From Convergence in Measure to Convergence in L1
Convergence in measure controls how much of the space has a large pointwise error, but it does not by itself control the total integral of that error. Uniform integrability supplies the missing control: it prevents substantial integral mass from concentrating in regions where the functions are large. Vitali’s theorem says that, on a finite measure space, these two conditions together imply convergence in L1.
Throughout this tutorial, \((X,\mathcal{M},\mu)\) is a measure space with \(\mu(X)<\infty\). The functions considered are real-valued and measurable, except that a limit may initially be specified only \(\mu\)-almost everywhere. A function \(f\) is integrable if \(\int_X |f|\,d\mu<\infty\). Convergence \(f_n\to f\) in measure means that for every \(\varepsilon>0\), \(\mu(\{|f_n-f|>\varepsilon\})\to0\). Convergence in L1 means \(\int_X|f_n-f|\,d\mu\to0\).
On a finite measure space, this tail condition also prevents the integrals \(\int|f_n|\,d\mu\) from becoming unbounded: for any \(K>0\), $$ \int_X|f_n|\,d\mu =\int_{\{|f_n|\le K\}}|f_n|\,d\mu+\int_{\{|f_n|>K\}}|f_n|\,d\mu \le K\mu(X)+\int_{\{|f_n|>K\}}|f_n|\,d\mu. $$ Finiteness of \(\mu(X)\) matters here. It is part of the setting of the theorem below.
Vitali’s Convergence Theorem
This is a standard form of Vitali’s convergence theorem. Its proof by truncation is developed in the later treatment of Vitali’s theorem. The point of the statement is the combination of two controls: convergence in measure controls errors away from large values, while uniform integrability controls the tails. Neither condition alone asserts L1 convergence.
Worked Example: Powers Converging to Zero
On \([0,1]\) with Lebesgue measure \(m\), let \(f_n(x)=x^n\), \(n\ge1\). Verify convergence in measure and uniform integrability, then compute the L1 limit.
For every \(x\in[0,1)\), \(x^n\to0\); at \(x=1\), the sequence equals \(1\). Thus the pointwise limit is \(0\) almost everywhere. To check convergence in measure directly, fix \(0<\varepsilon<1\). The strict error set is $$ \{x\in[0,1]:|x^n|>\varepsilon\}=(\varepsilon^{1/n},1], $$ whose measure is \(1-\varepsilon^{1/n}\), tending to \(0\). If \(\varepsilon\ge1\), the strict error set is empty because \(x^n\le1\). Hence \(f_n\to0\) in measure.
Also \(0\le f_n\le1\). For every \(K\ge1\), the set \(\{|f_n|>K\}\) is empty for every \(n\), so every tail integral is zero. The sequence is uniformly integrable. Direct integration gives $$ \int_0^1|f_n-0|\,dm=\int_0^1x^n\,dx=\frac{1}{n+1}\longrightarrow0. $$ Check: the computed L1 error is \(1/(n+1)\), which tends to zero and agrees with the conclusion of Vitali’s theorem.
Worked Example: Indicators of Shrinking Intervals
On \([0,1]\) with Lebesgue measure, set \(f_n=\chi_{(0,1/n)}\). Determine its limit in measure and in L1, and check uniform integrability.
For \(0<\varepsilon<1\), the strict error set \(\{|f_n|>\varepsilon\}\) is exactly \((0,1/n)\), so its measure is \(1/n\to0\). For \(\varepsilon\ge1\), the set is empty because \(f_n\) only takes the values \(0\) and \(1\). Thus \(f_n\to0\) in measure. Its integral error is $$ \int_0^1|f_n|\,dm=m((0,1/n))=\frac1n\longrightarrow0. $$ For \(K\ge1\), \(\{|f_n|>K\}\) is empty for every \(n\), so the sequence is uniformly integrable.
Check: the interval \((0,1/n)\) has measure \(1/n\), and the indicator equals \(1\) there, so its integral is also \(1/n\). The independent computation confirms the L1 error.
The Converse: What L1 Convergence Forces
When a sequence already converges in L1, it necessarily converges in measure. In this setting it also has uniformly integrable tails. The next proposition proves these implications directly; they explain why, for a sequence that converges in measure, uniform integrability is precisely the additional condition in Vitali’s theorem.
Proof. Fix \(\varepsilon>0\), and let \(E_n=\{|f_n-f|>\varepsilon\}\). On \(E_n\), \(|f_n-f|\ge\varepsilon\), so $$ \varepsilon\,\mu(E_n)\le\int_{E_n}|f_n-f|\,d\mu\le\int_X|f_n-f|\,d\mu. $$ The right side tends to zero by L1 convergence. Dividing by the positive number \(\varepsilon\) gives \(\mu(E_n)\to0\), which is convergence in measure. \(\square\)
Proof. Let \(\eta>0\). Since \(f\) is integrable, choose \(R\ge1\) so that $$ \int_{\{|f|>R\}}|f|\,d\mu<\frac{\eta}{8}. $$ Set \(\delta=\eta/(8R)\). For any measurable \(E\) with \(\mu(E)<\delta\), splitting into the parts where \(|f|>R\) and \(|f|\le R\) gives $$ \int_E|f|\,d\mu \le\int_{\{|f|>R\}}|f|\,d\mu+R\mu(E) <\frac{\eta}{8}+\frac{\eta}{8} =\frac{\eta}{4}. $$ By L1 convergence, there is an \(N\) such that \(\int_X|f_n-f|\,d\mu<\eta/4\) whenever \(n\ge N\).
For \(K>0\), put \(E_n(K)=\{|f_n|>K\}\). The triangle inequality shows $$ E_n(K)\subseteq\{|f|>K/2\}\cup\{|f_n-f|>K/2\}. $$ As \(K\to\infty\), \(\mu(\{|f|>K/2\})\to0\), since \(f\) is finite almost everywhere and the sets decrease to a null set. For \(n\ge N\), the elementary estimate used in the preceding proposition gives $$ \mu(\{|f_n-f|>K/2\})\le\frac{2}{K}\int_X|f_n-f|\,d\mu<\frac{\eta}{2K}. $$ Consequently, for all sufficiently large \(K\), \(\mu(E_n(K))<\delta\) for every \(n\ge N\). For such \(K\), $$ \int_{E_n(K)}|f_n|\,d\mu \le\int_X|f_n-f|\,d\mu+\int_{E_n(K)}|f|\,d\mu <\frac{\eta}{4}+\frac{\eta}{4} =\frac{\eta}{2} \qquad(n\ge N). $$ Here the first term bounds the integral of \(|f_n-f|\) over \(E_n(K)\), and the second is less than \(\eta/4\) by the choice of \(\delta\).
There are only finitely many indices \(n<N\). For each such \(n\), integrability gives \(\int_{\{|f_n|>K\}}|f_n|\,d\mu\to0\) as \(K\to\infty\): this follows from monotone convergence applied to \(|f_n|\chi_{\{|f_n|\le K\}}\uparrow |f_n|\), with the integral of the complementary tail tending to zero. Increase \(K\), if necessary, so each of these finitely many tail integrals is less than \(\eta\), while retaining the bound for \(n\ge N\). Then $$ \sup_{n\ge1}\int_{\{|f_n|>K\}}|f_n|\,d\mu\le\eta. $$ Since \(\eta>0\) was arbitrary, the tail integrals tend uniformly to zero. Thus \((f_n)\) is uniformly integrable. \(\square\)
Together, the theorem and these propositions give the useful equivalence: on a finite measure space, for a sequence converging in measure to an integrable function, L1 convergence is equivalent to uniform integrability of the sequence. The direction from L1 convergence also proves convergence in measure, so L1 convergence always supplies both conclusions.
Worked Example: L1 Convergence and Uniform Tails
On \([0,1]\) with Lebesgue measure, let \(f_n(x)=1+x/n\) and \(f(x)=1\). Compute the L1 error and verify uniform integrability directly.
For every \(x\in[0,1]\), \(|f_n(x)-f(x)|=x/n\le1/n\), so the sequence converges uniformly, hence also in measure. Its exact L1 error is $$ \int_0^1|f_n-f|\,dm=\int_0^1\frac{x}{n}\,dx=\frac{1}{2n}. $$ Also \(1\le f_n(x)\le2\) for all \(x\in[0,1]\) and all \(n\ge1\). Therefore, for every \(K\ge2\), the strict tail set \(\{|f_n|>K\}\) is empty and its integral is zero.
Check: integrating \(x/n\) gives \([x^2/(2n)]_0^1=1/(2n)\). This tends to zero, and the common threshold \(K=2\) makes every tail integral zero, as required.
A Pitfall About Limits in Measure
Convergence in measure does not say that a function is eventually equal to its limit outside small exceptional sets. In particular, a fixed error threshold gives a bound on the size of an error set, not equality of the functions. A concrete contradiction argument illustrates the correct use of those sets.
Worked Example: Alternating Constants Cannot Converge in Measure
On \([0,1]\) with Lebesgue measure, define \(f_n(x)=1\) for even \(n\) and \(f_n(x)=-1\) for odd \(n\). Suppose, for contradiction, that \(f_n\) converges in measure to a measurable real-valued \(h\).
The even subsequence is constantly \(1\). Its convergence in measure to \(h\), tested at threshold \(\varepsilon=1/2\), implies $$ m(\{|1-h|>1/2\})=0, $$ because this error set is the same for every even index, while its measure must tend to zero. The odd subsequence is constantly \(-1\), so the same reasoning gives \(m(\{|-1-h|>1/2\})=0\). Outside the union of these two null sets, both \(|1-h|\le1/2\) and \(|h+1|\le1/2\) hold. The triangle inequality would then give $$ 2=|1-(-1)|\le |1-h|+|h-(-1)|\le\frac12+\frac12=1, $$ which is impossible on a set of measure \(1\). Therefore the sequence does not converge in measure.
Correction to a common error: the threshold \(1/2\) does not imply \(h=1\) or \(h=-1\) almost everywhere. It implies the two distance bounds above, each outside a null set; those bounds are incompatible by the triangle inequality.
Vitali’s theorem is useful because it turns convergence in measure into convergence of absolute integral errors, which is the form needed to control expectations and other integral quantities. The key check is not merely whether each \(f_n\) is integrable, but whether one tail threshold \(K\) makes the tail integral small uniformly over all \(n\). On a finite measure space, convergence in measure plus this uniform tail control is enough; L1 convergence, in turn, forces both convergence in measure and uniform integrability.
Check Your Understanding
For each question, distinguish carefully between convergence in measure, convergence in L1, and uniform integrability.
- On \([0,1]\), let \(f_n(x)=x^n\). For \(0<\varepsilon<1\), what is the measure of \(\{|f_n|>\varepsilon\}\), and what is \(\int_0^1|f_n|\,dm\)?
- State the tail-integral definition of uniform integrability, including the quantifier that must be uniform in \(n\).
- Why does L1 convergence imply convergence in measure? Identify the inequality used for the set \(\{|f_n-f|>\varepsilon\}\).
- Explain why the sequence \(\chi_{(0,1/n)}\) is uniformly integrable and converges to zero in L1.
- For the alternating sequence with values \(1\) and \(-1\), explain why the two bounds \(|h-1|\le1/2\) and \(|h+1|\le1/2\) cannot both hold at a point.