A Map of the Main Modes
The previous tutorial defined convergence in measure by asking whether the sets on which the error exceeds a fixed threshold become small. This differs from uniform convergence, which controls the error at every point, and from convergence in \(L^p\), which controls an integral of the error. Almost-everywhere convergence instead describes the behavior at individual points, apart from a null set. These modes are related, but their implications depend on the measure space and on additional bounds.
In this tutorial, all functions are measurable and finite-valued. For \(1\leq p<\infty\), convergence in \(L^p\) means \(\|f_n-f\|_p\to0\). The basic implications to keep in view are these: uniform convergence always implies convergence in measure; on a finite-measure space, almost-everywhere convergence implies convergence in measure; and \(L^p\) convergence implies convergence in measure. In general, none of these statements should be reversed without further hypotheses.
Uniform Convergence Always Implies Convergence in Measure
Uniform convergence is a pointwise bound that works simultaneously at every point. Consequently, it controls each exceptional set in the definition of convergence in measure without requiring any assumption on \(\mu(X)\).
Proof. Fix \(\varepsilon>0\). Uniform convergence gives an index \(N\) such that, for every \(n\geq N\) and every \(x\in X\), \(|f_n(x)-f(x)|<\varepsilon\). Therefore $$ \{x\in X:|f_n(x)-f(x)|>\varepsilon\}=\varnothing \qquad(n\geq N). $$ The measure of this exceptional set is zero for every \(n\geq N\), so it tends to zero. This is convergence in measure. \(\square\)
The empty-set conclusion is important on an infinite-measure space: no estimate of the form “a small error times \(\mu(X)\)” is needed. Uniform convergence makes the error set empty once the uniform error falls below the chosen threshold.
Worked Example: Uniform Convergence on the Real Line
Fix \(1\leq p<\infty\), and on \(\mathbb{R}\) with Lebesgue measure define $$ f_n(x)=\frac{1}{n}\mathbf{1}_{[0,n^p]}(x), \qquad f(x)=0. $$ For every \(x\), \(|f_n(x)-f(x)|\leq1/n\), so \(f_n\to0\) uniformly. In particular, for any fixed \(\varepsilon>0\), if \(n>1/\varepsilon\), then the error is less than \(\varepsilon\) everywhere and the exceptional set is empty. Thus \(f_n\to0\) in measure.
Nevertheless, the \(L^p\) error does not tend to zero. Direct calculation gives $$ \|f_n-f\|_p^p =\int_{\mathbb{R}}|f_n(x)|^p\,dx =\frac{1}{n^p}\,n^p =1, \qquad \text{so}\qquad \|f_n-f\|_p=1. $$ Each \(f_n\) belongs to \(L^p(\mathbb{R})\), but the sequence does not converge to zero in \(L^p\). Uniform convergence therefore does not imply \(L^p\) convergence on an infinite-measure space.
Finite Measure Connects Almost-Everywhere and In-Measure Convergence
On a finite-measure space, almost-everywhere convergence implies convergence in measure. The proof was given in the previous tutorial: for a fixed threshold, the indicators of the error sets tend to zero almost everywhere, and the constant function \(1\) is integrable when \(\mu(X)<\infty\). The Dominated Convergence Theorem then shows that the measures of those sets tend to zero. The finite-measure hypothesis is what permits this argument.
Without that hypothesis, even pointwise convergence everywhere may fail to imply convergence in measure. The following example uses intervals of fixed length moving out along the real line.
Worked Example: Pointwise Convergence Without Convergence in Measure
On \(\mathbb{R}\) with Lebesgue measure, set \(f_n=\mathbf{1}_{[n,n+1]}\) and \(f=0\). A fixed point \(x\) can belong to at most two of the intervals \([n,n+1]\): only adjacent intervals can share an endpoint. Hence \(x\) belongs to only finitely many of them, and \(f_n(x)=0\) for all sufficiently large \(n\). Thus \(f_n(x)\to0\) for every \(x\).
For \(0<\varepsilon<1\), however, $$ \{x:|f_n(x)-f(x)|>\varepsilon\}=[n,n+1], \qquad \mu([n,n+1])=1. $$ The exceptional-set measures do not tend to zero, so \(f_n\) does not converge to zero in measure. This example shows why the finite-measure result cannot be applied on \(\mathbb{R}\).
Convergence in measure still has a useful relationship with almost-everywhere convergence on arbitrary spaces: the subsequence construction in the proof of the completeness theorem for convergence in measure in the previous tutorial shows that a sequence converging in measure has a subsequence converging almost everywhere. This is a subsequence conclusion, not a statement that the full sequence converges almost everywhere. The distinction matters in examples where small exceptional sets move around and repeatedly contain the same points.
Convergence in Measure and Lp Convergence
For finite \(p\), \(L^p\) convergence implies convergence in measure by the Finite-\(p\) Convergence Implies Convergence in Measure theorem from the tutorial “Convergence in Lp.” Thus the integral control supplied by the \(L^p\) norm is strong enough to make each fixed-threshold error set small. The converse does not hold in general, even on a space of finite measure.
Worked Example: Convergence in Measure Without Lp Convergence
On \([0,1]\) with Lebesgue measure, fix \(1\leq p<\infty\) and let $$ f_n(x)=n^{1/p}\mathbf{1}_{(0,1/n]}(x), \qquad f(x)=0. $$ For \(0<\varepsilon<n^{1/p}\), the error set is \((0,1/n]\), whose measure is \(1/n\); if \(\varepsilon\geq n^{1/p}\), the error set is empty because the defining inequality is strict. For any fixed \(\varepsilon>0\), eventually \(n^{1/p}>\varepsilon\), and then the error-set measure is \(1/n\to0\). Hence \(f_n\to0\) in measure.
The \(L^p\) norm is instead constant: $$ \|f_n\|_p^p =\int_0^1 |f_n(x)|^p\,dx =\int_{(0,1/n]} n\,dx =n\cdot\frac{1}{n} =1. $$ Therefore \(\|f_n\|_p=1\) for every \(n\), and there is no \(L^p\) convergence to zero. The supports shrink, but the function values grow enough to preserve the \(p\)th-power integral.
A useful additional condition reverses this failure: if the errors are uniformly bounded and the space has finite measure, convergence in measure does imply convergence in \(L^p\). This result makes precise how control of large values combines with control of exceptional sets.
Proof. If \(M=0\), then \(f_n=f\) almost everywhere for every \(n\), and the conclusion follows immediately. Suppose \(M>0\). Fix \(\delta>0\) and define \(A_n=\{x:|f_n(x)-f(x)|>\delta\}\). On \(X\setminus A_n\), the \(p\)th power of the error is at most \(\delta^p\); on \(A_n\), it is at most \(M^p\), apart from a null set. Consequently, $$ \int_X|f_n-f|^p\,d\mu \leq \delta^p\mu(X)+M^p\mu(A_n). $$ Since \(f_n\to f\) in measure, \(\mu(A_n)\to0\). Thus, for every fixed \(\delta>0\), $$ \limsup_{n\to\infty}\|f_n-f\|_p^p \leq \delta^p\mu(X). $$ This holds for every \(\delta>0\). Letting \(\delta\) decrease to zero gives \(\limsup_n\|f_n-f\|_p^p\leq0\). The norms are nonnegative, so \(\|f_n-f\|_p^p\to0\), and hence \(\|f_n-f\|_p\to0\). \(\square\)
The estimate separates the two sources of error: outside \(A_n\) the error is at most \(\delta\), while inside \(A_n\) it may be as large as \(M\), but that set has small measure. Finiteness of \(\mu(X)\) controls the contribution outside the exceptional set. The uniform bound \(M\) controls the contribution on it. Without either kind of control, the argument does not establish \(L^p\) convergence.
Why Lp Convergence Need Not Give Pointwise Convergence of the Full Sequence
The implications above do not mean that \(L^p\) convergence forces the full sequence to converge almost everywhere. The Almost-Everywhere Convergent Subsequence theorem from “Almost-Everywhere Convergence Versus Lp Convergence” gives an almost-everywhere convergent subsequence, but the full sequence can continue to revisit sets of small measure.
Worked Example: Lp Convergence with Persistent Pointwise Oscillation
On \([0,1]\) with Lebesgue measure, for each positive integer \(k\), partition the interval into \(k\) half-open intervals $$ I_{k,j}=\left[\frac{j-1}{k},\frac{j}{k}\right), \qquad j=1,\ldots,k, $$ and include the endpoint \(1\) in the final interval \(I_{k,k}\). Enumerate the functions \(\mathbf{1}_{I_{k,j}}\) in blocks: first all \(k\) indicators at level \(k=1\), then all \(k\) indicators at level \(k=2\), and so on. Let \(f_n\) denote this single enumerated sequence.
Every function in the level-\(k\) block has \(L^p\) norm \(k^{-1/p}\), since its support has measure \(1/k\). As the sequence advances through the blocks, \(k\to\infty\), so these norms tend to zero. Thus \(f_n\to0\) in \(L^p\).
For any fixed \(x\in[0,1]\), exactly one interval at each level contains \(x\). The corresponding indicator in that block takes value \(1\) at \(x\), while the indicators of the other intervals in the same block take value \(0\). There are infinitely many levels, so the values \(f_n(x)\) equal \(1\) infinitely often and \(0\) infinitely often. They do not converge at any \(x\). This example is consistent with the subsequence theorem: it rules out convergence of the full sequence, not the existence of an almost-everywhere convergent subsequence.
Putting the Implications Together
The main relationships can be summarized by recording both the implication and the condition attached to it. A missing condition can invalidate a conclusion, as the examples demonstrate.
| Starting mode | Conclusion | Condition or caution |
|---|---|---|
| Uniform convergence | Convergence in measure | Holds on any measure space |
| Almost-everywhere convergence | Convergence in measure | Holds when the whole space has finite measure |
| Convergence in Lp, for finite p | Convergence in measure | Holds on any measure space |
| Convergence in measure | An almost-everywhere convergent subsequence | Holds on any measure space; it need not apply to the full sequence |
| Convergence in measure | Convergence in Lp | Finite measure and uniformly bounded errors are sufficient |
A common pitfall is to treat these modes as interchangeable simply because each describes some form of small error. Uniform convergence makes every fixed-threshold exceptional set eventually empty, while convergence in measure only makes its measure small. \(L^p\) convergence controls an integral and therefore prevents the kind of persistent spike in the shrinking-support example. Almost-everywhere convergence controls individual points but, on an infinite-measure space, does not control the measures of error sets.
Check Your Understanding
Use the stated hypotheses carefully when deciding which implications apply.
- Why does uniform convergence imply convergence in measure even when the measure of the whole space is infinite?
- Which hypothesis allows almost-everywhere convergence to imply convergence in measure?
- In the shrinking-support example, why does the \(L^p\) norm remain equal to one even though the functions converge in measure to zero?
- What two bounds are used in the proof that bounded errors and convergence in measure imply \(L^p\) convergence?
- Why can a sequence converge in \(L^p\) while failing to converge pointwise at every point?
- What does the almost-everywhere subsequence theorem guarantee for a sequence converging in measure, and what does it not guarantee?