From Fatou’s Lemma to Monotone Convergence
Fatou’s lemma gives a one-sided bound for the integral of a pointwise lower limit. When a sequence of nonnegative functions increases at every point, that one-sided bound combines with a direct upper bound to give an exact limit. This is the Monotone Convergence Theorem, one of the main tools for constructing and evaluating nonnegative Lebesgue integrals.
The theorem permits the limit function, the integrals, or both to take the value \(+\infty\). No assumption that the underlying measure space has finite measure is needed. The important hypothesis is pointwise monotonicity: each term must be no larger than the next at every point.
For an increasing sequence, the pointwise limit exists in \([0,\infty]\) because each sequence of values is nondecreasing. Before integrating that limit, we check that it is measurable. This follows directly from the increasing structure: for every real \(a\), \( \{x:f(x)>a\}=\bigcup_{n=1}^{\infty}\{x:f_n(x)>a\}. \) Indeed, if the supremum of the values \(f_n(x)\) is greater than \(a\), at least one value is greater than \(a\); the converse is immediate. Each set on the right is measurable, so the set on the left is measurable. Thus the pointwise limit \(f\) is measurable.
The Monotone Convergence Theorem
Proof. The preceding level-set argument shows that \(f\) is measurable. We use two properties of the nonnegative Lebesgue integral. First, it is monotone: if \(0\leq u\leq v\), then \(\int_Xu\,d\mu\leq\int_Xv\,d\mu\). This follows from the definition of the integral as the supremum of integrals of nonnegative measurable simple functions below the function: every such simple function below \(u\) is also below \(v\). Second, Fatou’s lemma applies to any sequence of nonnegative measurable functions.
Since \(f_n\leq f_{n+1}\), monotonicity of the integral gives \( \int_X f_n\,d\mu\leq\int_X f_{n+1}\,d\mu \) for every \(n\). Thus the sequence of integrals is nondecreasing, and its limit exists in \([0,\infty]\); that limit equals its supremum. Also \(f_n\leq f\) for every \(n\), so \( \int_X f_n\,d\mu\leq\int_X f\,d\mu. \) Taking the supremum over \(n\) yields \( \lim_{n\to\infty}\int_X f_n\,d\mu\leq\int_X f\,d\mu. \)
For the reverse inequality, apply Fatou’s lemma, proved in the previous tutorial, to \((f_n)\). Its pointwise lower limit is \(f\): because the sequence increases, the infimum over the tail beginning at \(N\) is \(f_N(x)\), and the supremum of these tail infima is \(\sup_N f_N(x)=f(x)\). Therefore Fatou’s lemma gives \( \int_X f\,d\mu\leq\liminf_{n\to\infty}\int_X f_n\,d\mu. \) The integrals form a nondecreasing sequence, so their lower limit equals their limit, including when that limit is \(+\infty\). Combining this with the opposite inequality proves \( \int_X f\,d\mu=\lim_{n\to\infty}\int_X f_n\,d\mu. \) All inequalities are valid for extended nonnegative values, so the argument also covers an infinite integral. \(\square\)
The proof has a useful division of labor. The inequality \(f_n\leq f\) gives the upper bound for every approximating integral, while Fatou’s lemma gives the bound in the other direction. Monotonicity ensures that the integrals have a limit and that their lower limit is that same limit. Unlike an argument that attempts to pass a limit through an integral without hypotheses, this proof identifies exactly why increasing approximation works.
Worked Applications
Worked Example: Indicators of Increasing Intervals
On \([0,1]\) with Lebesgue measure, let \(E_n=[0,1-1/(n+1)]\) and \(f_n=\mathbf 1_{E_n}\). The sets increase, so \(f_n(x)\leq f_{n+1}(x)\) for every \(x\). Their union is \([0,1)\): every \(x<1\) eventually satisfies \(x\leq1-1/(n+1)\), while \(1\) belongs to none of the sets. Thus \(f_n(x)\) increases to \(\mathbf 1_{[0,1)}(x)\). For each \(n\), \( \int_{[0,1]} f_n\,d\mu=\mu(E_n)=1-\frac{1}{n+1}. \) These integrals converge to \(1\). The limit function also has integral \(\mu([0,1))=1\), so the theorem gives, and this computation verifies, \( \int_{[0,1]}\mathbf 1_{[0,1)}\,d\mu =\lim_{n\to\infty}\int_{[0,1]}f_n\,d\mu=1. \)
Worked Example: An Increasing Sequence with Infinite Limit
Let \(X=[0,1]\) with Lebesgue measure and set \(f_n(x)=n\) for every \(x\in X\). These are measurable nonnegative functions, and \(f_n(x)\leq f_{n+1}(x)\). At every point, \(f_n(x)\) tends to \(+\infty\), so the pointwise limit is the extended-valued function \(f(x)=+\infty\). Since the interval has measure \(1\), \( \int_{[0,1]}f_n\,d\mu=n\mu([0,1])=n. \) The integrals tend to \(+\infty\). The integral of the limit is also infinite: for each positive integer \(m\), the constant simple function with value \(m\) lies below \(f\), so \(\int f\,d\mu\geq m\). Since this holds for every \(m\), \(\int f\,d\mu=+\infty\), in agreement with the theorem.
Worked Example: Counting Measure
Take \(X=\mathbb N=\{1,2,3,\ldots\}\) with counting measure, for which the integral of a nonnegative function is the sum of its values over \(\mathbb N\). Define \(f_n(k)=1\) when \(k\leq n\), and \(f_n(k)=0\) when \(k>n\). For every fixed \(k\), \(f_n(k)\) is eventually \(1\), so \(f_n(k)\) increases to the constant function \(f(k)=1\). Each approximating integral is a finite sum: \( \int_{\mathbb N}f_n\,d\mu=\sum_{k=1}^{\infty}f_n(k)=\sum_{k=1}^{n}1=n. \) These integrals tend to \(+\infty\). The integral of \(f\) is \(\sum_{k=1}^{\infty}1=+\infty\), so the theorem applies even though neither side has a finite value.
Integrating a Nonnegative Series
The theorem also justifies exchanging an integral with a countable sum when every term is nonnegative. The partial sums increase pointwise, so the Monotone Convergence Theorem applies directly. No integrability of the individual terms beyond their measurability and nonnegativity is required; the common value may be infinite.
Proof. Define the partial sums \(s_n(x)=\sum_{k=1}^{n}g_k(x)\). Each \(s_n\) is measurable and nonnegative, and \(s_n(x)\leq s_{n+1}(x)\) because \(g_{n+1}(x)\geq0\). Their pointwise limit is \(s(x)=\sum_{k=1}^{\infty}g_k(x)\). By the Monotone Convergence Theorem, \( \int_Xs\,d\mu=\lim_{n\to\infty}\int_Xs_n\,d\mu. \) Finite additivity of the nonnegative integral gives \( \int_Xs_n\,d\mu=\sum_{k=1}^{n}\int_Xg_k\,d\mu. \) The right side is the \(n\)th partial sum of a series of nonnegative extended numbers. Taking its limit proves the stated identity. \(\square\)
Worked Example: Integrating a Series of Disjoint Indicators
On \([0,1]\), let \(A_k=(1/(k+1),1/k]\) and \(g_k=\mathbf 1_{A_k}\). These intervals are pairwise disjoint, and their union is \((0,1]\). Their lengths are \( \mu(A_k)=\frac1k-\frac1{k+1}=\frac{1}{k(k+1)}. \) At each \(x\), at most one \(g_k(x)\) is nonzero, so \(\sum_{k=1}^{\infty}g_k(x)=\mathbf 1_{(0,1]}(x)\). Its integral is \(1\). The series of integrals also sums to \(1\), since its \(n\)th partial sum telescopes: \( \sum_{k=1}^{n}\int_{[0,1]}g_k\,d\mu =\sum_{k=1}^{n}\left(\frac1k-\frac1{k+1}\right) =1-\frac{1}{n+1}\longrightarrow1. \) Thus both sides of the corollary equal \(1\).
When the Monotonicity Hypothesis Matters
The theorem does not say that the integral of a pointwise limit equals the limit of the integrals for every sequence. Its increasing hypothesis supplies a controlled approximation from below. If the functions oscillate, or if their values decrease, the argument above no longer applies as written; different hypotheses and theorems are needed to justify exchanging a limit and an integral.
It is also important to distinguish pointwise increase from an increase that holds only almost everywhere. The stated theorem assumes \(f_n(x)\leq f_{n+1}(x)\) at every point. In measure theory, hypotheses that hold almost everywhere can often be accommodated by modifying functions on a null set, but that is a separate step and should not be silently substituted into a proof.
These ensure that the pointwise limit and all the nonnegative integrals are defined.
Establish \(f_n(x)\leq f_{n+1}(x)\) for every point and index.
For an increasing sequence, it is the pointwise supremum, possibly \(+\infty\).
The integrals increase to the integral of that pointwise limit, with infinite values allowed.
Check Your Understanding
Use the theorem and its proof to answer these questions.
- Why is the pointwise limit of an increasing sequence of measurable functions measurable?
- Which inequality in the Monotone Convergence Theorem’s proof follows from \(f_n\leq f\), and which follows from Fatou’s lemma?
- Why does the sequence of integrals have a limit in \([0,\infty]\)?
- In the counting-measure example, why is the integral of the pointwise limit infinite?
- How does applying the theorem to partial sums justify integrating a nonnegative series term by term?