int_0^{1/n}x^{-1/2}\,dx =2n^{-1/2}\longrightarrow0. $$ The theorem applies even though the functions are not bounded by one common constant.
Why the Finite-Measure Hypothesis Matters
The proof shows exactly where finite total measure enters: on the set where \(|f_n-f|\leq\delta\), the integral of the difference is bounded by \(\delta\mu(X)\). On an infinite-measure space, this estimate need not be small, no matter how small \(\delta\) is. The next example shows that convergence in measure and uniform control of tails alone do not guarantee \(L^1\) convergence there.
Worked Example: Convergence in Measure Without \(L^1\) Convergence
On \(\mathbb{R}\) with Lebesgue measure, define $$ f_n(x)=\frac{1}{n}\mathbf{1}_{[0,n]}(x), \qquad f(x)=0. $$ For any fixed \(\varepsilon>0\), if \(n\geq1/\varepsilon\), then \(|f_n(x)|\leq\varepsilon\) everywhere. Consequently, $$ \mu(\{|f_n-f|>\varepsilon\})=0 $$ for all such \(n\), so \(f_n\to0\) in measure. Each function has integral $$ \int_{\mathbb{R}}|f_n|\,dx=\frac{1}{n}\,n=1. $$ The tail integrals are uniformly zero for \(K\geq1\), because \(|f_n|\leq1\); the integrals are also uniformly bounded. Nevertheless, $$ \int_{\mathbb{R}}|f_n-f|\,dx=1 $$ for every \(n\), so there is no \(L^1\) convergence. The supports spread across larger intervals while the function heights shrink: convergence in measure sees the heights, but the total integrals do not vanish.
This example also clarifies why hypotheses should not be moved from one step of a proof to another. Almost-everywhere subsequence extraction works on arbitrary measure spaces by choosing summable exceptional-set measures. In contrast, turning a small pointwise error into a small integral over its complement uses finite total measure. The two arguments have different requirements.
Check Your Understanding
Use the definitions and proof strategy in this tutorial to answer the following questions.
- How does the tail criterion for uniform integrability imply uniform control of integrals over sets of small measure?
- In the subsequence lemma, why does the sum of the exceptional-set measures being finite imply almost-everywhere convergence of the chosen subsequence?
- Where exactly does the finite-measure assumption enter the proof of the Vitali convergence theorem?
- Why does the sequence \(x^{-1/2}\mathbf{1}_{[1/n,1]}\) satisfy the uniform-integrability tail criterion?
- For the sequence \(n^{-1}\mathbf{1}_{[0,n]}\) on \(\mathbb{R}\), verify both convergence in measure to zero and failure of convergence in \(L^1\).