A Sequence with a Strict Fatou Inequality
Fatou’s lemma gives a one-sided bound: for nonnegative measurable functions, the integral of the pointwise lower limit cannot exceed the lower limit of the integrals. Equality need not hold. A basic way to see the gap is to concentrate a fixed amount of integral on smaller and smaller sets. At each fixed point the functions can eventually vanish, even though each function still has the same positive integral.
We first state the result being illustrated. On any measure space \((X,\mathcal{M},\mu)\), let \(f_n:X\to[0,\infty]\) be measurable. Fatou’s lemma states that the pointwise lower limit \(\liminf_{n\to\infty} f_n\) is measurable and \[ \int_X \liminf_{n\to\infty} f_n\,d\mu \le \liminf_{n\to\infty}\int_X f_n\,d\mu. \] The integrals and the lower limit of integrals may take the value \(+\infty\). No assumption that \(\mu\) is finite or sigma-finite is needed. Fatou’s lemma is a standard result established in the earlier material on Fatou’s lemma.
For a sequence of nonnegative real numbers \(a_n\), \(\liminf_n a_n\) is the limit of the nondecreasing sequence \(\inf_{k\ge n}a_k\). For functions, this definition is applied at each point. In particular, if \(f_n(x)=0\) for all sufficiently large \(n\), then \(\liminf_n f_n(x)=0\), regardless of the values at the finitely many earlier indices.
The Shrinking-Spike Calculation
Worked Example: \(n\) Times the Indicator of \((0,1/n)\)
Work on \((\mathbb{R},\mathcal{L},m)\), where \(m\) is Lebesgue measure, and define \(f_n(x)=n\chi_{(0,1/n)}(x)\) for \(n\ge1\). Each \(f_n\) is a nonnegative measurable real-valued function. We compute its pointwise lower limit and its integral separately.
Fix \(x\in\mathbb{R}\). If \(x\le0\), then \(x\notin(0,1/n)\) for every \(n\), so \(f_n(x)=0\) for every \(n\). If \(x>0\), choose an integer \(N>1/x\). For \(n\ge N\), we have \(1/n\le1/N<x\), so \(x\notin(0,1/n)\) and \(f_n(x)=0\). Thus at every fixed \(x\), the sequence is eventually zero, and \[ \liminf_{n\to\infty} f_n(x)=0. \] Consequently, \(\int_{\mathbb{R}}\liminf_n f_n\,dm=\int_{\mathbb{R}}0\,dm=0\).
For each \(n\), \(f_n\) equals \(n\) on \((0,1/n)\) and zero elsewhere. The interval has measure \(1/n\). By the integral formula for a nonnegative simple function, \[ \int_{\mathbb{R}} f_n\,dm =n\,m((0,1/n)) =n\cdot\frac1n =1. \] Therefore \(\liminf_n\int_{\mathbb{R}}f_n\,dm=1\), and Fatou’s inequality here is the strict inequality \(0<1\).
Check: The same integral follows by integrating the constant value \(n\) over an interval of length \(1/n\): \(n(1/n)=1\). The endpoint \(0\) and the endpoint \(1/n\) are excluded from the interval, but changing or adding finitely many endpoints would not change its Lebesgue measure.
The pointwise calculation depends on keeping \(x\) fixed while \(n\) grows. The interval \((0,1/n)\) still contains points for every \(n\), but its points move closer to zero. The integral records the height times the width, \(n\cdot(1/n)\), rather than the eventual value at any one fixed point.
Two General Ways to Make the Inequality Strict
Proof. Fix \(x\in X\). Since \(\bigcap_{n=1}^{\infty}E_n=\varnothing\), there is an index \(N\) such that \(x\notin E_N\). The sets decrease, so \(E_n\subseteq E_N\) for all \(n\ge N\), and therefore \(x\notin E_n\) for all \(n\ge N\). This gives \(g_n(x)=0\) for all \(n\ge N\), hence \(g_n(x)\to0\). For each \(n\), the integral of the indicator is \(\mu(E_n)\), so \[ \int_Xg_n\,d\mu=\frac{1}{\mu(E_n)}\int_X\chi_{E_n}\,d\mu =\frac{1}{\mu(E_n)}\mu(E_n)=1. \] The pointwise limit is zero everywhere, so its integral is zero. The lower limit of the constant sequence of integrals \(1,1,\ldots\) is \(1\), proving the asserted strict inequality. \(\square\)
The example on \(\mathbb{R}\) is exactly this construction: take \(E_n=(0,1/n)\). These sets decrease, their intersection is empty, and \(m(E_n)=1/n\). Thus \(\mu(E_n)^{-1}\chi_{E_n}=n\chi_{(0,1/n)}\).
Proof. Pairwise disjointness means that each \(x\in X\) belongs to at most one of the sets \(E_n\). Hence for each fixed \(x\), \(h_n(x)=0\) for all but at most one index \(n\), so \(h_n(x)\to0\). The simple-function integral gives \[ \int_Xh_n\,d\mu=\frac{1}{\mu(E_n)}\mu(E_n)=1 \] for every \(n\). Therefore the integral of the pointwise lower limit is zero, while the lower limit of the integrals is one. This proves strictness. \(\square\)
Worked Example: Spikes Moving Along the Real Line
Again take \((\mathbb{R},\mathcal{L},m)\), and define \(u_n=n\chi_{(n,n+1/n)}\). For each fixed \(x\), at most one of the intervals \((n,n+1/n)\) contains \(x\): for integer \(n\ge1\), its right endpoint is at most \(n+1\), and the next interval begins at \(n+1\), which is excluded. Thus \(u_n(x)=0\) for all but at most one \(n\), and \(u_n(x)\to0\).
The interval \((n,n+1/n)\) has length \(1/n\), so \[ \int_{\mathbb{R}}u_n\,dm =n\,m((n,n+1/n)) =n\cdot\frac1n =1. \] It follows that \(\int_{\mathbb{R}}\liminf_nu_n\,dm=0\), whereas \(\liminf_n\int_{\mathbb{R}}u_n\,dm=1\). Check: the intervals are pairwise disjoint, and the computed integral is the height \(n\) multiplied by the length \(1/n\), so the disjoint-set proposition applies.
Worked Example: Moving a Unit Mass Across Counting Measure
Let \(X=\mathbb{N}=\{1,2,\ldots\}\), let \(\mathcal{M}\) be the set of all subsets of \(\mathbb{N}\), and let \(\mu\) be counting measure. This measure space is sigma-finite: \(\mathbb{N}\) is a countable union of singleton sets, each of measure one. Define \(v_n(k)=1\) when \(k=n\), and \(v_n(k)=0\) otherwise.
Fix \(k\in\mathbb{N}\). Then \(v_n(k)=0\) for every \(n\ne k\), so \(v_n(k)\) is eventually zero and \(\liminf_nv_n(k)=0\). The integral with respect to counting measure is the sum over all points: \[ \int_{\mathbb{N}}v_n\,d\mu=\sum_{k=1}^{\infty}v_n(k)=1, \] because exactly one summand, the one with \(k=n\), equals one. Hence \(\int_{\mathbb{N}}\liminf_nv_n\,d\mu=0\), while \(\liminf_n\int_{\mathbb{N}}v_n\,d\mu=1\). Check: the supports are the pairwise disjoint singletons \(\{n\}\), each of counting measure one, so this is also an instance of the disjoint-set proposition.
Why There Is No Conflict with Monotone Convergence
The Monotone Convergence Theorem applies to a nondecreasing sequence of nonnegative measurable functions: if \(f_n\le f_{n+1}\) pointwise for all \(n\), then \(\int\lim_n f_n\,d\mu=\lim_n\int f_n\,d\mu\). The shrinking-spike sequence does not satisfy that order condition. For example, at \(x=3/4\), \(f_1(x)=1\) because \(3/4\in(0,1)\), but \(f_2(x)=0\) because \(3/4\notin(0,1/2)\). Thus \(f_1(x)>f_2(x)\), and the sequence is not pointwise nondecreasing. Fatou’s lemma still applies, since it only requires nonnegative measurability, but its conclusion is an inequality rather than equality.
Common mistake. One might reason, “Every integral is \(1\), so the integral of the limit is \(1\).” This incorrectly interchanges a pointwise limit and an integral. In this example, the actual pointwise calculation gives \(\liminf_n f_n(x)=0\) at every \(x\), and therefore \(\int\liminf_n f_n\,dm=0\). The separate calculation gives \(\liminf_n\int f_n\,dm=1\). Fatou’s lemma compares these two quantities; it does not assert that they are equal.
The examples show two different ways for fixed points to lose the mass: supports can shrink toward a set that no point ever belongs to, or supports can move through pairwise disjoint locations. In both cases, each function has integral one, but the pointwise lower limit is zero. This is the basic mechanism behind strictness in Fatou’s lemma.
Check Your Understanding
For each question, identify the measure space and compute both sides of Fatou’s inequality before deciding whether equality holds.
- On \(\mathbb{R}\) with Lebesgue measure, define \(a_n=n\chi_{(-1/n,0)}\). What is the pointwise limit, and what are the integrals?
- Suppose \(E_n\) decrease, have finite positive measure, and have nonempty intersection. Which step in the decreasing-set proposition’s proof may fail to show pointwise convergence to zero?
- On \(\mathbb{N}\) with counting measure, define \(w_n=2\chi_{\{n\}}\). Compute \(\int\liminf_nw_n\,d\mu\) and \(\liminf_n\int w_n\,d\mu\).
- Why does the example \(n\chi_{(0,1/n)}\) not meet the pointwise monotonicity hypothesis of the Monotone Convergence Theorem?
- In the shrinking-spike example, which pointwise calculation establishes that the lower limit is zero for every fixed \(x>0\)?