A Decision Rule for Limit Problems
In a limit problem, nonnegativity alone does not decide whether the integral of a pointwise limit equals the limit of the integrals. The first question is whether the functions increase pointwise: if they do, the Monotone Convergence Theorem gives equality. Without that ordering, Fatou’s lemma generally gives only a lower bound. The examples below use the same checks repeatedly: specify the measure space, compute the pointwise limit or liminf, compute the integrals, and compare the result with the theorem that applies.
We use the following standard results, established in the earlier material on the Monotone Convergence Theorem and Fatou’s lemma. In each statement, \((X,\mathcal{M},\mu)\) is an arbitrary measure space, not necessarily finite or sigma-finite, and the functions are measurable with values in \([0,\infty]\). Convergence and inequalities in the hypotheses hold pointwise; versions holding \(\mu\)-a.e. give the corresponding a.e. conclusions.
Thus the selection step is not “Are the functions nonnegative?” Both results require nonnegativity. The extra question for exact equality is whether the sequence is increasing. If it is not, compute the two sides rather than assuming equality; Fatou still supplies a lower bound.
Name the measure space, verify measurability, and check that the functions are nonnegative.
To use monotone convergence, verify \(f_n(x)\le f_{n+1}(x)\) for every \(n\) and \(x\), or a.e. if using its a.e. version.
For monotone convergence, find the pointwise limit. For Fatou, find the pointwise liminf and the liminf of the integrals.
Monotone convergence yields equality under its hypotheses; Fatou yields an inequality, which can be strict.
Increasing Functions: Equality from Monotone Convergence
Worked Example: An Increasing Sequence on a Finite Interval
On \(X=[0,1]\) with the restricted Lebesgue measure \(m\), define \(f_n(x)=1-(1-x)^n\) for \(n\ge1\). These are real-valued measurable functions, and \(0\le f_n\le1\). We check monotonicity directly: for \(x\in[0,1]\), \(0\le1-x\le1\), so \((1-x)^{n+1}\le(1-x)^n\). Subtracting from \(1\) gives \(f_n(x)\le f_{n+1}(x)\).
At \(x=0\), \(f_n(0)=0\) for all \(n\). At each \(x\in(0,1]\), \(0\le1-x<1\), so \((1-x)^n\to0\), and hence \(f_n(x)\to1\). Therefore the pointwise limit is \(f=\chi_{(0,1]}\), whose integral is \(m((0,1])=1\). The endpoint \(0\) has measure zero, but the pointwise limit there is also recorded.
The functions are measurable, nonnegative, and increasing, so Monotone Convergence applies and \(\lim_n\int f_n\,dm=\int f\,dm=1\). Check: substituting \(n=1,2,3\) into \(n/(n+1)\) gives \(1/2,2/3,3/4\), and these values approach \(1\). Directly, \(\int\chi_{(0,1]}\,dm=1\), in agreement with the theorem.
The monotonicity check is essential even when the limit and each integral are easy to compute. A sequence can consist entirely of nonnegative functions and still have integrals that fail to converge to the integral of its pointwise limit.
Fatou Bounds for Moving Mass and Indicators
Worked Example: A Moving Spike on the Real Line
On \((\mathbb{R},\mathcal{L},m)\), where \(m\) is Lebesgue measure and the measure space is sigma-finite but not finite, let \(f_n(x)=n\chi_{(0,1/n)}(x)\). Each \(f_n\) is nonnegative and measurable. For each fixed \(x\le0\), \(f_n(x)=0\). For each fixed \(x>0\), once \(n\ge1/x\), the strict condition \(x<1/n\) fails, so \(f_n(x)=0\) from then on. Thus \(f_n(x)\to0\) at every real \(x\), and \(\int\liminf_n f_n\,dm=0\).
The interval \((0,1/n)\) has measure \(1/n\), so the integral at each index is
Consequently, Fatou’s lemma gives the computed bound \(0\le1\), and the inequality is strict: the integral of the pointwise limit is \(0\), while the liminf of the integrals is \(1\). Check: the calculation uses the open interval specified in the definition; its endpoints would not change the measure, but the pointwise statement at \(x=1/n\) follows from the strict membership condition \(x<1/n\).
This example shows exactly what monotonicity would have ruled out: the positive values move toward \(0\) while their heights grow, so no fixed positive \(x\) retains mass in the limit. The functions are not increasing. For instance, at \(x=3/4\), \(f_1(x)=1\) whereas \(f_2(x)=0\). Nonnegativity therefore permits Fatou’s bound but does not permit replacing it by equality.
Proof. For each \(x\in X\), \(\liminf_n\chi_{A_n}(x)=1\) exactly when \(\chi_{A_n}(x)=1\) for all sufficiently large \(n\), which is exactly when \(x\in\liminf_n A_n\). If this eventual membership fails, infinitely many indicators equal \(0\), so their liminf is \(0\). Thus \(\liminf_n\chi_{A_n}=\chi_{\liminf_n A_n}\) pointwise. Since each indicator is nonnegative and measurable, Fatou’s lemma gives \(\int\chi_{\liminf_n A_n}\,d\mu\le\liminf_n\int\chi_{A_n}\,d\mu\). Using \(\int\chi_E\,d\mu=\mu(E)\) for measurable \(E\) proves the claim, including when a measure is infinite. \(\square\)
Worked Example: Alternating Indicator Sets
Take \(X=[0,1]\) with restricted Lebesgue measure \(m\). For each \(n\ge1\), set \(A_n=[0,3/4]\) when \(n\) is even and \(A_n=[0,1/4]\) when \(n\) is odd, and let \(f_n=\chi_{A_n}\). The functions are nonnegative, measurable, and bounded by \(1\), but they are not increasing: for example, \(f_1(1/2)=0\) and \(f_2(1/2)=1\).
Every odd-indexed set is \([0,1/4]\) and every even-indexed set is \([0,3/4]\). Both types occur after every index \(N\). A point belongs to all sets from some index onward precisely when it belongs to the smaller interval, so \(\liminf_n A_n=[0,1/4]\). The endpoint \(1/4\) is included, as indicated by the closed interval.
The liminf of the integrals is \(1/4\), since the value \(1/4\) occurs at arbitrarily large odd indices and all the values are at least \(1/4\). Thus Fatou gives \(1/4\le1/4\), with equality in this example. Check: the set calculation gives the same answer as the pointwise indicator calculation: \(\liminf_n f_n\) equals \(1\) on \([0,1/4]\) and \(0\) on \((1/4,1]\), so its integral is \(1\cdot(1/4)+0\cdot(3/4)=1/4\).
The indicator proposition gives a convenient interpretation of Fatou’s lower bound: the measure of points that eventually remain in every set cannot exceed the liminf of the measures. It does not say that the measures of the sets must converge, nor does it say that the bound is always strict.
Nonnegative Series: Partial Sums and Escaping Terms
Proof. For \(N\ge1\), let \(s_N=\sum_{k=1}^{N}g_k\). Each \(s_N\) is measurable and nonnegative, and \(s_N\le s_{N+1}\) because \(g_{N+1}\ge0\). Its pointwise limit is \(s=\sum_{k=1}^{\infty}g_k\). By finite additivity of the integral for nonnegative functions, \(\int s_N\,d\mu=\sum_{k=1}^{N}\int g_k\,d\mu\). Applying the Monotone Convergence Theorem to \(s_N\uparrow s\) gives \(\int s\,d\mu=\lim_{N\to\infty}\sum_{k=1}^{N}\int g_k\,d\mu=\sum_{k=1}^{\infty}\int g_k\,d\mu\), where the series on the right is defined as the limit of its increasing partial sums. \(\square\)
Worked Example: Integrating a Geometric Series of Indicators
On \((\mathbb{R},\mathcal{L},m)\), define \(g_k=2^{-k}\chi_{[k-1,k)}\) for \(k\ge1\). Each \(g_k\) is nonnegative and measurable. The finite partial sums \(s_N=\sum_{k=1}^N g_k\) increase pointwise, because \(s_{N+1}=s_N+g_{N+1}\ge s_N\). Each interval \([k-1,k)\) has Lebesgue measure \(1\), so \(\int g_k\,dm=2^{-k}\).
The series proposition, or equivalently monotone convergence applied to \(s_N\), gives
For a finite check, the first three integrals are \(1/2,1/4,1/8\), whose sum is \(7/8\). In general the \(N\)-th partial sum is \(1-2^{-N}\), which increases to \(1\). The intervals are disjoint, so the pointwise sum has value \(2^{-k}\) on \([k-1,k)\) and its integral is also the sum of the areas of these disjoint pieces, namely \(\sum_{k\ge1}2^{-k}=1\). This independently checks the result.
Worked Example: A Series Whose Mass Moves with the Index
Let \(X=\mathbb{N}=\{1,2,\ldots\}\) with counting measure \(\#\), which assigns to a set its number of elements and is sigma-finite but not finite. Define \(a_n(k)=1\) if \(k=n\), and \(a_n(k)=0\) otherwise. Each \(a_n\) is measurable and nonnegative. For every fixed \(k\), \(a_n(k)\) is \(1\) only at \(n=k\) and is \(0\) for all \(n>k\); hence \(a_n(k)\to0\).
Under counting measure, the integral of a nonnegative function is its sum over \(\mathbb{N}\). At each index \(n\), exactly one term of the series is nonzero, so \(\int a_n\,d\#=\sum_{k=1}^{\infty}a_n(k)=1\). In contrast, \(\sum_{k=1}^{\infty}\liminf_n a_n(k)=\sum_{k=1}^{\infty}0=0\). Fatou’s lemma therefore gives the exact bound \(0\le1\), strictly.
Check: there is no hidden tail in the sum for a fixed \(n\): its \(k=n\) term is \(1\), and every \(k\ne n\) term is \(0\), so the sum is \(1\). The pointwise limit sequence is identically zero, whose counting-measure integral is \(0\). This verifies both sides independently and shows why one cannot interchange a pointwise limit and an infinite series here.
A Common Error and Its Correction
A common incorrect calculation is to reason, “The functions are nonnegative, so the integral of the limit equals the limit of the integrals.” For the moving spike \(f_n=n\chi_{(0,1/n)}\), that would give \(0=1\): the pointwise limit is zero, but every integral is one. The corrected statement is that nonnegativity allows Fatou’s lemma, which says \(0\le1\) here. To obtain equality from Monotone Convergence, one must additionally verify that \(f_n\le f_{n+1}\) pointwise (or a.e. for the a.e. version). The spike sequence fails that test, as its value at \(x=3/4\) decreases from \(1\) at \(n=1\) to \(0\) at \(n=2\).
For a series problem, the parallel check is to use its finite partial sums: partial sums of nonnegative terms increase, so monotone convergence applies to them. For a general sequence of functions, however, there may be no such partial-sum structure. Compute the liminf and use Fatou’s inequality unless the required monotonicity or another applicable theorem has been verified.
Check Your Understanding
For each problem, identify the measure space and check the hypotheses before computing the integral or bound.
- On \([0,1]\) with Lebesgue measure, let \(f_n(x)=x^n\). Is the sequence increasing or decreasing pointwise? Find its pointwise limit and compare the limit of the integrals with the integral of the limit.
- On \(\mathbb{R}\) with Lebesgue measure, let \(u_n=n\chi_{(1,1+1/n)}\). Compute \(\liminf_n u_n(x)\) at each fixed \(x\), the integral of that liminf, and \(\liminf_n\int u_n\,dm\).
- For measurable sets \(A_n\) in a measure space, explain why \(\liminf_n\chi_{A_n}=\chi_{\liminf_n A_n}\), and state the resulting measure inequality.
- On \(\mathbb{N}\) with counting measure, set \(b_n(k)=2^{-k}\) for \(k\le n\) and \(b_n(k)=0\) for \(k>n\). Verify monotonicity in \(n\), then compute the pointwise limit and the limit of the integrals.
- Give an example from this tutorial where Fatou’s inequality is strict, and identify the condition that prevents applying monotone convergence.