Why Pointwise Limits Do Not Control Integrals
Pointwise convergence describes the values of a sequence at each fixed point. An integral, however, accounts for both the size of the values and the measure of the sets on which they occur. A sequence can therefore converge pointwise to zero while maintaining a fixed positive integral: its values may grow as its support shrinks, or its support may move out of every fixed bounded region. These examples do not contradict the Dominated Convergence Theorem. They show why its integrable-domination hypothesis matters.
Throughout, functions in the examples are measurable and real-valued, and integrals are Lebesgue integrals. The Dominated Convergence Theorem says that on a measure space \((X,\mathcal{M},\mu)\), if \(f_n\to f\) \(\mu\)-a.e. and \(|f_n|\le g\) \(\mu\)-a.e. for every \(n\), where \(g\) is integrable, then \(f\) is integrable and \(\int f_n\,d\mu\to\int f\,d\mu\). The examples below violate the common integrable bound; pointwise convergence alone is not a replacement for it.
Shrinking Supports and Moving Spikes
Worked Example: A Spike on a Finite Measure Space
Take \(X=[0,1]\), the Lebesgue sigma-algebra \(\mathcal{L}\) restricted to \(X\), and Lebesgue measure \(m\). For each integer \(n\ge1\), set \(f_n(x)=n\chi_{(0,1/n)}(x)\). The strict interval membership means \(f_n(0)=0\); the endpoint \(1/n\) is also outside the support.
For \(x=0\), every \(f_n(x)=0\). For fixed \(x\in(0,1]\), choose an integer \(N>1/x\). Whenever \(n\ge N\), \(1/n<x\), so \(x\notin(0,1/n)\) and \(f_n(x)=0\). Thus \(f_n(x)\to0\) at every \(x\in[0,1]\). Each support has measure \(m((0,1/n))=1/n\), and the function has value \(n\) on that support, so
The limit function is zero everywhere, so its integral is \(0\), whereas every integral in the sequence is \(1\). In fact, for each fixed \(\varepsilon>0\) and every integer \(n>\varepsilon\), \(\{x\in X:|f_n(x)|>\varepsilon\}=(0,1/n)\), whose measure is \(1/n\to0\). Hence \(f_n\to0\) in measure as well as pointwise, but \(\int_X|f_n|\,dm=1\) for every \(n\), so the sequence does not converge to zero in \(L^1(m)\).
Check: the support length is \(1/n\), the height is \(n\), and their product is \(n(1/n)=1\), agreeing with the computed integral. A common integrable majorant cannot exist; the proposition below explains this directly from the shrinking supports.
Proof. Suppose, to the contrary, that a measurable integrable \(g:X\to[0,\infty]\) satisfies \(|f_n|\le g\) \(\mu\)-a.e. for every \(n\). Then \(c\le\int_{E_n}|f_n|\,d\mu\le\int_{E_n}g\,d\mu\). We show that the last quantity tends to zero. Given \(\varepsilon>0\), integrability of \(g\) gives a finite \(K>0\) with \(\int_{\{g>K\}}g\,d\mu<\varepsilon/2\), by integrable tail control. On the remainder of \(E_n\), \(g\le K\), so \(\int_{E_n}g\,d\mu\le\int_{\{g>K\}}g\,d\mu+K\mu(E_n)<\varepsilon/2+K\mu(E_n)\). Since \(\mu(E_n)\to0\), for all sufficiently large \(n\), \(K\mu(E_n)<\varepsilon/2\). Consequently \(\int_{E_n}g\,d\mu<\varepsilon\) for all such \(n\). This contradicts \(c\le\int_{E_n}g\,d\mu\), since choosing \(\varepsilon=c/2\) would imply both \(\int_{E_n}g\,d\mu\ge c\) and \(\int_{E_n}g\,d\mu<c/2\) for sufficiently large \(n\). Therefore no such \(g\) exists. \(\square\)
In the preceding example, \(E_n=(0,1/n)\) has measure \(1/n\to0\), and \(\int_{E_n}|f_n|\,dm=1\). The proposition applies with \(c=1\). This identifies the precise obstruction: a single integrable function cannot supply at least this much integral on every set whose measure tends to zero.
Supports That Escape Every Fixed Region
Worked Example: Unit Intervals Moving to Infinity
On \((\mathbb{R},\mathcal{L},m)\), define \(u_n=\chi_{[n,n+1)}\) for integers \(n\ge1\). The half-open intervals are pairwise disjoint, and each has Lebesgue measure \(1\). For every fixed \(x\in\mathbb{R}\), at most one of these intervals contains \(x\); thus \(u_n(x)=0\) for all but at most one \(n\), and \(u_n(x)\to0\) everywhere.
Each \(u_n\) is nonnegative and integrable, and
The pointwise limit has integral zero. A common integrable majorant is impossible: if \(g\ge u_n\) \(m\)-a.e. for every \(n\), then on each \([n,n+1)\), \(g\ge1\) almost everywhere. For every positive integer \(N\), disjointness and monotonicity of the integral give
This holds for every \(N\), so \(\int_{\mathbb{R}}g\,dm=\infty\), contrary to integrability. Check: the first \(N\) supports have total measure \(N\), and \(g\) must be at least \(1\) on their union, so the lower bound on its integral is exactly \(N\).
The distinction from the shrinking-spike example is useful. In the spike example, the supports become small and the heights grow. Here the height stays equal to \(1\), while the supports remain of measure \(1\) and move to the right. Pointwise convergence sees only that any fixed \(x\) eventually lies outside the support; it does not see the total mass being carried farther along the real line.
Proof. If an integrable \(g:X\to[0,\infty]\) were a common majorant, then for each positive integer \(N\), the almost-everywhere inequalities on the finitely many sets \(E_1,\ldots,E_N\) and their disjointness would imply \(\int_Xg\,d\mu\ge\sum_{n=1}^{N}\int_{E_n}g\,d\mu\ge\sum_{n=1}^{N}a_n\mu(E_n)\). The finite partial sums on the right are unbounded as \(N\to\infty\), by the assumed divergence. No finite value of \(\int_Xg\,d\mu\) can be at least all of them, contradicting integrability. Thus a common integrable majorant does not exist. \(\square\)
Worked Example: Escaping Points Under Counting Measure
Let \(X=\mathbb{N}=\{1,2,\ldots\}\), let \(\mathcal{M}\) be its power set, and let \(\mu\) be counting measure. Define \(v_n(k)=1\) if \(k=n\), and \(v_n(k)=0\) otherwise. At each fixed \(k\), \(v_n(k)=0\) for every \(n\ne k\), so \(v_n(k)\to0\). Counting measure gives \(\int_Xv_n\,d\mu=\sum_{k=1}^{\infty}v_n(k)=1\), since exactly the single term \(k=n\) equals \(1\). The zero limit has integral zero.
If \(g\) were a common integrable majorant, then \(g(k)\ge1\) for every \(k\), because \(v_k(k)=1\). Hence for each positive integer \(N\), \(\int_Xg\,d\mu=\sum_{k=1}^{\infty}g(k)\ge\sum_{k=1}^{N}g(k)\ge N\). This forces the integral to be infinite. The disjoint-support proposition gives the same conclusion with \(E_n=\{n\}\), \(a_n=1\), and \(\mu(E_n)=1\); its divergent sum is \(\sum_{n=1}^{\infty}1=\infty\). Check: each individual integral is a one-point sum equal to \(1\), while the lower bound from the first \(N\) points is \(N\), so no integrable \(g\) can dominate all the functions.
Cancellation Can Hide the Failure
Worked Example: Signed Spikes with Zero Integrals
On \((\mathbb{R},\mathcal{L},m)\), for \(n\ge1\) set \(w_n=n\chi_{(0,1/n)}-n\chi_{[1/n,2/n)}\). The two intervals are disjoint, have length \(1/n\), and together form \((0,2/n)\). For fixed \(x>0\), choose \(N>2/x\); if \(n\ge N\), then \(2/n<x\), so \(w_n(x)=0\). At \(x=0\), the value is always zero. Thus \(w_n\to0\) everywhere.
Both terms are integrable. By linearity of the Lebesgue integral, the positive interval contributes \(n(1/n)=1\) and the negative interval contributes \(-n(1/n)=-1\). Therefore \(\int_{\mathbb{R}}w_n\,dm=1-1=0\) for each \(n\), matching the integral of the limit. But the two values of \(w_n\) have absolute value \(n\) on disjoint sets of length \(1/n\), so
Thus \(w_n\) does not converge to zero in \(L^1(m)\), even though its signed integrals happen to agree with the integral of its pointwise limit. The shrinking-support proposition applied to \(E_n=(0,2/n)\) and \(|w_n|\) rules out a common integrable majorant, since \(m(E_n)=2/n\to0\) and \(\int_{E_n}|w_n|\,dm=2\). Check: the positive and negative contributions cancel in the signed integral, but their absolute contributions add to \(1+1=2\).
What These Counterexamples Do—and Do Not—Show
For each sequence above, every \(f_n\) is integrable and the pointwise limit is integrable. The failure is not that individual integrals are undefined. It is that pointwise convergence, even pointwise convergence everywhere, does not by itself control the amount of integral concentrated on moving sets. In the moving-spike example, convergence in measure also holds, but convergence in \(L^1\) does not. The signed example adds a further warning: convergence of the signed integrals can occur through cancellation without convergence of the integrals of the absolute differences.
The Dominated Convergence Theorem does not assert that integrals fail to converge whenever no common integrable majorant exists. Rather, it provides a sufficient condition for passing the limit through the integral. The examples establish that pointwise convergence alone is insufficient, and they exhibit concrete sequences for which the conclusion fails. To apply the theorem, one must check its hypotheses, including a single integrable \(g\) that dominates \(|f_n|\) for all \(n\), up to a.e. equality. In these examples, the propositions show exactly why such a \(g\) cannot exist.
Common mistake: “Since \(f_n(x)\to0\) for every fixed \(x\), the integral must tend to zero.” For \(f_n=n\chi_{(0,1/n)}\) on \([0,1]\), that conclusion would give \(\lim_n\int f_n\,dm=0\). The actual substitution into the integral formula gives \(\int f_n\,dm=n\,m((0,1/n))=n(1/n)=1\) for every \(n\), so the limit of the integrals is \(1\), not \(0\). The correction is to check a convergence theorem’s hypotheses rather than infer integral convergence from pointwise convergence.
Check Your Understanding
For each question, state the measure space and distinguish the pointwise limit from the behavior of the integrals.
- For \(f_n=n\chi_{(0,1/n)}\) on \([0,1]\) with Lebesgue measure, compute \(\int f_n\,dm\) and \(\int|f_n|\,dm\), and explain why the sequence converges in measure to zero.
- For \(u_n=\chi_{[n,n+1)}\) on \(\mathbb{R}\) with Lebesgue measure, why does pointwise convergence to zero not imply convergence of the integrals?
- State the shrinking-support proposition’s conditions and identify \(E_n\) and \(c\) for the moving-spike sequence.
- Under counting measure on \(\mathbb{N}\), why can no integrable function dominate all \(v_n=\chi_{\{n\}}\)?
- For the signed sequence \(w_n\), compare \(\int w_n\,dm\) with \(\int|w_n|\,dm\). What does this show about cancellation?