The Tail-Infimum Idea
Fatou’s lemma compares the integral of a pointwise lower limit with the lower limit of a sequence of integrals. The pointwise lower limit can be difficult to integrate directly, so the proof replaces it with a sequence of more manageable functions: the infimum over each tail of the original sequence. These tail infima increase, which is exactly the structure needed to use the Monotone Convergence Theorem from the previous tutorial.
Throughout, \((X,\mathcal A,\mu)\) is a measure space, and all functions take values in \([0,\infty]\). This allows both function values and integrals to be infinite. We use the standard pointwise lower limit \( \liminf_{n\to\infty} f_n(x)=\sup_{N\geq1}\inf_{k\geq N}f_k(x). \) The key is to understand what the infimum over a tail does: it records a lower bound that every sufficiently late term must satisfy.
For each \(x\), enlarging the starting index of a tail removes some of the values over which the infimum is taken. Removing values cannot decrease an infimum, so \(g_N(x)\leq g_{N+1}(x)\). Thus the tail-infimum functions increase pointwise. Their supremum is the pointwise lower limit of \(f_n\), by definition.
Measurability and the Tail-Infimum Lemma
Before applying a convergence theorem, we verify measurability. A countable infimum of measurable extended-valued functions is measurable. In this case, for every real number \(a\), \( \{x:g_N(x)<a\}=\bigcup_{k\geq N}\{x:f_k(x)<a\}. \) Indeed, an infimum is less than \(a\) exactly when at least one of the values being infimized is less than \(a\). Each set on the right is measurable, so \(g_N\) is measurable. Since \(g_N\) increases pointwise, its limit is measurable as well; that limit is \(\liminf_n f_n\).
Proof. The measurability argument above applies to every \(g_N\). For every \(x\), the set of indices \(k\geq N+1\) is contained in the set of indices \(k\geq N\). Therefore its infimum is at least as large, and \(g_N(x)\leq g_{N+1}(x)\). Also \(g_N(x)\leq f_k(x)\) for every \(k\geq N\). Monotonicity of the nonnegative integral gives \( \int_X g_N\,d\mu\leq\int_X f_k\,d\mu \) for each such \(k\). A number that is at most every member of a set is at most its infimum, proving the stated bound. Finally, the pointwise supremum of the increasing sequence \(g_N\) is \( \sup_{N\geq1}\inf_{k\geq N}f_k(x)=\liminf_{n\to\infty}f_n(x). \) This proves all the claims. \(\square\)
Proof of Fatou’s Lemma
Proof. Let \( g_N(x)=\inf_{k\geq N}f_k(x). \) By the tail-infimum lemma, each \(g_N\) is measurable, the sequence \((g_N)\) increases, and \( g_N(x)\longrightarrow\liminf_{n\to\infty}f_n(x). \) The Monotone Convergence Theorem therefore gives \( \displaystyle \int_X\liminf_{n\to\infty}f_n\,d\mu =\lim_{N\to\infty}\int_Xg_N\,d\mu. \) For each \(N\), the tail-infimum integral bound yields \( \displaystyle \int_Xg_N\,d\mu\leq\inf_{k\geq N}\int_Xf_k\,d\mu. \) Taking the limit in \(N\) on the left and the limit of the tail infima on the right gives \( \displaystyle \int_X\liminf_{n\to\infty}f_n\,d\mu \leq\lim_{N\to\infty}\inf_{k\geq N}\int_Xf_k\,d\mu =\liminf_{n\to\infty}\int_Xf_n\,d\mu. \) The last equality is the definition of the lower limit of a sequence. All steps use inequalities and limits in \([0,\infty]\), so no finiteness assumption is needed. \(\square\)
The proof has two separate stages. The Monotone Convergence Theorem identifies the integral of the pointwise lower limit with the limit of the integrals of the tail infima. The pointwise inequality \(g_N\leq f_k\) for every \(k\geq N\) then bounds each of those integrals by the infimum of the corresponding tail of integrals. Keeping these stages distinct helps avoid a common error: Fatou’s lemma does not assert that the integral of a pointwise limit equals the limit of the integrals.
Worked Applications
Worked Example: Alternating Constant Functions
Take \(X=[0,1]\) with Lebesgue measure, and define \(f_n(x)=1\) when \(n\) is odd and \(f_n(x)=3\) when \(n\) is even. For every \(x\), every tail contains both an odd and an even index. Its infimum is therefore \(1\), so \( \liminf_{n\to\infty} f_n(x)=1. \) Consequently, \( \displaystyle \int_{[0,1]}\liminf_{n\to\infty}f_n\,d\mu=1\cdot\mu([0,1])=1. \) The integrals alternate between \(1\) and \(3\), since the interval has measure \(1\). The infimum of every tail of these integrals is \(1\), giving \( \displaystyle \liminf_{n\to\infty}\int_{[0,1]}f_n\,d\mu=1. \) Thus Fatou’s inequality holds with equality. The sequence of integrals does not converge, but its lower limit is still well-defined.
Worked Example: Narrow Spikes with Strict Inequality
On \([0,1]\) with Lebesgue measure, let \( f_n(x)=n\mathbf 1_{(0,1/n)}(x). \) For \(x=0\), every value is zero. For any fixed \(x>0\), choose an integer \(N>1/x\). Then \(1/n<x\) for all \(n\geq N\), so \(x\notin(0,1/n)\) and \(f_n(x)=0\) for all \(n\geq N\). Hence \(f_n(x)\) is eventually zero at every point, and \( \liminf_{n\to\infty}f_n(x)=0. \) Its integral is therefore zero. On the other hand, the interval \((0,1/n)\) has measure \(1/n\), so for every \(n\), \( \displaystyle \int_{[0,1]}f_n\,d\mu=n\mu((0,1/n))=n\cdot\frac1n=1. \) It follows that the lower limit of the integrals is \(1\), and Fatou’s inequality reads \(0\leq1\), strictly. The heights increase while the supports shrink, so the positive integrals do not force a positive pointwise lower limit.
Worked Example: A Pointwise Limit with Infinite Integral
Let \(X=\mathbb N\) with counting measure, and define \(f_n(k)=1\) when \(k\leq n\) and \(f_n(k)=0\) when \(k>n\). Fix \(k\). For every \(n\geq k\), \(f_n(k)=1\), so \( \liminf_{n\to\infty}f_n(k)=1. \) The integral of this pointwise lower limit is the counting-measure sum \( \displaystyle \int_{\mathbb N}1\,d\mu=\sum_{k=1}^{\infty}1=+\infty. \) For each \(n\), exactly \(n\) terms of \(f_n\) equal \(1\), giving \( \displaystyle \int_{\mathbb N}f_n\,d\mu=\sum_{k=1}^{n}1=n. \) Thus the lower limit of the integrals is \(+\infty\), and both sides of Fatou’s inequality are infinite. This example shows why the theorem must allow extended values rather than assuming all integrals are finite.
What the Hypotheses Do—and Do Not—Give
Nonnegativity is what allows the Monotone Convergence Theorem to be used on the tail infima and ensures that the integrals have an order compatible with pointwise inequalities. Without nonnegativity, the integrals may involve cancellation or may not be defined as extended values, and this proof does not apply. A separate argument with additional hypotheses is needed in that setting.
Fatou’s inequality is one-sided. The narrow-spike example shows that its two sides need not agree, even when every function has a finite integral. Nor does the lemma require the original functions to increase or decrease: monotonicity is used only for the auxiliary tail-infimum functions. To apply the result, identify the pointwise lower limit first, then compare its integral with the lower limit—not necessarily the ordinary limit—of the integrals.
Each \(f_n\) must be measurable and nonnegative.
Set \(g_N=\inf_{k\geq N}f_k\); these functions are measurable and increase pointwise.
The increasing limit of \(g_N\) is \(\liminf_n f_n\).
Use the Monotone Convergence Theorem for \(g_N\), then use \(g_N\leq f_k\) for every \(k\geq N\).
Check Your Understanding
Use the tail-infimum proof to answer these questions.
- Why does \(g_N(x)\leq g_{N+1}(x)\) hold for every \(x\)?
- How does the identity \(\{x:g_N(x)<a\}=\bigcup_{k\geq N}\{x:f_k(x)<a\}\) establish measurability?
- At which step is the Monotone Convergence Theorem used in the proof of Fatou’s lemma?
- In the narrow-spike example, why is the pointwise lower limit zero even though every integral equals one?
- Does Fatou’s lemma guarantee equality, or even convergence of the sequence of integrals?