A Counterexample Is a Hypothesis Test
A useful counterexample does more than show that a conclusion can fail. It identifies which assumption made the conclusion possible. To do that, first write down the hypotheses of the result being tested. Then keep all but one of them, construct explicit measurable sets or functions, and check both the retained assumptions and the failed conclusion. In measure theory, the checks usually involve measurability, the size of an exceptional set, convergence at each point, and the exact value of an integral.
We will use Lebesgue measure \(m\). On \(\mathbb{R}\), \(m\) is sigma-finite but not finite. On \([0,1]\), the restricted Lebesgue measure is finite and complete. The functions in the examples are real-valued and measurable; statements about convergence will specify whether convergence is pointwise or almost everywhere. A function \(G\) is an integrable dominator for \(f_n\) if \(G\ge 0\), \(\int G\,dm<\infty\), and \(|f_n|\le G\) almost everywhere for every \(n\).
Two standard results provide a useful framework. Egorov’s theorem says that on a finite-measure space, if measurable real-valued functions \(f_n\) converge to \(f\) almost everywhere, then for every \(\varepsilon>0\) there is a measurable set \(E\) with measure less than \(\varepsilon\) such that the convergence is uniform outside \(E\). The standard Dominated Convergence Theorem says that if \(f_n\to f\) almost everywhere and \(|f_n|\le G\) almost everywhere for one integrable \(G\), then \(\int f_n\,d\mu\to\int f\,d\mu\). We use these named standard results to identify exactly which assumptions the examples test.
Removing the Finite-Measure Hypothesis
Worked Example: Pointwise Convergence on an Infinite-Measure Space
On \((\mathbb{R},\mathcal{L},m)\), define \(f_n=\chi_{[n,\infty)}\) for integers \(n\ge1\). Each \(f_n\) is measurable and real-valued. For a fixed \(x\in\mathbb{R}\), choose an integer \(N>x\). Whenever \(n\ge N\), \(x<n\), so \(x\notin[n,\infty)\) and \(f_n(x)=0\). Thus \(f_n(x)\to0\) at every \(x\).
Nevertheless, no measurable set \(E\) of finite measure can be removed to make this convergence uniform. Fix such an \(E\). For every integer \(n\ge1\), the set \([n,\infty)\) has infinite measure, whereas \(E\) has finite measure. Therefore \([n,\infty)\setminus E\) is nonempty: if it were empty, then \([n,\infty)\subseteq E\) would imply \(\infty=m([n,\infty))\le m(E)<\infty\). Choose \(x_n\in[n,\infty)\setminus E\). Then \(x_n\notin E\) and \(f_n(x_n)=1\). Hence \(\sup_{x\in E^c}|f_n(x)|=1\) for every \(n\), so the sequence cannot converge uniformly on \(E^c\). Check: for the uniform-convergence threshold \(1/2\), every proposed index \(N\) fails, since \(n=N\) already has a point outside \(E\) where \(|f_n-0|=1>1/2\).
Proof. Fix \(x\in\mathbb{R}\). There is an integer \(N>x\), and for all \(n\ge N\) we have \(x<n\), so \(f_n(x)=0\). This proves pointwise convergence to \(0\). Now fix a measurable \(E\) with \(m(E)<\infty\). For each \(n\), if \([n,\infty)\setminus E\) were empty, then \(m([n,\infty))\le m(E)<\infty\), contradicting \(m([n,\infty))=\infty\). Thus some \(x\in E^c\) satisfies \(x\ge n\), and at that point \(|f_n(x)-0|=1\). The uniform error on \(E^c\) is therefore at least \(1\) for every \(n\); in particular it never becomes less than \(1/2\). This proves the proposition. \(\square\)
The example isolates the finite-measure assumption in Egorov’s theorem: pointwise convergence alone does not ensure uniform convergence after deleting a set of arbitrarily small finite measure when the whole space has infinite measure. It does not contradict Egorov’s theorem, whose finite-measure hypothesis is absent here.
Removing the Integrable-Dominator Hypothesis
Worked Example: Pointwise Convergence Without Convergence of Integrals
On the finite measure space \(([0,1],\mathcal{L}|_{[0,1]},m)\), define \(f_n(x)=n\chi_{(0,1/n)}(x)\). Each \(f_n\) is nonnegative, measurable, and integrable. For every \(x\in[0,1]\), \(f_n(x)\to0\): at \(x=0\) the indicator is always zero, and for \(x>0\), once \(n\ge1/x\), we have \(x\notin(0,1/n)\). Thus the convergence holds everywhere, not merely almost everywhere.
The interval \((0,1/n)\) has measure \(1/n\), so direct integration gives
The pointwise limit is the zero function, whose integral is \(0\), while every integral in the sequence equals \(1\). Check: at \(n=4\), the function has height \(4\) on an interval of length \(1/4\); its integral is \(4\cdot(1/4)=1\), in agreement with the formula. The mismatch shows that pointwise convergence and integrability of every \(f_n\) do not suffice for convergence of the integrals.
Proof. Suppose such a function \(G\) exists. For each \(n\), the inequality \(G\ge n\) holds almost everywhere on \((0,1/n)\). Since there are only countably many \(n\), the union of the exceptional null sets is null; outside that union, all these inequalities hold at once. For \(0<x\le1/2\), set \(y=1/(2x)\ge1\) and \(n=\lfloor y\rfloor\). Then \(n\ge y/2=1/(4x)\): if \(1\le y<2\), \(n=1\ge y/2\), and if \(y\ge2\), \(\lfloor y\rfloor\ge y-1\ge y/2\). Also \(nx\le1/2<1\), so \(x\in(0,1/n)\). Consequently, outside the null exceptional set, \(G(x)\ge n\ge1/(4x)\) for almost every \(x\in(0,1/2]\).
For every integer \(k\ge1\), the interval \([2^{-(k+1)},2^{-k}]\) lies in \((0,1/2]\), and
These intervals have disjoint interiors. Summing their integrals over \(k=1,\ldots,K\) gives \(K(\ln 2)/4\), which is unbounded as \(K\to\infty\). Thus \(\int G\,dm=\infty\), contradicting the assumed integrability of \(G\). No such integrable dominator exists. \(\square\)
This proof checks more than the fact that no convenient bound has been found: it rules out every integrable common dominator. Each \(f_n\) is integrable, but the sequence has no single integrable function controlling all its members. That is the missing hypothesis relevant to the Dominated Convergence Theorem.
Why “Almost Everywhere” Cannot Mean “Except on a Large Set”
Worked Example: A Positive-Measure Set of Nonconvergence
Again use \(([0,1],\mathcal{L}|_{[0,1]},m)\). Let \(f_n(x)=1\) for every \(x\in[0,1]\) and every \(n\), and define \(f=\chi_{(1/2,1]}\). Outside \(A=[0,1/2]\), the sequence converges to \(f\): if \(x\in(1/2,1]\), then \(f_n(x)=1=f(x)\) for every \(n\). On \(A\), however, \(f(x)=0\) while \(f_n(x)=1\) for every \(n\). The set of nonconvergence is exactly \(A\), and \(m(A)=1/2>0\). Thus convergence outside \(A\) is not convergence almost everywhere.
The functions are dominated by the integrable function \(G=1\), since \(|f_n(x)|=1=G(x)\) everywhere, and \(\int G\,dm=1\). But direct calculation gives
The interval \((1/2,1]\) has length \(1-1/2=1/2\), so the two values differ. Check: at \(x=3/4\), \(f_n(x)=1=f(x)\); at \(x=1/4\), \(f_n(x)=1\ne0=f(x)\). These checks confirm both the convergence set and its complement. The integrable bound is present, but the required almost-everywhere convergence is not.
An exceptional set in an almost-everywhere statement must have measure zero. A set of measure \(1/2\), however small it might seem in another setting, is not negligible for this purpose. This example does not dispute the Dominated Convergence Theorem: one of its hypotheses fails on a positive-measure set.
How to Diagnose a Counterexample
Before accepting a proposed example, verify it against a short checklist. A counterexample is conclusive only when it satisfies the hypotheses it is meant to retain and demonstrably violates the targeted condition.
| Hypothesis being tested | What to verify | Failure exhibited here |
|---|---|---|
| Finite measure | Check the measure of the whole space and every proposed exceptional set | On \(\mathbb{R}\), every finite-measure deletion leaves points in each tail |
| Integrable domination | Check whether one integrable \(G\) bounds every \(|f_n|\) | The proposed bound would have to dominate \(1/(4x)\) near zero |
| Almost-everywhere convergence | Measure the set where convergence actually fails | The failure set has measure \(1/2\), not zero |
Common mistake. For the sequence \(f_n=n\chi_{(0,1/n)}\), one might write, “Since \(f_n\to0\), \(\int f_n\,dm\to\int0\,dm=0\).” That step interchanges a limit and an integral without checking a theorem’s hypotheses. The correction is to compute first: \(\int f_n\,dm=n\cdot(1/n)=1\) for every \(n\), whereas the integral of the pointwise limit is \(0\). The missing integrable dominator is not a technical detail; its absence permits precisely this discrepancy.
In practice, changing one assumption at a time makes an example easier to interpret. If the intended test is domination, retain measurability and almost-everywhere convergence, then show that the integrals fail to converge and verify that no integrable common bound exists. If the test is finite measure, retain pointwise convergence and show directly why every finite-measure exceptional set is insufficient. If the test is almost-everywhere convergence, identify the nonconvergence set and compute its measure. These checks turn examples into precise diagnostics rather than suggestive pictures.
Check Your Understanding
For each question, identify the hypothesis being tested and justify the relevant convergence or integral calculation.
- For \(f_n=\chi_{[n,\infty)}\) on \(\mathbb{R}\), prove that \(f_n(x)\to0\) for every fixed \(x\), even though \(m([n,\infty))=\infty\) for every \(n\).
- For \(f_n=n\chi_{(0,1/n)}\) on \([0,1]\), compute \(\int f_n\,dm\) and \(\int\lim_n f_n\,dm\). Which condition needed for the Dominated Convergence Theorem is not available?
- Explain why an integrable \(G\) dominating every \(n\chi_{(0,1/n)}\) would have to satisfy \(G(x)\ge1/(4x)\) for almost every \(x\in(0,1/2]\).
- In the positive-measure exception example, identify the exact set where \(f_n\) fails to converge to \(f\), and compute its measure.
- When checking a counterexample to an almost-everywhere convergence hypothesis, why is it not enough to say that convergence fails only on a set that “looks small”?