From Fatou’s Lemma to an Upper Bound
Fatou’s lemma controls the integral of a pointwise lower limit from below by the lower limit of the integrals. A related upper bound for upper limits is possible when the functions have a common integrable majorant. The key is to apply Fatou’s lemma not to the original functions, but to their differences from that majorant. No finiteness or sigma-finiteness assumption on the measure space is needed.
Throughout, \((X,\mathcal{M},\mu)\) is a measure space, with no assumption that \(\mu(X)\) is finite or that the space is sigma-finite or complete. All functions in the result below are measurable and take nonnegative real values. We use Fatou’s lemma in its established form: for any sequence of nonnegative measurable functions \(h_n\), \(\int\liminf_n h_n\,d\mu\le\liminf_n\int h_n\,d\mu\), with infinite values allowed.
Proof. First suppose \(f_n\le g\) everywhere. Define \(h_n=g-f_n\). Each \(h_n\) is measurable and nonnegative. Since \(0\le f_n\le g\) and \(\int g\,d\mu<\infty\), monotonicity of the nonnegative integral gives \(\int f_n\,d\mu\le\int g\,d\mu<\infty\). Also \(0\le h_n\le g\), so each \(\int h_n\,d\mu\) is finite. Additivity for nonnegative measurable functions gives \(\int h_n\,d\mu=\int g\,d\mu-\int f_n\,d\mu\): indeed, \(g=f_n+h_n\), and all three integrals are finite.
Write \(F=\limsup_n f_n\). Since \(0\le f_n\le g\), we have \(0\le F\le g\). At each \(x\), \(g(x)\) is finite and fixed, so the elementary relation between the lower and upper limits of a bounded real sequence gives \(\liminf_n h_n(x)=g(x)-F(x)\). Fatou’s lemma applied to \((h_n)\) therefore yields \[ \int(g-F)\,d\mu =\int\liminf_n h_n\,d\mu \le\liminf_n\int h_n\,d\mu =\int g\,d\mu-\limsup_n\int f_n\,d\mu. \] Here the last equality follows by setting \(b=\int g\,d\mu<\infty\) and using \(\liminf_n(b-a_n)=b-\limsup_n a_n\) for \(a_n=\int f_n\,d\mu\in[0,b]\). Since \(g=F+(g-F)\) and both summands are nonnegative, additivity gives \(\int(g-F)\,d\mu=\int g\,d\mu-\int F\,d\mu\). Substituting this into the inequality and subtracting the finite quantity \(\int g\,d\mu\) proves the result. Both sides are at most \(\int g\,d\mu\). If the majorization holds only \(\mu\)-a.e. for each \(n\), the union of the countably many exceptional null sets is a measurable null set. On its complement the proof just given applies; altering values on that null set does not change any of the integrals or the asserted a.e. conclusion. \(\square\)
The proof uses the majorant twice: it makes every difference \(g-f_n\) nonnegative, and it ensures all the quantities being subtracted are finite. Without the second property, the subtraction in the proof may be undefined; without the first, Fatou’s lemma cannot be applied to the differences in this way.
Worked Applications
Worked Example: A Strict Reverse Inequality on the Unit Interval
Let \(X=[0,1]\), let \(\mathcal{M}\) be the Lebesgue sigma-algebra, and let \(\mu=m\). Define \(A=[0,\tfrac12]\) and \(B=(\tfrac12,1]\). For even \(n\), let \(f_n=\chi_A\); for odd \(n\), let \(f_n=\chi_B\). Take \(g=1\). These functions are measurable and satisfy \(0\le f_n\le g\), while \(\int g\,dm=m([0,1])=1<\infty\).
For each \(n\), \(m(A)=\tfrac12-0=\tfrac12\) and \(m(B)=1-\tfrac12=\tfrac12\). Thus \(\int f_n\,dm=\tfrac12\) for every \(n\), so \(\limsup_n\int f_n\,dm=\tfrac12\). Every point of \(A\) belongs to the support of every even-indexed function, and every point of \(B\) belongs to the support of every odd-indexed function. Since \(A\cup B=[0,1]\), each point has \(f_n(x)=1\) for infinitely many \(n\), and hence \(\limsup_n f_n(x)=1\) everywhere. Consequently \(\int\limsup_n f_n\,dm=\int_0^1 1\,dm=1\).
Check: the theorem gives \(\tfrac12\le1\), and the inequality is strict. In the proof, \(h_n=1-f_n\) alternates between \(\chi_B\) and \(\chi_A\). Every point has \(h_n(x)=0\) for infinitely many \(n\), so \(\liminf_n h_n=0\) everywhere and its integral is \(0\). Meanwhile \(\int h_n\,dm=1-\tfrac12=\tfrac12\) for each \(n\). Fatou’s inequality for these differences reads \(0\le\tfrac12\), consistent with the proof.
Worked Example: Powers on the Unit Interval
On \([0,1]\) with Lebesgue measure, let \(f_n(x)=x^n\) for \(n\ge1\), and take \(g(x)=1\). For each \(x\in[0,1]\), \(0\le x^n\le1\), and \(\int g\,dm=1\). If \(0\le x<1\), then \(x^n\to0\); at \(x=1\), \(x^n=1\) for every \(n\). Thus \(\limsup_n f_n=\chi_{\{1\}}\), whose integral is \(m(\{1\})=0\).
For every \(n\ge1\), direct integration gives \(\int_0^1x^n\,dm=[x^{n+1}/(n+1)]_0^1=1/(n+1)\). Hence \(\limsup_n\int f_n\,dm=0\), and the reverse Fatou inequality is equality in this example. To check the difference argument, \(h_n=1-x^n\), so \(\int h_n\,dm=1-1/(n+1)=n/(n+1)\), which tends to \(1\). Pointwise, \(\liminf_n h_n=1\) for \(x<1\) and \(0\) at \(x=1\), so its integral is \(1\). Fatou’s lemma gives \(1\le1\), as required.
Worked Example: A Finite-Support Sequence under Counting Measure
Let \(X=\mathbb{N}=\{1,2,\ldots\}\), \(\mathcal{M}\) be all subsets of \(\mathbb{N}\), and \(\mu\) be counting measure. Define \(g(1)=g(2)=1\) and \(g(j)=0\) for \(j\ge3\). For even \(n\), set \(f_n(1)=1\) and \(f_n(j)=0\) for \(j\ne1\); for odd \(n\), set \(f_n(2)=1\) and \(f_n(j)=0\) for \(j\ne2\). These functions are measurable and \(0\le f_n\le g\). The majorant has integral \(\int g\,d\mu=g(1)+g(2)=1+1=2\).
Each \(f_n\) has exactly one nonzero value, equal to \(1\), so \(\int f_n\,d\mu=1\) and \(\limsup_n\int f_n\,d\mu=1\). At \(j=1\), the value \(1\) occurs for every even \(n\); at \(j=2\), it occurs for every odd \(n\); and for \(j\ge3\), the value is always \(0\). Therefore \(\limsup_n f_n\) equals \(1\) at \(1\) and \(2\), and \(0\) elsewhere. Its counting-measure integral is \(1+1=2\).
Check: the asserted inequality is \(1\le2\). For the differences, \(h_n=g-f_n\) has exactly one nonzero value, also equal to \(1\), so \(\int h_n\,d\mu=1\). At each of \(1\) and \(2\), \(h_n\) is zero infinitely often, and it is always zero for \(j\ge3\). Thus \(\int\liminf_n h_n\,d\mu=0\le1=\liminf_n\int h_n\,d\mu\), verifying the Fatou step directly.
A Useful Consequence and a Common Pitfall
Proof. Since \(f_n\ge0\), each integral \(a_n=\int f_n\,d\mu\) is nonnegative. The theorem gives \(\limsup_n a_n\le\int\limsup_n f_n\,d\mu=0\). Nonnegativity also gives \(0\le\liminf_n a_n\le\limsup_n a_n\le0\). Thus the lower and upper limits are both zero, which means \(a_n\to0\). \(\square\)
Common mistake: applying Fatou’s lemma to the original functions to obtain an upper bound. Fatou’s lemma says \(\int\liminf_n f_n\,d\mu\le\liminf_n\int f_n\,d\mu\); it does not give an upper bound for \(\limsup_n\int f_n\,d\mu\). In the alternating-halves example, \(\liminf_n f_n=0\) everywhere, so Fatou gives only \(0\le\tfrac12\). The valid upper-bound argument instead uses \(h_n=g-f_n\), applies Fatou to this nonnegative sequence, and then uses the finite integral of \(g\) to rearrange the resulting inequality.
The integrable-majorant condition is essential. On \(\mathbb{R}\) with Lebesgue measure, define \(f_n=n\chi_{(0,1/n)}\). At \(x=0\), every \(f_n(x)=0\); at each \(x>0\), \(f_n(x)=0\) for all integers \(n\ge1/x\); and for \(x\le0\), every \(f_n(x)=0\). Thus \(f_n(x)\to0\) everywhere and \(\int\limsup_n f_n\,dm=0\). But \(\int f_n\,dm=n\,m((0,1/n))=n(1/n)=1\) for every \(n\), so the proposed upper bound would read \(1\le0\), which is false.
There cannot be an integrable nonnegative majorant for this sequence. If a measurable \(g\) majorizes every \(f_n\), then for each integer \(k\ge1\), on the interval \((2^{-(k+1)},2^{-k})\) we have \(x<2^{-k}\), so \(f_{2^k}(x)=2^k\) and \(g(x)\ge2^k\). These intervals are disjoint and each has length \(2^{-k}-2^{-(k+1)}=2^{-(k+1)}\). Therefore the integral of \(g\) over each such interval is at least \(2^k2^{-(k+1)}=\tfrac12\). Summing this lower bound over the first \(N\) intervals gives \(\int g\,dm\ge N/2\) for every \(N\), so \(\int g\,dm=\infty\). The failing sequence has no integrable majorant, exactly the hypothesis needed for the reverse Fatou argument.
Check Your Understanding
Use the theorem and its proof to answer the following questions.
- Which part of the proof fails if \(g-f_n\) is not nonnegative?
- For the alternating-halves sequence, compute \(\liminf_n f_n(x)\) at a point in \(A\) and at a point in \(B\).
- On counting measure, change the sequence in the finite-support example so that \(f_n\) is always supported at \(1\). What are the integral of its pointwise limsup and the limsup of its integrals?
- Explain why \(\int(g-f_n)\,d\mu=\int g\,d\mu-\int f_n\,d\mu\) is safe under the theorem’s assumptions.
- For \(f_n=n\chi_{(0,1/n)}\), verify the lower bound on \(g\) over each interval \((2^{-(k+1)},2^{-k})\) if \(g\) is assumed to majorize every \(f_n\).