Prove a Measure-Theoretic Convergence Theorem

From Pointwise Limits to Integral Bounds

Pointwise convergence describes what happens at each individual point. It does not, by itself, guarantee that integrals converge: functions can move their mass across the space or concentrate it on smaller and smaller sets. A useful result in measure theory gives a one-sided conclusion that survives these changes. Fatou’s lemma says that for a sequence of nonnegative measurable functions, the integral of the pointwise lower limit is no larger than the lower limit of the integrals.

The proof relies on two basic ideas. First, a measurable set that is the increasing union of sets has measure equal to the limit of their measures. Second, the nonnegative Lebesgue integral can be approximated from below by integrals of simple functions. We make the first idea explicit before applying it to the second.

The lower limit is not necessarily an ordinary pointwise limit. For example, a sequence may keep oscillating while its lower limit still exists at each point. The definition by tail infima is particularly suited to Fatou’s lemma: if the lower limit is at least \(a\) at a point, then all sufficiently late terms are at least \(a-\delta\), for any fixed \(\delta>0\).

A Measure-Theoretic Continuity Principle

Proof. Define \(B_1=A_1\), and for \(N\geq2\) define \(B_N=A_N\setminus A_{N-1}\). The sets \(B_N\) are pairwise disjoint. Moreover, for every positive integer \(m\), \( A_m=\bigcup_{N=1}^{m}B_N \), and the full union of the \(B_N\) is the full union of the \(A_N\). By finite additivity and then countable additivity, \( \mu(A_m)=\sum_{N=1}^{m}\mu(B_N) \) and \( \mu\left(\bigcup_{N=1}^{\infty}A_N\right)=\sum_{N=1}^{\infty}\mu(B_N) \). The finite partial sums on the right are nondecreasing and converge to the infinite series, including when it diverges to infinity. Therefore \(\lim_{m\to\infty}\mu(A_m)=\mu(\bigcup_{N=1}^{\infty}A_N)\). \(\square\)

This theorem will be applied to sets that grow as the starting point of a tail moves farther out. At each point where the lower limit is at least a specified value, the point eventually belongs to every such tail-threshold set. Continuity from below then ensures that these sets recover the whole relevant level set in the limit.

Fatou’s Lemma

Proof. Set \(g=\liminf_{n\to\infty}f_n\). We use the defining approximation property of the nonnegative Lebesgue integral: \(\int_X g\,d\mu\) is the supremum of \(\int_X\phi\,d\mu\) over nonnegative measurable simple functions \(\phi\leq g\). It is therefore enough to show that, for every such \(\phi\), \( \int_X\phi\,d\mu\leq\liminf_{n\to\infty}\int_Xf_n\,d\mu. \) The argument includes simple functions whose integrals are infinite.

Write \(\phi\) using its distinct positive values on disjoint measurable level sets: \( \phi=\sum_{j=1}^{m}a_j\mathbf 1_{A_j} \), where each \(a_j>0\), the sets \(A_j\) are pairwise disjoint, and \(\phi=0\) outside their union. The case \(\phi=0\) is immediate, so suppose there is at least one positive value. Fix \(\delta\) with \(0<\delta<\min_{1\leq j\leq m}a_j\). For each \(j\) and positive integer \(N\), define \( E_{j,N}=A_j\cap\bigcap_{n\geq N}\{x\in X:f_n(x)\geq a_j-\delta\}. \) These sets are measurable. For fixed \(j\), they increase as \(N\) increases.

Since \(\phi\leq g\), every \(x\in A_j\) satisfies \(g(x)\geq a_j\). By the definition of the lower limit, there is a tail on which \(f_n(x)\geq a_j-\delta\). Thus \(x\) belongs to \(E_{j,N}\) for some \(N\), and \( \bigcup_{N=1}^{\infty}E_{j,N}=A_j. \) By continuity from below, \( \lim_{N\to\infty}\mu(E_{j,N})=\mu(A_j). \)

For any fixed \(N\) and every \(n\geq N\), the function \(f_n\) is at least \(a_j-\delta\) on \(E_{j,N}\). The sets \(E_{1,N},\ldots,E_{m,N}\) are disjoint because they lie in the disjoint sets \(A_1,\ldots,A_m\). Monotonicity of the nonnegative integral therefore gives \( \int_X f_n\,d\mu\geq\sum_{j=1}^{m}(a_j-\delta)\mu(E_{j,N})\qquad(n\geq N). \) Consequently, \( \liminf_{n\to\infty}\int_X f_n\,d\mu\geq\sum_{j=1}^{m}(a_j-\delta)\mu(E_{j,N}) \) for every \(N\). Letting \(N\to\infty\) and using continuity from below yields \( \liminf_{n\to\infty}\int_X f_n\,d\mu\geq\sum_{j=1}^{m}(a_j-\delta)\mu(A_j). \) This remains valid if one of the measures on the right is infinite.

Finally, let \(\delta\) decrease to zero. The finite sum on the right increases to \(\sum_{j=1}^{m}a_j\mu(A_j)=\int_X\phi\,d\mu\), with the same conclusion if that integral is infinite. We have proved the required bound for every nonnegative simple \(\phi\leq g\). Taking the supremum over these functions gives \( \int_Xg\,d\mu\leq\liminf_{n\to\infty}\int_Xf_n\,d\mu. \) This is Fatou’s lemma. \(\square\)

The proof’s order of choices is important. We first fix a simple function below the lower limit and a positive tolerance \(\delta\). For that tolerance, each level set is recovered by an increasing union of tail-threshold sets. Only after applying continuity from below do we let the tolerance shrink to zero.

A Set Version

Proof. Apply Fatou’s lemma to the indicator functions \(f_n=\mathbf 1_{E_n}\). At a point \(x\), the lower limit of \(\mathbf 1_{E_n}(x)\) is \(1\) exactly when \(x\in E_n\) for all sufficiently large \(n\), and is \(0\) otherwise. Hence \(\liminf_n\mathbf 1_{E_n}=\mathbf 1_{\liminf_n E_n}\). Since \(\int_X\mathbf 1_E\,d\mu=\mu(E)\) for every measurable set \(E\), Fatou’s inequality gives the stated result. \(\square\)

Worked Applications

What the Inequality Does—and Does Not—Say

Fatou’s lemma provides a lower bound, not a general rule that limits can be moved inside an integral. The moving-interval and concentrated-function examples both have pointwise lower limit zero but integrals equal to one at every stage. The points carrying the integral can change with \(n\), so examining the eventual behavior at each fixed point does not capture all the integral at each stage.

The hypotheses also matter. Nonnegativity ensures that the integrals and lower limits are defined without cancellation between positive and negative parts. Measurability is needed both for the integrals and for the lower-limit function. No finite-measure assumption and no separate integrability assumption are needed: the conclusion is valid with extended nonnegative integrals, including \(+\infty\).