From Sequence Tails to a Lower Bound
Pointwise convergence does not by itself allow an integral to pass through a limit. A useful statement that does hold for nonnegative measurable functions is a one-sided bound: the integral of the pointwise lower limit cannot exceed the lower limit of the integrals. The key construction is to look, at each stage, at the smallest value a function takes anywhere in the remaining tail of the sequence. These tail infima increase, so the Monotone Convergence Theorem applies.
Let \((X,\mathcal{M},\mu)\) be any measure space; no finiteness, sigma-finiteness, or completeness assumption is needed. Let \(f_n:X\to[0,\infty]\) be measurable for each \(n\). Integrals below are extended nonnegative Lebesgue integrals and may equal \(+\infty\). The lower limit of a sequence of extended nonnegative numbers \(a_n\) is \(\liminf_n a_n=\sup_n\inf_{k\ge n}a_k\).
For each \(n\), write \(g_n(x)=\inf_{k\ge n}f_k(x)\). Each \(g_n\) is no larger than any function in its tail, while removing the first term of a tail cannot decrease its infimum. Thus \(g_n\le f_k\) for \(k\ge n\), and \(g_n\le g_{n+1}\). We first check carefully that these functions are measurable.
Proof. Fix \(n\) and a real number \(a\). By the definition of an infimum, \(g_n(x)<a\) holds exactly when at least one \(f_k(x)<a\) for \(k\ge n\). Therefore \(\{x:g_n(x)<a\}=\bigcup_{k\ge n}\{x:f_k(x)<a\}\), which belongs to \(\mathcal{M}\), since each \(f_k\) is measurable and \(\mathcal{M}\) is closed under countable unions. This proves measurability of \(g_n\). For any real \(a\), \(\{x:\sup_n g_n(x)>a\}=\bigcup_n\{x:g_n(x)>a\}\), also a countable union of measurable sets. Hence \(f=\sup_n g_n\) is measurable. \(\square\)
Fatou’s Lemma
Proof. Define \(g_n=\inf_{k\ge n}f_k\) and \(f=\sup_n g_n=\liminf_n f_n\). By the proposition, these functions are measurable. Also \(0\le g_n\le g_{n+1}\), so the Monotone Convergence Theorem applies to \((g_n)\), giving \(\int_X f\,d\mu=\lim_n\int_X g_n\,d\mu\). For every \(n\) and every \(k\ge n\), the pointwise inequality \(g_n\le f_k\) gives \(\int_X g_n\,d\mu\le\int_X f_k\,d\mu\) by monotonicity of the nonnegative integral. Consequently \(\int_X g_n\,d\mu\le\inf_{k\ge n}\int_X f_k\,d\mu\). The right-hand side increases to \(\liminf_n\int_X f_n\,d\mu\) as \(n\to\infty\). Taking limits in the inequality and using the Monotone Convergence identity proves the result. All quantities lie in \([0,\infty]\), so the argument also includes infinite values. \(\square\)
The proof uses two distinct lower limits: first, the pointwise lower limit of the functions; second, the numerical lower limit of their integrals. The tail infimum \(g_n\) is bounded above by every \(f_k\) with \(k\ge n\), which is what compares these two. There is no step that interchanges a limit and an integral without justification: the limit of the integrals of \(g_n\) is identified by Monotone Convergence, whose hypotheses hold because the \(g_n\) are measurable, nonnegative, and increasing.
Worked Example: A Decreasing Error on a Finite Interval
On \(X=[0,1]\) with its Borel sigma-algebra and Lebesgue measure \(m\), let \(f_n(x)=x+1/n\). Each function is measurable and nonnegative. For each fixed \(x\), the sequence decreases to \(x\), so its pointwise lower limit is \(f(x)=x\). Direct integration gives $$ \int_{[0,1]} f_n\,dm =\int_0^1\left(x+\frac1n\right)\,dx =\frac12+\frac1n. $$ Thus \(\liminf_n\int f_n\,dm=\lim_n(1/2+1/n)=1/2\), while \(\int_{[0,1]}f\,dm=\int_0^1x\,dx=1/2\). Fatou’s inequality holds with equality in this calculation.
Check: at \(n=2\), the formula gives \(\int f_2\,dm=1/2+1/2=1\). Integrating \(x+1/2\) directly gives \([x^2/2+x/2]_0^1=1\), confirming the value. The lower-limit integral is also \(1/2\), as \([x^2/2]_0^1=1/2\).
Worked Example: Moving Spikes on the Real Line
Take \(X=\mathbb{R}\), \(\mathcal{M}=\mathcal{L}\), and \(\mu=m\), Lebesgue measure. Define \(f_n(x)=n\chi_{(0,1/n)}(x)\). These are nonnegative measurable functions. At \(x=0\) the value is always zero. If \(x<0\), it is always zero. If \(x>0\), choose an integer \(N>1/x\); for every \(n\ge N\), \(1/n<x\), so \(x\notin(0,1/n)\) and \(f_n(x)=0\). Hence \(f_n(x)\to0\) for every \(x\in\mathbb{R}\), and the pointwise lower limit is identically zero.
The interval \((0,1/n)\) has Lebesgue measure \(1/n\), so for every \(n\), $$ \int_{\mathbb{R}} f_n\,dm =n\,m((0,1/n)) =n\cdot\frac1n =1. $$ Therefore \(\int_{\mathbb{R}}\liminf_n f_n\,dm=0\le1=\liminf_n\int_{\mathbb{R}}f_n\,dm\). Check: substituting \(n=4\), the function has height \(4\) on an interval of length \(1/4\), and its integral is \(4(1/4)=1\), as the general calculation predicts.
Worked Example: Tail Infima on a Two-Point Space
Let \(X=\{a,b\}\), \(\mathcal{M}\) be the collection of all subsets of \(X\), and define the finite measure by \(\mu(\{a\})=2\), \(\mu(\{b\})=1\). For even \(n\), set \(f_n(a)=1\) and \(f_n(b)=2\); for odd \(n\), set \(f_n(a)=2\) and \(f_n(b)=3\). Every function on this finite measurable space is measurable and these functions are nonnegative.
Every tail contains both an even and an odd index. Thus the infimum at \(a\) over any tail is \(1\), and the infimum at \(b\) over any tail is \(2\). The pointwise lower limit is consequently \(f(a)=1,\ f(b)=2\), and its integral is \(1\cdot2+2\cdot1=4\). For even \(n\), \(\int f_n\,d\mu=1\cdot2+2\cdot1=4\); for odd \(n\), \(\int f_n\,d\mu=2\cdot2+3\cdot1=7\). Hence the lower limit of the integrals is \(4\), and Fatou’s inequality reads \(4\le4\).
Check: for \(n=1\), the first tail contains values \((2,3)\) at \(a,b\) and also the next even values \((1,2)\); its pointwise infimum is \((1,2)\), whose integral is \(1\cdot2+2\cdot1=4\). This agrees with the tail-infimum calculation.
Using the Bound Correctly
Fatou’s lemma is most useful when the pointwise lower limit is accessible but the sequence of integrals is not known to converge. It supplies a lower bound for the lower limit of those integrals, not an equality in general. If the functions converge pointwise to a measurable \(f\), then their pointwise lower limit is \(f\), and the conclusion becomes \(\int f\,d\mu\le\liminf_n\int f_n\,d\mu\), provided all the functions are nonnegative and measurable on the same measure space. If convergence is only almost everywhere, the same conclusion follows by applying the lemma to the pointwise lower limit, which agrees with \(f\) outside the measurable null set where convergence fails; nonnegative functions equal almost everywhere have equal integrals.
Common mistake: replacing the lower bound by an equality. For the moving spikes above, it is tempting to reason that \(f_n\to0\) pointwise and then conclude that \(\int f_n\,dm\to\int0\,dm=0\). That calculation is invalid: pointwise convergence alone does not justify interchanging the limit and integral. In fact, for every \(n\), the height is \(n\), the interval length is \(1/n\), and the integral is \(n(1/n)=1\). Fatou’s lemma correctly gives only \(0\le1\). A stronger conclusion needs additional hypotheses and a different convergence theorem.
Check Your Understanding
In each question, identify the measure space and verify the nonnegativity and measurability hypotheses before applying Fatou’s lemma.
- On \([0,1]\) with Lebesgue measure, let \(f_n(x)=x+2/n\). Find the pointwise lower limit and both sides of Fatou’s inequality.
- For nonnegative measurable \(f_n\), define \(g_n(x)=\inf_{k\ge n}f_k(x)\). Explain why \(g_n\le g_{n+1}\) and why \(g_n\le f_k\) whenever \(k\ge n\).
- For \(f_n=n\chi_{(0,1/n)}\) on \(\mathbb{R}\), determine the pointwise limit and compute each integral. Which invalid inference would assert equality of the two sides?
- On a finite space with two points of masses \(2\) and \(1\), explain how to compute the integral of a nonnegative function from its two values.
- Which step in the proof of Fatou’s lemma uses the Monotone Convergence Theorem, and what properties of the tail infima verify its hypotheses?