Here are three sequences of functions on the real line. The first is a spike, \(n\mathbf 1_{(0,1/n)}\) : height \(n\) on an interval of length \(\tfrac1n\) . The second is a block that slides to the right, \(\mathbf 1_{[n,n+1]}\) . The third is a block that flattens, \(\tfrac1n\mathbf 1_{[0,n]}\) : height \(\tfrac1n\) on an interval of length \(n\) . Each function has integral exactly \(1\) . Each sequence converges to \(0\) at every point. So in all three cases
The unit of mass has not been destroyed; it has gone somewhere that pointwise convergence cannot follow. In the first sequence it concentrates at a point and escapes to infinite height. In the second it escapes to infinity along the axis. In the third it spreads over ever larger sets and escapes to infinite width. Figure 5.1 shows the three.
Two questions organize the chapter. What can be said about the integrals of a pointwise convergent sequence with no further hypothesis at all? And what hypothesis rules out all three kinds of escape at once? The answers are Fatou’s lemma and the dominated convergence theorem. Monotone convergence, from Chapter 4, is the engine for both.
5.1 The inequality that always holds
Theorem 5.1 (Fatou’s lemma). Let \(f_1,f_2,\dots\) be nonnegative measurable functions on a measure space \((X,\mathcal A,\mu)\) . Then
Proof. Put \(g_n=\inf_{k\ge n}f_k\) . These functions are measurable and nonnegative, they increase with \(n\) , and by definition their limit is \(\liminf_nf_n\) . Monotone convergence gives \(\int\liminf_nf_n\,d\mu=\lim_n\int g_n\,d\mu\) . Since \(g_n\le f_n\) , we have \(\int g_n\,d\mu\le\int f_n\,d\mu\) for each \(n\) , and taking the lower limit of both sides,
Nothing is assumed except measurability and nonnegativity; the sequence need not converge, and the integrals may be infinite. The direction of the inequality is the direction of the three examples: in the limit, mass can be lost but never created. If you forget which way it goes, test it on the spike, where it reads \(0\le1\) .
The most frequent use of Fatou’s lemma is to prove that a limit function is integrable.
Corollary 5.2 (integrability of a limit). Let \(f_n\) and \(f\) be measurable with \(f_n\to f\) a.e., and suppose \(\int\lvert f_n\rvert\,d\mu\le M\) for all \(n\) . Then \(f\) is integrable and \(\int\lvert f\rvert\,d\mu\le M\) .
Proof. Apply Theorem 5.1 to the nonnegative functions \(\lvert f_n\rvert\) , whose lower limit is \(\lvert f\rvert\) a.e.
Example 5.1: Fatou’s inequality on four sequences
For the spike \(f_n=n\mathbf 1_{(0,1/n)}\) , the sliding block \(f_n=\mathbf 1_{[n,n+1]}\) , and the flattening block \(f_n=\tfrac1n\mathbf 1_{[0,n]}\) , the integrals are
and \(\liminf_nf_n=0\) everywhere. Fatou’s lemma says \(0\le1\) , with strict inequality.
The inequality can be strict even when there is no escape to infinity and no convergence. On \([0,1]\) let \(f_n=\mathbf 1_{[0,1/2]}\) for even \(n\) and \(f_n=\mathbf 1_{(1/2,1]}\) for odd \(n\) . Every point is outside infinitely many of the sets, so \(\liminf_nf_n=0\) , while \(\int f_n\,d\lambda=\tfrac12\) for every \(n\) . Fatou’s lemma reads \(0\le\tfrac12\) . Here the mass does not leave; it alternates between two places, and the pointwise lower limit sees neither.
Nonnegativity cannot be dropped. For \(f_n=-\mathbf 1_{[n,n+1]}\) the left side of Fatou’s inequality is \(0\) and the right side is \(-1\) .
5.2 Dominated convergence
Theorem 5.3 (dominated convergence). Let \(f_n\) and \(f\) be measurable functions with \(f_n\to f\) a.e. Suppose there is an integrable function \(g\) such that \(\lvert f_n\rvert\le g\) a.e. for every \(n\) . Then \(f\) is integrable,
Proof. The hypotheses hold outside a countable union of null sets, which is null. Redefining all functions to be \(0\) there changes no integral, so we may assume that \(f_n\to f\) and \(\lvert f_n\rvert\le g<\infty\) everywhere. Then \(\lvert f\rvert\le g\) , so \(f\) is integrable.
Since \(\lvert f_n-f\rvert\le2g\) , the functions \(2g-\lvert f_n-f\rvert\) are nonnegative, and they converge pointwise to \(2g\) . By Fatou’s lemma,
The number \(\int2g\,d\mu\) is finite and may be cancelled, leaving \(\limsup_n\int\lvert f_n-f\rvert\,d\mu\le0\) . The final claim follows from \(\lvert\int f_n\,d\mu-\int f\,d\mu\rvert\le\int\lvert f_n-f\rvert\,d\mu\) .
In these statements the limit \(f\) is assumed measurable. On a complete measure space that is automatic, by Proposition 3.7; in general an almost-everywhere limit agrees a.e. with a measurable function, and the theorems apply to that function. The function \(g\) is called a dominating function, or dominator. Three features of the hypothesis deserve emphasis, because each is a place where applications go wrong. The dominator must be a single function, the same for all \(n\) . It must be integrable, which on a space of infinite measure excludes constants. And the conclusion is stronger than convergence of the integrals: the functions converge in the sense that the integral of \(\lvert f_n-f\rvert\) tends to zero.
The proof also shows how domination defeats each kind of escape. A function \(g\) that lies above the spike for every \(n\) must be large near \(0\) ; one that lies above every sliding block must be at least \(1\) on the whole half-line; one that lies above every flattening block must decay no faster than \(\tfrac1x\) . In each case \(g\) would have infinite integral. The next example makes this exact.
Example 5.2: The smallest candidate dominator of each escaping sequence
If any dominator exists, then \(g^*=\sup_n\lvert f_n\rvert\) is one, and it is the smallest. So the test for whether a sequence is dominated is to compute this supremum and integrate it.
Spike. For \(x\in[\tfrac1{k+1},\tfrac1k)\) the functions \(n\mathbf 1_{(0,1/n)}\) with \(n\le k\) are equal to \(n\) at \(x\) and those with \(n>k\) vanish there. So \(g^*=k\) on that interval, whose length is \(\tfrac1k-\tfrac1{k+1}\) , and
Sliding block. Here \(g^*=\mathbf 1_{[1,\infty)}\) , with integral \(\infty\) .
Flattening block. For \(x\in(k-1,k]\) the functions \(\tfrac1n\mathbf 1_{[0,n]}\) with \(n\ge k\) are positive at \(x\) , the largest being \(\tfrac1k\) . So \(g^*=\tfrac1k\) on \((k-1,k]\) and \(\int g^*\,d\lambda=\sum_k\tfrac1k=\infty\) .
In all three cases the smallest candidate fails to be integrable, by a harmonic series or worse. None of the three sequences is dominated, as must be the case, since for each of them the conclusion of the theorem is false.
Corollary 5.4 (bounded convergence). Let \(\mu(X)<\infty\) and let \(f_n\) and \(f\) be measurable with \(\lvert f_n\rvert\le M\) for all \(n\) and \(f_n\to f\) a.e. Then \(\int f_n\,d\mu\to\int f\,d\mu\) .
Proof. The constant \(M\) is integrable because \(\int M\,d\mu=M\mu(X)<\infty\) ; apply Theorem 5.3.
Common mistake. Calling a uniformly bounded sequence “dominated” on a space of infinite measure. The sliding block and the flattening block are both bounded by \(1\) , and both violate the conclusion. A dominator has to be named and its integral has to be shown finite; a bound is not a dominator until that is done.
Example 5.3: A limit under the integral sign
Evaluate \(\displaystyle\lim_{n\to\infty}\int_0^\infty\frac{e^{-x}}{(1+\frac xn)^n}\,dx\) .
Pointwise limit. For each \(x\ge0\) , \((1+\tfrac xn)^n\to e^x\) , so the integrand converges to \(e^{-x}\cdot e^{-x}=e^{-2x}\) .
Dominator. Since \((1+\tfrac xn)^n\ge1\) for \(x\ge0\) , the integrand is at most \(g(x)=e^{-x}\) for every \(n\) , and \(\int_0^\infty e^{-x}\,dx=1<\infty\) .
Conclusion. By dominated convergence the limit is \(\int_0^\infty e^{-2x}\,dx=\tfrac12\) .
Numerically the integrals are \(0.5963\) for \(n=1\) , \(0.5122\) for \(n=10\) , and \(0.5012\) for \(n=100\) . The sequence of integrands is in fact decreasing, so the decreasing form of monotone convergence (Exercise 4.4) would also apply; but the dominated convergence argument did not need to notice that, and it would survive a perturbation, such as a factor \(\cos(x/n)\) , that destroys monotonicity.
Applied to the partial sums of a series, dominated convergence gives the signed counterpart of Theorem 4.5.
Theorem 5.5 (term-by-term integration of series). Let \(f_n\) be measurable with \(\sum_{n=1}^\infty\int\lvert f_n\rvert\,d\mu<\infty\) . Then the series \(\sum_nf_n\) converges absolutely a.e., its sum is integrable, and
Proof. Let \(G=\sum_n\lvert f_n\rvert\) . By Theorem 4.5, \(\int G\,d\mu=\sum_n\int\lvert f_n\rvert\,d\mu<\infty\) , so \(G<\infty\) a.e. by Proposition 4.7(c). At every point where \(G\) is finite the series \(\sum_nf_n\) converges absolutely. Its partial sums are dominated by the integrable function \(G\) , and Theorem 5.3 applies to them.
Example 5.4: A series with terms of both signs
For \(x>0\) ,
The terms change sign, so Theorem 4.5 does not apply. Check the hypothesis of Theorem 5.5 using \(\lvert\sin x\rvert\le x\) :
So the series may be integrated term by term. Two integrations by parts give \(\int_0^\infty e^{-kx}\sin x\,dx=\frac1{k^2+1}\) , and therefore
the closed form of the series being a standard result quoted from analysis. The justification consisted of one inequality and one convergent comparison series, written down before the interchange was made.
5.3 Integrals that depend on a parameter
Continuity and differentiability are limits, so dominated convergence governs when they pass through an integral.
Theorem 5.6 (continuity and differentiation under the integral). Let \(I\) be an open interval and let \(f:X\times I\to\mathbb R\) be such that \(x\mapsto f(x,t)\) is integrable for each \(t\in I\) . Put \(F(t)=\int f(x,t)\,d\mu(x)\) .
(a) If \(t\mapsto f(x,t)\) is continuous on \(I\) for every \(x\) , and there is an integrable \(g\) with \(\lvert f(x,t)\rvert\le g(x)\) for all \(x\) and \(t\) , then \(F\) is continuous on \(I\) .
(b) If \(\partial f/\partial t\) exists at every point of \(X\times I\) , and there is an integrable \(g\) with \(\lvert\partial f/\partial t\,(x,t)\rvert\le g(x)\) for all \(x\) and \(t\) , then \(F\) is differentiable on \(I\) and
Proof. (a) Let \(t_n\to t\) in \(I\) . Then \(f(x,t_n)\to f(x,t)\) for every \(x\) and \(\lvert f(x,t_n)\rvert\le g(x)\) , so \(F(t_n)\to F(t)\) by dominated convergence. Since the sequence was arbitrary, \(F\) is continuous at \(t\) .
(b) Fix \(t\in I\) and let \(t_n\to t\) with \(t_n\ne t\) . The difference quotients
are measurable in \(x\)
and converge to \(\partial f/\partial t\,(x,t)\)
, which is therefore measurable in \(x\)
. By the mean value theorem (math.real-analysis.mean-value-theorem), applied to the differentiable function \(s\mapsto f(x,s)\)
on the interval between \(t\)
and \(t_n\)
, each \(h_n(x)\)
equals \(\partial f/\partial t\,(x,s)\)
for some \(s\)
in that interval, so \(\lvert h_n(x)\rvert\le g(x)\)
. Dominated convergence gives
Both conditions are local in \(t\) : to prove continuity or differentiability at a point it is enough to find a dominator that works for \(t\) in a neighbourhood of that point.
Example 5.5: An integral evaluated by differentiating under the sign
For real \(t\) let
The integrand is integrable for each \(t\) because \(\lvert\sin(tx)/x\rvert\le\lvert t\rvert\) . Its derivative with respect to \(t\) is \(e^{-x}\cos(tx)\) , and
which is integrable on \((0,\infty)\) and does not depend on \(t\) . By Theorem 5.6(b),
the last integral by two integrations by parts. Since \(F(0)=0\) , integrating gives \(F(t)=\arctan t\) . In particular
The integral was never computed directly. What made the method legitimate was a dominator for the derivative, uniform in the parameter.
5.4 Sequences without a dominator
When no dominator exists the theorem is silent, and the integrals may or may not converge to the integral of the limit.
Example 5.6: Two sequences that look alike
On \([0,1]\) compare
Both tend to \(0\) at every \(x>0\) , and \(u_n(0)=0\) .
For \(u_n\) : since \(1+n^2x^2\ge2nx\) , we have \(0\le u_n\le\tfrac12\) . The measure of \([0,1]\) is finite, so bounded convergence applies and \(\int_0^1u_n\,dx\to0\) . Indeed
For \(v_n\) : the maximum of \(\frac n{1+n^2x^2}\) over \(n\) , for fixed \(x\) , is attained near \(n=\tfrac1x\) and is about \(\tfrac1{2x}\) , which is not integrable near \(0\) . There is no dominator, and in fact
The two formulas differ by a factor of \(x\) . That factor is exactly what tames the concentration at the origin: \(v_n\) is a spike in disguise, with height \(n\) and width of order \(\tfrac1n\) , while \(u_n\) has height \(\tfrac12\) at \(x=\tfrac1n\) and never grows.
Domination is sufficient for convergence of the integrals but not necessary. Exercise 5.13 gives spikes of height \(n/\ln n\)
on \((0,\tfrac1n)\)
, whose integrals \(1/\ln n\)
tend to \(0\)
although no integrable function dominates them. The condition that is both necessary and sufficient on a finite measure space, uniform integrability, is developed with Vitali’s theorem in Measure-Theoretic Probability (math.measure-theoretic-probability.uniform-integrability-vitali). For nonnegative functions there is a clean statement that needs no new concept.
Proposition 5.7 (no lost mass means convergence in the mean). Let \(f_n\ge0\) be integrable and \(f\) measurable with \(f_n\to f\) a.e., and suppose \(\int f_n\,d\mu\to\int f\,d\mu<\infty\) . Then \(\int\lvert f_n-f\rvert\,d\mu\to0\) .
Proof. The functions \((f-f_n)^+\) satisfy \(0\le(f-f_n)^+\le f\) , because \(f_n\ge0\) , and they tend to \(0\) a.e. As \(f\) is integrable, dominated convergence gives \(\int(f-f_n)^+\,d\mu\to0\) . Then
and \(\lvert f_n-f\rvert=(f-f_n)^++(f-f_n)^-\) .
So for nonnegative functions the gap in Fatou’s inequality is precisely the amount of mass that escaped. If the gap is zero, nothing escaped, and the convergence is as good as dominated convergence would have given.
5.5 Which theorem?
There are now three theorems for moving a limit through an integral. They are not interchangeable, and choosing among them is a matter of reading hypotheses.
| Theorem | Requires | Delivers | Does not require |
|---|---|---|---|
| Monotone convergence | \(f_n\ge0\) and \(f_n\le f_{n+1}\) | \(\int f_n\,d\mu\to\int f\,d\mu\) , possibly infinite | any bound on the integrals |
| Fatou’s lemma | \(f_n\ge0\) | \(\int\liminf_nf_n\,d\mu\le\liminf_n\int f_n\,d\mu\) | convergence of \(f_n\) |
| Dominated convergence | \(f_n\to f\) a.e. and \(\lvert f_n\rvert\le g\) with \(\int g\,d\mu<\infty\) | \(\int\lvert f_n-f\rvert\,d\mu\to0\) | any sign or monotonicity |
A workable order of questions is this. Are the functions nonnegative and increasing? Then monotone convergence applies and no estimate is needed. Is only an upper bound for the integral of the limit wanted, or a proof that the limit is integrable? Then Fatou’s lemma is enough. Otherwise look for a dominator: compute or estimate \(\sup_n\lvert f_n\rvert\) and integrate it. If that integral is infinite, the mass may be escaping, and the three pictures of Figure 5.1 tell you where to look.
Example 5.7: One family, three verdicts
On \([0,1]\) consider \(x^n\) and \(nx^n\) . Both tend to \(0\) at every \(x<1\) .
The sequence \(x^n\) . It is decreasing, not increasing, so monotone convergence in the form of Theorem 4.4 does not apply. Fatou’s lemma gives only \(0\le\liminf_n\int_0^1x^n\,dx\) , which is true and useless. Dominated convergence with the dominator \(g=1\) , integrable on \([0,1]\) , gives \(\int_0^1x^n\,dx\to0\) , in agreement with the value \(\tfrac1{n+1}\) .
The sequence \(nx^n\) . Here \(\int_0^1nx^n\,dx=\tfrac n{n+1}\to1\) , while the pointwise limit is \(0\) on \([0,1)\) . So the conclusion of dominated convergence is false and no dominator can exist. To see the failure directly, maximize \(nx^n\) over real \(n\) for fixed \(x\in(0,1)\) : the maximum is at \(n=-1/\ln x\) and equals \(\frac{1}{e\,\lvert\ln x\rvert}\) , which behaves like \(\frac1{e(1-x)}\) as \(x\to1\) and is not integrable there. The factor \(n\) turns a harmless sequence into a spike at the right endpoint.
Fatou’s lemma is consistent with both: \(0\le0\) in the first case and \(0\le1\) in the second.
The convergence theorems concern one variable of integration and one limit. An iterated integral is two integrations, and exchanging them is another interchange of limits. When is the integral over a product space equal to an integral of integrals, in either order?
Exercises
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State Fatou’s lemma for indicator functions \(f_n=\mathbf 1_{A_n}\) and show that it reduces to the inequality \(\mu(\liminf_nA_n)\le\liminf_n\mu(A_n)\) of Exercise 1.8.
Answer
For indicators, \(\liminf_n\mathbf 1_{A_n}=\mathbf 1_{\liminf_nA_n}\) by Exercise 3.17, so the left side of Fatou’s inequality is \(\mu(\liminf_nA_n)\) and the right side is \(\liminf_n\mu(A_n)\) . The proof of Fatou’s lemma specializes to the proof given in Exercise 1.8(a): \(g_n=\inf_{k\ge n}\mathbf 1_{A_k}\) is the indicator of \(\bigcap_{k\ge n}A_k\) .
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Evaluate \(\displaystyle\lim_{n\to\infty}\int_0^n\Bigl(1+\frac xn\Bigr)^ne^{-2x}\,dx\) , naming a dominating function and verifying that it is integrable.
Solution
Let \(f_n(x)=(1+\tfrac xn)^ne^{-2x}\mathbf 1_{[0,n]}(x)\) on \([0,\infty)\) .
Pointwise limit. For fixed \(x\ge0\) the indicator equals \(1\) once \(n\ge x\) , and \((1+\tfrac xn)^n\to e^x\) . So \(f_n(x)\to e^xe^{-2x}=e^{-x}\) .
Dominator. From \(1+u\le e^u\) with \(u=\tfrac xn\) we get \((1+\tfrac xn)^n\le e^x\) , hence \(0\le f_n(x)\le e^{x}e^{-2x}=e^{-x}=g(x)\) for every \(n\) and every \(x\ge0\) . The dominator is integrable: \(\int_0^\infty e^{-x}\,dx=1\) .
Conclusion. By dominated convergence,
\[ \lim_{n\to\infty}\int_0^n\Bigl(1+\frac xn\Bigr)^ne^{-2x}\,dx=\int_0^\infty e^{-x}\,dx=1 . \]Numerically the integrals are about \(0.5808\) , \(0.8856\) , and \(0.9817\) for \(n=1,5,50\) . The sequence is also increasing, so monotone convergence would serve; with the dominator in hand there is no need to prove that.
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Evaluate \(\displaystyle\lim_{n\to\infty}\int_0^1\frac{1+nx^2}{(1+x^2)^n}\,dx\) and justify the interchange of limit and integral.
Solution
Let \(f_n(x)=\dfrac{1+nx^2}{(1+x^2)^n}\) on \([0,1]\) .
Bound. Expanding by the binomial theorem, all terms being nonnegative, \((1+x^2)^n\ge1+nx^2\) . Hence \(0<f_n\le1\) for all \(n\) , and the constant \(1\) is integrable on \([0,1]\) .
Pointwise limit. For \(n\ge2\) the expansion gives the sharper bound \((1+x^2)^n\ge1+nx^2+\tfrac{n(n-1)}2x^4\) , so for \(x>0\)
\[ f_n(x)\le\frac{1+nx^2}{1+nx^2+\tfrac{n(n-1)}2x^4}\longrightarrow0, \]because the denominator grows like \(n^2\) and the numerator like \(n\) . At \(x=0\) the value is \(1\) for all \(n\) , but a single point is a null set.
Conclusion. By bounded convergence (Corollary 5.4) the limit of the integrals is \(\int_0^10\,dx=0\) . The convergence is slow, the integrals being about \(0.681\) , \(0.191\) , and \(0.060\) for \(n=5,50,500\) , because near \(x=0\) the integrand stays close to \(1\) on an interval whose length shrinks only like \(n^{-1/2}\) .
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For each sequence on \((0,1)\) either produce an integrable dominator or prove that none exists: (a) \(f_n(x)=\dfrac{n^{3/2}x}{1+n^2x^2}\) ; (b) \(f_n(x)=ne^{-nx}\) .
Answer
(a) Dominated. With \(u=nx\) , \(f_n(x)=x^{-1/2}\dfrac{u^{3/2}}{1+u^2}\le Cx^{-1/2}\) , where \(C=\max_{u\ge0}\frac{u^{3/2}}{1+u^2}=\frac{3^{3/4}}4\approx0.570\) , attained at \(u=\sqrt3\) . The function \(Cx^{-1/2}\) is integrable on \((0,1)\) , so \(\int f_n\,d\lambda\to0\) . (b) Not dominated: \(f_n\to0\) on \((0,1)\) but \(\int_0^1ne^{-nx}\,dx=1-e^{-n}\to1\) , so the conclusion of dominated convergence fails.
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Let \(f_n\to f\) and \(g_n\to g\) a.e., with \(\lvert f_n\rvert\le g_n\) , all \(g_n\) and \(g\) integrable, and \(\int g_n\,d\mu\to\int g\,d\mu\) . Prove that \(\int\lvert f_n-f\rvert\,d\mu\to0\) .
Solution
After discarding a null set we may assume all the convergence and inequalities hold everywhere. Passing to the limit in \(\lvert f_n\rvert\le g_n\) gives \(\lvert f\rvert\le g\) , so \(f\) is integrable, and
\[ \lvert f_n-f\rvert\le\lvert f_n\rvert+\lvert f\rvert\le g_n+g . \]So the functions \(h_n=g_n+g-\lvert f_n-f\rvert\) are nonnegative, and \(h_n\to2g\) pointwise. Fatou’s lemma gives
\[ \int2g\,d\mu\le\liminf_n\int h_n\,d\mu=\lim_n\int(g_n+g)\,d\mu-\limsup_n\int\lvert f_n-f\rvert\,d\mu=\int2g\,d\mu-\limsup_n\int\lvert f_n-f\rvert\,d\mu, \]where the hypothesis \(\int g_n\,d\mu\to\int g\,d\mu\) was used to evaluate the limit. Since \(\int2g\,d\mu<\infty\) , cancelling it leaves \(\limsup_n\int\lvert f_n-f\rvert\,d\mu\le0\) .
This is the proof of Theorem 5.3 with a moving dominator. The fixed dominator is the special case \(g_n=g\) ; what the general version needs in exchange for letting the dominator move is that the dominators themselves lose no mass.
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Let \(f_n\) be measurable with \(f_n\le g\) for an integrable \(g\) . Prove the reverse inequality \(\limsup_n\int f_n\,d\mu\le\int\limsup_nf_n\,d\mu\) , and show with the sliding block that it fails without the upper bound \(g\) .
Hint
Apply Fatou’s lemma to the nonnegative functions \(g-f_n\) and use \(\liminf_n(g-f_n)=g-\limsup_nf_n\) ; the finite number \(\int g\,d\mu\) cancels. For the sliding block \(\limsup_n\int f_n\,d\lambda=1\) while \(\limsup_nf_n=0\) .
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Suppose \(f_n\to f\) a.e. and \(\sup_n\int\lvert f_n\rvert\,d\mu<\infty\) . Prove that \(f\) is integrable with \(\int\lvert f\rvert\,d\mu\le\liminf_n\int\lvert f_n\rvert\,d\mu\) , and give an example in which \(\int f_n\,d\mu\) does not converge to \(\int f\,d\mu\) .
Solution
Let \(M=\sup_n\int\lvert f_n\rvert\,d\mu<\infty\) . The functions \(\lvert f_n\rvert\) are nonnegative and measurable, and \(\lvert f_n\rvert\to\lvert f\rvert\) a.e., so \(\liminf_n\lvert f_n\rvert=\lvert f\rvert\) a.e. By Fatou’s lemma,
\[ \int\lvert f\rvert\,d\mu=\int\liminf_n\lvert f_n\rvert\,d\mu\le\liminf_n\int\lvert f_n\rvert\,d\mu\le M<\infty . \]So \(f\) is integrable, with the stated bound. (If the measure space is not complete, \(f\) is a.e. equal to a measurable function, to which the argument applies.)
For the example take the spike \(f_n=n\mathbf 1_{(0,1/n)}\) on \((0,1)\) : the integrals \(\int\lvert f_n\rvert\,d\lambda=1\) are bounded, \(f_n\to0=f\) pointwise, and \(\int f_n\,d\lambda=1\) does not converge to \(\int f\,d\lambda=0\) . Bounded integrals guarantee an integrable limit and nothing more. The hypothesis controls how much mass there is, not where it goes.
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A student writes: “On \(\mathbb R\) the functions \(\tfrac1n\mathbf 1_{[0,n]}\) satisfy \(\lvert f_n\rvert\le1\) , so by dominated convergence their integrals tend to \(0\) .” Say exactly what is wrong and what the integrals actually do.
Answer
The constant function \(1\) is not integrable on \(\mathbb R\) , so it is not a dominating function in the sense of Theorem 5.3. The integrals are all equal to \(1\) and so converge to \(1\) , not \(0\) . Example 5.2 shows that the smallest function above every \(f_n\) has infinite integral, so no choice of dominator can rescue the argument.
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Prove that \(\displaystyle\int_0^1\frac{\ln(1/x)}{1+x}\,dx=\sum_{k=0}^\infty\frac{(-1)^k}{(k+1)^2}\) by expanding \(\frac1{1+x}\) and verifying the hypothesis of Theorem 5.5; the sum equals \(\tfrac{\pi^2}{12}\) .
Solution
For \(0<x<1\) , \(\dfrac1{1+x}=\sum_{k=0}^\infty(-1)^kx^k\) . Put \(f_k(x)=(-1)^kx^k\ln(1/x)\) , so that the integrand is \(\sum_kf_k\) on \((0,1)\) . To apply Theorem 5.5 compute, integrating by parts on \([\varepsilon,1]\) and letting \(\varepsilon\to0\) ,
\[ \int_0^1x^k\ln(1/x)\,dx=\Bigl[\frac{x^{k+1}}{k+1}\ln(1/x)\Bigr]_0^1+\frac1{k+1}\int_0^1x^k\,dx=\frac1{(k+1)^2}. \]Hence \(\sum_k\int_0^1\lvert f_k\rvert\,dx=\sum_k\frac1{(k+1)^2}<\infty\) , which is the hypothesis of Theorem 5.5. The theorem gives
\[ \int_0^1\frac{\ln(1/x)}{1+x}\,dx=\sum_{k=0}^\infty\int_0^1f_k\,dx=\sum_{k=0}^\infty\frac{(-1)^k}{(k+1)^2}=\frac{\pi^2}{12}\approx0.8225 . \]The value of the alternating sum follows from \(\sum1/k^2=\pi^2/6\) by subtracting twice the even-indexed terms: \(\tfrac{\pi^2}6-2\cdot\tfrac14\cdot\tfrac{\pi^2}6=\tfrac{\pi^2}{12}\) . Theorem 4.5 could not have been used, because the terms alternate in sign; the absolute series is what has to be checked.
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Let \(F(t)=\int_0^\infty e^{-x^2}\cos(tx)\,dx\) . Show that differentiation under the integral is justified, derive \(F'(t)=-\tfrac t2F(t)\) , and conclude that \(F(t)=F(0)e^{-t^2/4}\) .
Solution
Let \(f(x,t)=e^{-x^2}\cos(tx)\) for \(x\ge0\) and \(t\in\mathbb R\) . For each \(t\) it is integrable in \(x\) , since \(\lvert f\rvert\le e^{-x^2}\) . Its partial derivative in \(t\) is \(-xe^{-x^2}\sin(tx)\) , and
\[ \Bigl\lvert\frac{\partial f}{\partial t}(x,t)\Bigr\rvert\le xe^{-x^2}=g(x),\qquad\int_0^\infty xe^{-x^2}\,dx=\tfrac12<\infty . \]The dominator does not depend on \(t\) , so Theorem 5.6(b) gives \(F'(t)=-\int_0^\infty xe^{-x^2}\sin(tx)\,dx\) . Integrate by parts, with \(-xe^{-x^2}\) the derivative of \(\tfrac12e^{-x^2}\) :
\[ F'(t)=\Bigl[\tfrac12e^{-x^2}\sin(tx)\Bigr]_0^\infty-\frac t2\int_0^\infty e^{-x^2}\cos(tx)\,dx=-\frac t2F(t). \]The boundary term vanishes at both ends. The differential equation \(F'=-\tfrac t2F\) has the solutions \(F(t)=F(0)e^{-t^2/4}\) : indeed the derivative of \(F(t)e^{t^2/4}\) is \((F'+\tfrac t2F)e^{t^2/4}=0\) . With the value \(F(0)=\int_0^\infty e^{-x^2}\,dx=\tfrac{\sqrt\pi}2\) from Example 6.5, \(F(t)=\tfrac{\sqrt\pi}2e^{-t^2/4}\) ; for instance \(F(2)=\tfrac{\sqrt\pi}{2e}\approx0.3260\) .
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Prove that \(\Gamma(t)=\int_0^\infty x^{t-1}e^{-x}\,dx\) is finite and continuous for \(t>0\) , by finding for each compact interval \([a,b]\subset(0,\infty)\) a dominator valid for all \(t\in[a,b]\) .
Hint
For \(t\in[a,b]\) with \(0<a<b\) , use \(x^{t-1}\le x^{a-1}\) on \((0,1]\) and \(x^{t-1}\le x^{b-1}\) on \((1,\infty)\) . The function equal to \(x^{a-1}e^{-x}\) on \((0,1]\) and \(x^{b-1}e^{-x}\) on \((1,\infty)\) is integrable, by Exercise 4.10 near \(0\) and by comparison with \(e^{-x/2}\) at infinity. Apply Theorem 5.6(a) on \((a,b)\) .
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Let \(f_n\ge0\) with \(\int f_n\,d\mu=1\) for all \(n\) , \(f_n\to f\) a.e., and \(\int f\,d\mu=1\) . Prove that \(\sup_{E\in\mathcal A}\bigl\lvert\int_Ef_n\,d\mu-\int_Ef\,d\mu\bigr\rvert\to0\) .
Hint
For any measurable \(E\) , \(\lvert\int_Ef_n\,d\mu-\int_Ef\,d\mu\rvert\le\int\lvert f_n-f\rvert\,d\mu\) , and the right side does not depend on \(E\) . Proposition 5.7 shows that it tends to \(0\) .
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For \(n\ge2\) let \(f_n=\dfrac{n}{\ln n}\mathbf 1_{(0,1/n)}\) on \((0,1)\) . Show that \(f_n\to0\) pointwise and \(\int f_n\,d\lambda\to0\) , and prove that nevertheless no integrable function dominates the sequence.
Solution
Pointwise convergence. For fixed \(x\in(0,1)\) , \(f_n(x)=0\) as soon as \(n\ge1/x\) .
Integrals. \(\int f_n\,d\lambda=\dfrac n{\ln n}\cdot\dfrac1n=\dfrac1{\ln n}\to0\) . So the integrals do converge to the integral of the limit.
No dominator. Any dominator is at least \(g^*=\sup_{n\ge2}f_n\) . Fix \(k\ge3\) and \(x\in[\tfrac1{k+1},\tfrac1k)\) . The function \(f_k\) is positive at \(x\) , since \(x<\tfrac1k\) , so \(g^*(x)\ge f_k(x)=\dfrac k{\ln k}\) . Therefore
\[ \int_{(0,1)}g^*\,d\lambda\ \ge\ \sum_{k=3}^\infty\frac k{\ln k}\Bigl(\frac1k-\frac1{k+1}\Bigr)=\sum_{k=3}^\infty\frac1{(k+1)\ln k}. \]This series diverges, by comparison with \(\sum\frac1{(k+1)\ln(k+1)}\) and the integral test: \(\int\frac{dx}{x\ln x}=\ln\ln x\to\infty\) . So \(g^*\) is not integrable and no integrable dominator exists.
The sequence is a spike whose mass, \(1/\ln n\) , tends to zero. Dominated convergence cannot detect that, because the dominator must cover all the spikes at once, and their envelope is just barely non-integrable. Convergence of the integrals here follows from the direct computation, not from any theorem of this chapter.
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Assume Fatou’s lemma has been proved by some independent route. Deduce the monotone convergence theorem from it in two lines.
Answer
Let \(0\le f_n\uparrow f\) . Since \(f_n\le f\) , \(\limsup_n\int f_n\,d\mu\le\int f\,d\mu\) . Since \(f=\liminf_nf_n\) , Fatou gives \(\int f\,d\mu\le\liminf_n\int f_n\,d\mu\) . The two inequalities force the limit to exist and equal \(\int f\,d\mu\) .
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State the dominated convergence theorem for counting measure on \(\{0,1,2,\dots\}\) as a theorem about series whose terms depend on \(n\) , and use it to prove that \((1+\tfrac xn)^n\to e^x\) for every real \(x\) .
Solution
For counting measure on \(\{0,1,2,\dots\}\) an integrable function is an absolutely summable sequence and the integral is the sum (Exercise 4.2). Dominated convergence becomes: if \(a_{n,k}\to a_k\) as \(n\to\infty\) for each \(k\) , and \(\lvert a_{n,k}\rvert\le b_k\) for all \(n\) with \(\sum_kb_k<\infty\) , then \(\sum_ka_{n,k}\to\sum_ka_k\) .
Fix real \(x\) . By the binomial theorem \((1+\tfrac xn)^n=\sum_{k=0}^\infty a_{n,k}\) with
\[ a_{n,k}=\binom nk\frac{x^k}{n^k}=\frac{x^k}{k!}\cdot\frac{n(n-1)\cdots(n-k+1)}{n^k}\quad(k\le n),\qquad a_{n,k}=0\quad(k>n). \]For fixed \(k\) the last fraction is a product of \(k\) factors each tending to \(1\) , so \(a_{n,k}\to\frac{x^k}{k!}\) . Each factor lies in \([0,1]\) , so \(\lvert a_{n,k}\rvert\le\frac{\lvert x\rvert^k}{k!}=b_k\) , and \(\sum_kb_k=e^{\lvert x\rvert}<\infty\) . The theorem gives
\[ \Bigl(1+\frac xn\Bigr)^n\longrightarrow\sum_{k=0}^\infty\frac{x^k}{k!}=e^x . \]The dominating sequence does not depend on \(n\) , and that is the whole point: termwise convergence alone would not justify taking the limit inside an infinite sum.
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Prove Corollary 5.4 directly from Egorov’s theorem, without using Fatou’s lemma: split \(X\) into a set on which convergence is uniform and a set of small measure.
Hint
Given \(\varepsilon>0\) choose, by Egorov’s theorem, a set \(E\) with \(\mu(X\setminus E)<\varepsilon\) on which \(f_n\to f\) uniformly. Then \(\int\lvert f_n-f\rvert\,d\mu\le\mu(E)\sup_E\lvert f_n-f\rvert+2M\varepsilon\) , using \(\lvert f\rvert\le M\) a.e. Let \(n\to\infty\) and then \(\varepsilon\to0\) .
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Evaluate \(\displaystyle\lim_{n\to\infty}\int_0^\infty\frac{\sin(x^n)}{x^n}\,dx\) by treating \((0,1)\) and \((1,\infty)\) separately and giving a dominator valid for all \(n\ge2\) .
Answer
The limit is \(1\) . On \((0,1)\) , \(x^n\to0\) and \(\frac{\sin u}{u}\to1\) as \(u\to0\) , so the integrand tends to \(1\) , and it is bounded by \(1\) . On \((1,\infty)\) the integrand is at most \(x^{-n}\le x^{-2}\) for \(n\ge2\) and tends to \(0\) . The dominator equal to \(1\) on \((0,1]\) and \(x^{-2}\) on \((1,\infty)\) is integrable, so the limit is \(\int_0^11\,dx+0=1\) .