From Pointwise Lower Limits to Integral Bounds
A sequence of nonnegative functions can have contributions that move through the space as the index changes. Such contributions may affect every integral while disappearing at each fixed point in the limit. Fatou’s lemma gives a reliable conclusion in this setting: the integral of the pointwise lower limit cannot exceed the lower limit of the integrals. In applications, the main work is often to compute the pointwise lower limit correctly and to keep track of where each piece of the formula is supported.
Throughout, functions in the general statements are measurable maps \(f_n:X\to[0,\infty]\) on a measure space \((X,\mathcal{M},\mu)\). No finiteness, completeness, or sigma-finiteness assumption on \(\mu\) is needed. The lower limit is the extended-real-valued measurable function \(\liminf_n f_n=\sup_N\inf_{n\ge N}f_n\), and the integrals and inequalities below may take the value \(+\infty\).
The useful feature for explicit lower bounds is that Fatou’s lemma can be combined with a pointwise estimate. If a nonnegative measurable function \(h\) lies below \(\liminf_n f_n\), then monotonicity of the nonnegative Lebesgue integral gives \(\int h\,d\mu\le\int\liminf_n f_n\,d\mu\). Fatou’s lemma then bounds \(\int h\,d\mu\) by the lower limit of the integrals of the sequence.
Proof. In the everywhere case, monotonicity gives \(\int h\,d\mu\le\int\liminf_n f_n\,d\mu\), and Fatou’s lemma gives \(\int\liminf_n f_n\,d\mu\le\liminf_n\int f_n\,d\mu\). For an inequality holding only \(\mu\)-a.e., the nonnegative measurable functions \(h\) and \(\liminf_n f_n\) have the same integral comparison: changing their values on a measurable null set does not change their Lebesgue integrals. Chaining the two inequalities proves the claim, including when an integral is infinite. \(\square\)
One particularly direct application is to a finite collection of disjoint measurable regions. If the pointwise lower limit is at least \(c_j\) on the region \(E_j\), its integral is at least the sum of the corresponding contributions \(c_j\mu(E_j)\). Here \(c_j\ge0\), and a zero coefficient contributes zero even if \(\mu(E_j)=\infty\).
Proof. Define \(h=\sum_{j=1}^r c_j\chi_{E_j}\). This is a nonnegative measurable simple function. On each \(E_j\), the hypothesis gives \(h(x)=c_j\le\liminf_n f_n(x)\). Outside the union of the \(E_j\), \(h(x)=0\le\liminf_n f_n(x)\), since every \(f_n\) is nonnegative. Thus \(h\le\liminf_n f_n\) everywhere. The corollary applies, and the definition of the integral of a nonnegative simple function gives \(\int h\,d\mu=\sum_{j=1}^r c_j\mu(E_j)\). Combining these equalities and inequalities proves the proposition. \(\square\)
Worked Example: A Fixed Piece and a Moving Interval
On \(X=[0,1]\) with Lebesgue measure \(m\), define, for each integer \(n\ge1\), $$ f_n=2\chi_{[0,1/2]}+5\chi_{[1/n,\,2/n]\cap[0,1]}. $$ Compute the pointwise lower limit and the lower limit of the integrals, taking care with \(n=1\).
For any fixed \(x\in[0,1]\), the moving interval contains \(x\) only when \(1/n\le x\le2/n\). If \(x=0\), this never happens. If \(x>0\), these inequalities imply \(1/x\le n\le2/x\), which allows only finitely many integer values of \(n\). Thus the moving term is eventually zero at each fixed point, and $$ \liminf_{n\to\infty}f_n(x)=2\chi_{[0,1/2]}(x) \quad\text{for every }x\in[0,1]. $$
The first interval has measure \(1/2\), so its contribution to each integral is \(2(1/2)=1\). When \(n=1\), the moving interval intersects \([0,1]\) in the singleton \(\{1\}\), which has measure zero; hence \(\int_0^1 f_1\,dm=1\). For \(n\ge2\), the interval \([1/n,2/n]\) is contained in \([0,1]\) and has measure \(2/n-1/n=1/n\). The integral of the sum is the sum of the two nonnegative contributions, even if the supports overlap, so $$ \int_0^1 f_n\,dm=1+\frac{5}{n}\quad(n\ge2). $$
Therefore \(\liminf_n\int_0^1 f_n\,dm=1\), while \(\int_0^1\liminf_n f_n\,dm=2(1/2)=1\). Check: for \(n=2\), the formula gives \(1+5/2=7/2\); for \(n=3\), it gives \(1+5/3=8/3\). These values, together with the separately computed value \(1\) at \(n=1\), have lower limit \(1\), as required.
Moving Contributions and Strict Inequality
Fatou’s inequality can be strict. A sequence may carry a fixed amount of integral in a piece whose location moves away, so that every fixed point eventually stops seeing that piece. The pointwise lower limit retains the fixed part of the functions but not the moving contribution. The next example places the moving intervals on the half-line, where their supports do not drift back toward the fixed interval.
Worked Example: A Contribution That Escapes to the Right
On \(X=[0,\infty)\) with Lebesgue measure \(m\), let \(f_n=\chi_{[0,1]}+3\chi_{[n,n+1]}\) for integers \(n\ge1\). Find the pointwise lower limit and compare the two sides of Fatou’s inequality.
Fix \(x\ge0\). The condition \(x\in[n,n+1]\) can hold for at most two positive integers \(n\): if it holds, then \(x-1\le n\le x\), an interval of length one. Consequently the moving indicator is eventually zero at this fixed \(x\). It follows that \(\liminf_n f_n(x)=\chi_{[0,1]}(x)\) everywhere.
The first interval has measure \(1\), so its integral is \(1\). The interval \([n,n+1]\) also has measure \(1\), so the moving term contributes \(3\). At \(n=1\), the two intervals meet at \(x=1\); this singleton overlap has measure zero and does not alter the sum of the integrals. Hence \(\int_0^\infty f_n\,dm=1+3=4\) for every \(n\). On the other hand, \(\int_0^\infty\liminf_n f_n\,dm=m([0,1])=1\). Thus Fatou’s inequality reads \(1\le4\), and it is strict.
Check: the moving interval has length \((n+1)-n=1\), so its integral is \(3\cdot1=3\) for every \(n\); the lower limit of the sequence of integrals is therefore \(4\). The pointwise limit has integral \(1\), confirming the strict inequality directly.
When the lower bound is described by several pieces rather than a single interval, the finite-piece proposition is a convenient way to organize the calculation. The regions must be measurable, the pointwise lower estimate must hold on each region, and the functions must remain nonnegative so that Fatou’s lemma applies.
Worked Example: A Lower Bound on Disjoint Regions
On \(X=[0,1]\) with Lebesgue measure \(m\), define \(f_n\) by $$ f_n(x)= \begin{cases} 3,&0\le x<1/4,\\ 2,&1/4\le x\le1, \end{cases} \quad+\quad 4\chi_{[1/n,\,2/n]\cap[0,1]}(x). $$ Compute an explicit Fatou lower bound.
For every fixed \(x\), the moving interval appears only finitely often: for \(x=0\) it never appears, and for \(x>0\) its membership requires \(1/x\le n\le2/x\). Thus the pointwise lower limit is the fixed piecewise function, with value \(3\) on \([0,1/4)\) and \(2\) on \([1/4,1]\). Its integral is $$ 3\cdot\frac14+2\cdot\frac34=\frac34+\frac32=\frac94. $$
For \(n\ge2\), the moving interval has measure \(1/n\), so its integral contribution is \(4/n\). The fixed part has integral \(9/4\), and additivity for nonnegative functions gives \(\int_0^1 f_n\,dm=9/4+4/n\). For \(n=1\), the moving interval intersects \([0,1]\) only in \(\{1\}\), so its contribution is zero and \(\int_0^1 f_1\,dm=9/4\). Hence the lower limit of the integrals is \(9/4\), equal to the integral of the pointwise lower limit.
Check: at \(n=2\), the moving interval is \([1/2,1]\), of length \(1/2\), and its contribution is \(4(1/2)=2\). The total is \(9/4+2=17/4\), matching \(9/4+4/2\). The exceptional \(n=1\) value is \(9/4\), not \(9/4+4\), because the intersected moving set is a singleton.
Two Checks Before Applying Fatou’s Lemma
Track membership at a fixed point. If a piece depends on \(n\), its measure at each index does not determine its pointwise lower limit. For an interval such as \([1/n,2/n]\), fix \(x\) first and solve \(1/n\le x\le2/n\). This shows that a fixed positive \(x\) belongs to only finitely many such intervals. The moving piece therefore disappears pointwise, even though its integral can be positive at every index.
Do not use the nonnegative version for signed functions. On \([0,1]\) with Lebesgue measure, set \(g_n=-n\chi_{(0,1/n)}\). Each \(g_n\) is real-valued and integrable, and \(g_n(x)\to0\) at every \(x\in[0,1]\): at \(x=0\) every value is zero, and for fixed \(x>0\), the indicator is eventually zero. But \(\int_0^1g_n\,dm=-n(1/n)=-1\) for every \(n\). The inequality that would result from applying Fatou’s lemma without its nonnegativity hypothesis is \(0\le-1\), which is false. The issue is not a pointwise-limit calculation; the required hypothesis has been dropped.
A common calculation error is to treat every moving interval as though its full stated length lies in the domain. In the first example, \([1/n,2/n]\) has length \(1/n\) when \(n\ge2\), but at \(n=1\) its intersection with \([0,1]\) is \(\{1\}\), of measure zero. The correction is to calculate the intersection with the domain at each exceptional index before integrating. Fatou’s lemma concerns the lower limit as \(n\to\infty\), so a finite exceptional index does not change that lower limit, but it still must not be assigned an incorrect integral.
Check Your Understanding
For each question, identify the pointwise lower limit before comparing integrals.
- On \([0,1]\) with Lebesgue measure, let \(f_n=1+2\chi_{[1/n,\,2/n]\cap[0,1]}\). Find \(\liminf_n f_n\), compute \(\int_0^1 f_n\,dm\) for \(n=1\) and for \(n\ge2\), and verify Fatou’s inequality.
- On \([0,\infty)\) with Lebesgue measure, let \(f_n=2\chi_{[n,n+1]}\). Compute the pointwise lower limit and the lower limit of the integrals. Is Fatou’s inequality strict?
- Suppose on a measure space \((X,\mathcal{M},\mu)\), \(f_n\ge0\) are measurable and \(\liminf_n f_n\ge3\) on a measurable set \(E\). What lower bound does Fatou’s lemma give in terms of \(\mu(E)\)?
- For \(g_n=-n\chi_{(0,1/n)}\) on \([0,1]\), compute the pointwise limit and every integral. Which hypothesis prevents using Fatou’s lemma?
- In the first worked example, why does the fact that \([1,2]\cap[0,1]=\{1\}\) change the integral at \(n=1\) but not the lower limit of the sequence of integrals?