The Proof Through Tail Infima
Fatou’s Lemma is driven by a simple construction: at each point, take the infimum over a tail of the sequence. These tail infima increase as the tails become shorter, and their limit is the pointwise liminf. The Monotone Convergence Theorem then turns this pointwise increase into a statement about integrals.
The other part of the argument is a comparison. A tail infimum is bounded above by every function in its tail, so monotonicity of the integral bounds its integral by each of those functions’ integrals. Combining these two facts gives the direction of Fatou’s inequality.
Tail Infima and Their Integral Bound
Let \(f_n:X\to[0,\infty]\) be measurable functions on a measure space \((X,\mathcal{F},\mu)\). For each positive integer \(m\), define the \(m\)-th tail infimum by
The function \(u_m\) is measurable. Indeed, for every real number \(a\),
The equality holds because an infimum is strictly below \(a\) exactly when at least one value in the set being infimized is strictly below \(a\). The right-hand side is a countable union of measurable sets. The threshold characterization of measurability therefore gives measurability of \(u_m\). Also, \(u_m\leq u_{m+1}\), because the infimum over the shorter tail \(\{k:k\geq m+1\}\) cannot be smaller than the infimum over the tail that also includes \(k=m\).
Proof. Fix \(m\). For every \(k\geq m\), the definition of \(u_m\) gives \(u_m(x)\leq f_k(x)\) for every \(x\in X\). Both functions are nonnegative and measurable, so the Monotonicity of the Lebesgue Integral gives $$ \int_X u_m\,d\mu\leq\int_X f_k\,d\mu. $$ This inequality holds for every \(k\geq m\). Its left-hand side is therefore a lower bound for all the numbers \(\int_X f_k\,d\mu\) with \(k\geq m\), and hence $$ \int_X u_m\,d\mu \leq \inf_{k\geq m}\int_X f_k\,d\mu. $$ No subtraction or assumption of finite integrals is involved, so the argument also applies when some or all of the integrals are infinite.
Fatou’s Lemma: Complete Proof
Proof. Define \(u_m=\inf_{k\geq m}f_k\) as above. The sequence \((u_m)\) is measurable, nonnegative, and increasing pointwise. By the definition of the pointwise liminf, $$ \lim_{m\to\infty}u_m(x)=\liminf_{n\to\infty}f_n(x) $$ for every \(x\in X\). The Monotone Convergence Theorem applies to \((u_m)\) and gives $$ \int_X\liminf_{n\to\infty}f_n\,d\mu = \lim_{m\to\infty}\int_X u_m\,d\mu. $$ By the Integral Bound for Tail Infima, for each \(m\), $$ \int_X u_m\,d\mu \leq \inf_{k\geq m}\int_X f_k\,d\mu. $$ As \(m\) increases, the infimum on the right is taken over shorter tails, so these tail infima increase to the lower limit of the sequence of integrals. Taking limits in the inequality yields $$ \lim_{m\to\infty}\int_X u_m\,d\mu \leq \lim_{m\to\infty}\inf_{k\geq m}\int_X f_k\,d\mu = \liminf_{n\to\infty}\int_X f_n\,d\mu. $$ Together with the Monotone Convergence identity, this is Fatou’s inequality. All quantities lie in \([0,\infty]\); the proof does not subtract extended values or require any integral to be finite.
Reading the Two Limits in the Proof
There are two distinct sequences of tail infima in the argument. The functions \(u_m(x)=\inf_{k\geq m}f_k(x)\) are formed pointwise before integration. Their increasing limit is the pointwise liminf function. Separately, the numbers \(\inf_{k\geq m}\int_X f_k\,d\mu\) are formed from the integrals; their increasing limit is the liminf of the integrals.
The comparison between these two processes is only an inequality. For each fixed \(m\), \(u_m\leq f_k\) for every \(k\geq m\), which controls \(\int_Xu_m\,d\mu\) by each integral in that tail. It does not say that the integral of an infimum equals the infimum of the integrals. Confusing these expressions can lead to an incorrect equality in place of Fatou’s inequality.
Worked Example: Alternating Functions with a Strict Bound
On \([0,1]\) with Lebesgue measure, define \(f_n(x)=x\) when \(n\) is odd and \(f_n(x)=1-x\) when \(n\) is even. Both functions are nonnegative and measurable. At each \(x\), every tail contains both types of term, so $$ \inf_{k\geq m}f_k(x)=\min(x,1-x). $$ Thus the pointwise liminf is \(\min(x,1-x)\), and its integral is $$ \int_0^1\min(x,1-x)\,dx = \int_0^{1/2}x\,dx+\int_{1/2}^1(1-x)\,dx = \frac18+\frac18 = \frac14. $$ For odd \(n\), \(\int_0^1 f_n(x)\,dx=\int_0^1x\,dx=1/2\). For even \(n\), \(\int_0^1 f_n(x)\,dx=\int_0^1(1-x)\,dx=1/2\). Therefore the liminf of the integrals is \(1/2\), and Fatou’s inequality reads $$ \frac14\leq\frac12. $$ In this example the inequality is strict. The pointwise tail infimum keeps the smaller value at each \(x\), while each individual function has integral \(1/2\).
Worked Example: A Moving Unit Mass on the Counting Measure Space
Let \(X=\mathbb{N}\) with counting measure, and define \(f_n(k)=1\) if \(k=n\) and \(f_n(k)=0\) if \(k\neq n\). For any fixed \(k\), \(f_n(k)=0\) for all \(n>k\). Consequently, \(\liminf_n f_n(k)=0\) for every \(k\), and the integral of the pointwise liminf is \(0\).
For each \(n\), exactly one value of \(f_n\) on \(\mathbb{N}\) is \(1\), so $$ \int_{\mathbb{N}}f_n\,d\mu = \sum_{k=1}^{\infty}f_n(k) = 1. $$ Hence the liminf of the integrals is \(1\), and Fatou’s inequality gives \(0\leq1\). The tail infima make the pointwise behavior explicit: for fixed \(k\), every tail has a term after \(k\) where \(f_n(k)=0\), so \(\inf_{n\geq m}f_n(k)=0\) for every \(m\). The integrals retain the unit mass even though its location changes with \(n\).
Worked Example: A Convergent Sequence with Equality
On \([0,1]\) with Lebesgue measure, let \(f_n(x)=x+1/n\). These functions are nonnegative and measurable, and \(f_n(x)\to x\) at every point. For each \(m\) and each \(x\), $$ \inf_{k\geq m}\left(x+\frac1k\right)=x, $$ because \(1/k\) tends to zero along the tail. Therefore every tail infimum \(u_m\) equals the limit function \(x\), even though no finite \(k\) in the tail attains that infimum. The integrals are $$ \int_0^1 f_n(x)\,dx = \int_0^1x\,dx+\int_0^1\frac1n\,dx = \frac12+\frac1n. $$ They converge to \(1/2\), which is also \(\int_0^1x\,dx\). Thus Fatou’s inequality is an equality: $$ \int_0^1\liminf_{n\to\infty}f_n(x)\,dx = \frac12 = \liminf_{n\to\infty}\int_0^1f_n(x)\,dx. $$ The example also shows why a tail infimum need not be a minimum.
Why Nonnegativity and the Direction Matter
Nonnegativity is built into this proof in two places. First, it allows us to apply the Monotone Convergence Theorem directly to the increasing tail infima. Second, all integrals and their tail infima are well-defined in \([0,\infty]\), so the comparison remains valid even when a value is infinite. For general signed functions, tail infima and integrals can involve undefined differences or cancellation, and this proof does not apply without additional hypotheses.
The conclusion is a lower bound, not a general interchange rule between limits and integrals. The alternating-function example shows that the bound can be strict; the convergent example shows that equality can occur. The proof identifies exactly what is guaranteed: each tail infimum lies below every function in its tail, and the increasing limit of those infima is the pointwise liminf. Nothing in the comparison forces equality between the integral of that limit and the liminf of the integrals.
Check Your Understanding
Use the proof and examples above to answer these questions.
- Why is the sequence of tail infima \(u_m\) increasing pointwise?
- Which inequality follows from \(u_m\leq f_k\) when \(k\geq m\), and which integral theorem justifies it?
- How does the Monotone Convergence Theorem identify the integral of the pointwise liminf?
- In the alternating-functions example, why is the integral of the pointwise liminf \(1/4\) while each function has integral \(1/2\)?
- Can a tail infimum fail to be attained by any function in the tail? Explain using the sequence \(x+1/n\).