limits of integrals

LIMITS OF INTEGRALS

BY RAbPH PALMER AGNEW

1. Introduction. Let integrals over finite intervals be Lebesgue integrals,and let integrals over infinite intervals be Cauchy-Lebesgue integrals defined by

(1.1) lim limA A,B A

In case g(t) is integrable over ech finite intervM, the identity

) g(t)]dt f+x f+xf. [g(t + g(t)dt- g(-t)dt(1.2)J--A--k OB OA

implies that the equMity

dt lim /+x d lim[ [(+ x (01 () (-)(1.a) dA A A A

holds whenever either of the two members exists. For this nd other resons,fets relating to

A+X

(1.4) lira f(t) dtA eA

are of interest.Perhaps our most striking result is that if the limit in (1.4) exists for ech

in some set having positive measure, then the limit exists for each rel andthe convergence is uniform over each finite interval. Some applications of thisresult are given in 4 and 5.

2. A preliminary theorem. Our first step in the study of (1.4) is to prove thefollowing theorem.

THEOREM 2.1. If

(2.2) L(X) lim f(g) d

exists for each X in some set having positive measure, then L(X) exists for each real Xand L(h) hL where L L(1).The hypothesis implies existence of a number a such that f(t) is integrable

over b _-< c provided b and c are greater than a; all limits of integrationwhich we use are assumed to be greater than a. The identity

(2.3) f,+x=_Xl f,+x=.4 A d (A+2--,1)

Received May 10, 1941; presented to the American Mathematical Society, May 2, 1941.

10

LIMITS OF INTEGRALS 11

with integrand f(t), implies that L(h2 },1) exists and

(2.4) L(h2- hi) L(h.) L(}l)whenever L(),) and L(h.) both exist. Since the set E of values of h for whichL(},) exists has positive measure, there is positive number ti such that eachnumber },0 for which h01 is representable in the form ho h: M whereh. and h are points of E. Hence L(h) exists when I},l < ti. The identity

A A "(A+Xl)

with integrand f(t), implies that L(}, + h2) exists and

(2.6)

provided L() and L(h) both exist. It is now easy to show that L(h) existsfor each real h, and that (2.4) and (2.6) hold whenever },1 and h are real. From(2.4) we see that L(},) is continuous everywhere or discontinuous everywhereaccording as L(h) is continuous or discontinuous at h 0. Since L(},), beingthe limit of the sequence of continuous functions obtained by giving integervalues to A, cannot be discontinuous everywhere it must be continuous every-where. The functional equation (2.5) implies, in a familiar and simple way, thatL(r) rL(1) when r is rational; continuity of L(),) then implies that L(),)hL(1) for each and Theorem 2.1 is proved.

3. Uniformity of the convergence. In this section, we prove the followingtheorem.

THEOREM 3.1. IfA+X

(3.2) lim f(t) dt ,L, < h < ,then the convergence in (3.2) is uniform over each finite interval -a <= h <= a.

Let a be a fixed positive number, let E1 denote the interval -a =< h =< a,and let the measure of E1 be denoted by] E so that E 2a. By a theoremof Egoroff, the convergence in (3.2) is essentially uniform over E that is, toeach 0 > 0 there corresponds a subset E of E such that ]E] > ]E 0 andthe convergence is uniform over E. Let t and E be fixed such that 0 < 1/2aand accordingly E > a. Let A (e) be a function, defined for e > 0, such that

A+X

(3.3) f(t) dt XL < 1/2, E, A > A().

This fact, first proved by Steinhaus, Fundamenta Mathematicae, vol. 1(1920), pp.93-104, has since received very simple proofs. One chooses a point at which the density isgreater than 1/2 and applies the idea following equation (3.4) below. It is an interestingfact that some ses of measure 0, notably the Can,or middle-third set, hve the essentialproperty which we are using.

12 RALPH PALER AGNEW

Let X0 represent any point in the interval -a =< X =< a.of the set E exist such that

Then points ), and

(3.4) X0 X2- Xl.

To prove this, we observe that if such a representation of k0 were impossible,then the set E could have no points in common with the set E0 obtained bytranslating the set E to the right I),01 units;-this would lead to the absurd con-clusion that E and E0 are two disjoint subsets of the interval -a <= <- 2aeach having measure greater than a.Use of the representation (3.4) and the inequality (3.3) gives, when A >

a -F A(e) and -a =< X0 =< a,

f;+x0 f(t) dt o L f(t) dt (X.

f(t) dt kko(t+x.-x)

f(t) dt Xt

and the uniform convergence is established.Because of the identity (1.2), it is a consequence of Theorems 2.1 and 3.1

that ifB

lim f_ [g(t 4r- X)- g(t)] dtA,B-’*o A

exists for each }, in some set having positive measure, then there is a constant Msuch that the limit is hM uniformly over each finite interval of values of .

4. Two theorems of Iyengar. In this section we prove two theorems which,as may be seen by making an exponential change of variable and suitable changesin notation, imply and are implied by a theorem and other results of Iyengar.The proofs of Iyengar are ingenious; by making use of Theorem 3.1 we obtainsimpler proofs.

THEOREM 4.1. A necessary and suicient condition thatA+X

(4.11) lim f(t) dt

K. S. K. Iyengar, On Frullani integrals, Proc. Cambridge Philos. Soc., vol. 37(1941),pp. 9-13. The condition

(u)lim du B log p, p > 0,

of Iyengar becomes (4.21) when we set

log u-x, A log -x, X log p-X, L -B, f(t) q(e-t).

The author must confess that these theorems seem strange to him; he and some ofhis colleagues feel it to be incredible that no Tauberian conditions are involved in thetheorems.


exist for each real h is that

(4.12)

exist.

THEOREM 4.2.

(4.21)

iS that

(4.22)

lim ea f(t)e’-’ dt

A necessary and su2cient condition that

lim f(t) dt hi,

lim et f(t)e-’ dt L.

Because of the equality

(4.23)f(t)e-’ dt f; f(t)e-(-) dt

fo y(t -at- A)e-tdt,

--<h<,

(4.25) fl(t) f(t) L,

Let

(4.24)

so that (4.24) may be writtent+k

(4.26) fl(t) dt < 1/2(1 e-a), A > A0, 0 _-< h _<- a.aA

Let x and A be momentarily fixed such that x > A > A0, and choose an index Nsuch that

(4.27) A q- Na < x <__ A q- (N - 1)a.

We are going to use the inequality

(4.28) -<- + + +a+(N-,),

+A Aq-a

f(t) dt-hL < 1/2,(1-e-a), A>Ao, O-<h <=a.

which holds whenever any one of the integrals exists, the conditions (4.12) and(4.22) can be put in different forms.Using Theorem 2.1, we can see that Theorem 4.1 is a corollary of Theorem

4.2. We prove Theorem 4.2. To prove necessity, let a be a fixed positivenumber; we could take a 1. Then, by Theorem 3.1, to each e > 0 corre-sponds a number A0 A0(e) > 0 such that

14 RALPH PALMER AGNEW

with integrand e’4fl(t)e-t. Using the second mean value theorem we obtainfor each n 1, 2, N

A+na

(4.29) .A+(n_l)afl(t)e--t dt e (t) dt,+(n--1)

where is a properly chosen point between A + (n 1)a and A ha. Using(4.29) and (4.26) we obtain

A+na

(4.30) 1< e(1 e-)e-(-1).

Likewise

(4.31) e (t)e-t dt < (1 e-a)e-v.-t-Na

From (4.28), (4.30), and (4.31) we obtain

(4.32) e j’t(t)e-tdt < , x :> A > Ao(e).

Since A0(e) was chosen greter than 0, this implies that

f(t)e-t dt < e, x > A > A0(e);

and hence the Cauchy criterion for convergence implies existence of

ff(t)e-t dt(4.34)B

for each sufficiently great constant B.(4.32) to obtain

Hence we can let x become infinite in

and therefore

(4.36)

-< e, A > A0(e),ea f(t)e-t dt

=< e, A > A0(e).e" f(t)e- dt L

This completes proof of necessity for Theorem 4.2.To prove sufficiency, let

F(t) e f(u)e du;

In case f(t) is complex valued, we obtain the results separately for the real and imagi-nary parts of f(t) and Theorem 4.2 then follows.


our hypothesis (4.22) then becomes

(4.38) lim F(t) L.

Differentiating (4.37) gives for almost all (that is, for all except those in aset of measure 0)

(4.39) F’(t) F(t) f(t).

Integrating over the interval with end points at A and A -t- k gives

f+ f+(4.40) F(A - k) F(A f(t) dt f(t) dr.A A

Because of (4.38), we can let A become infinite in (4.40) to obtain (4.21). Thiscompletes proof of Theorems 4.1 and 4.2.

5. A Tauberian theorem. We now use Theorem 4.2 to prove the followingtheorem which could be easily generalized by replacing F(t) by F(t) L,L being a constant.

TEo 5.1. If F(t) is absolutely continuous over each finite interval andATh

(5.11) lim [F(t)- F’(t)]dt O, - < ) < ,A-

and

(5.2)

then

(5.3)

To prove this theorem, put

(5.14)

Then, by Theorem 4.2,

lim e-t F(t) O,

(5.15) lim e f(u)e-" du O.

f(t) F(t)- F’(t).

Multiplying (5.14) by the integrating factor e-t gives

d_

(5.16) d--t e F(t) -f(t)e-and, since (5.15) implies existence of the integral on the right,

(5.17) e- F(t) c - f(u)e du.

lim F(t) O.


The result of letting become infinite shows that c 0. Hence

F(0 e’ f( )e

and our resulg follows from (5.15).That ghe Tauberian condition (5.1) eannog be removed from Theorem 15.1

is an obvious consequence of ghe fact thag the funegion () e satisfies (15.11)but not (5.13).

6. Bounded functions f(t). It was pointed out to the author by R. P. Boasthat, in case f(t) is bounded and measurable and the equality

(6.01 lim f(t) dt

in which L is a constant, holds for two values kl and X2 of X for which X1/,2 isirrational, an application of a Tauberian theorem of Wiener establishes theequality

(6.02) lim e" f(t)e-’ dt L;

in this case (6.01) holds also for all values of . It is a consequence of a Tau-berian theorem of Wiener that if K(t), K(t), K(t) are three functions havingabsolutely convergent integrals over - < < , if

(6.11) K() d 1 j 1,

if ghe ourier gransforms of K() and K() have no common eros, and ifis a bounded measurable funegion for which

(6.12) lim K(- Ag()d L

when j 1, 2, ghen (6.12) holds when j a. 0n segging, when j 1, 2,

andKs(t) 0

for0-< <= h,otherwise,

(6.13)K3(t) e-t fort >= 0,

K3(t) 0 fort < 0,

we obtain the conclusion (6.02). If we set, for k > 0,

K(t) X- for 0

___-< X,

Ka(t) 1 otherwise,

N. Wiener, The Fourier Integral, Cambridge (1933), p. 75, Theorem 6.

LIMITS OF NTEGRALS 17

we obtain (6.01) for k > 0; and it then follows easily that (6.01) holds whenk

_0. That (6.02) implies (6.01) follows, in case f(x) is bounded, from Theo-

rem 4, p. 73, of Wiener’s book and the fact that the Fourier transform of thefunction K3(t) in (6.13) has no zeros.Even when f(t) is not assumed bounded, the hypothesis that (6.01) holds

when }, and when is no less general than the hypothesis that the leftmember of (6.01) exists when k hi and when ) h. This is a consequenceof the following theorem.

THORE 6.2. If

(6.21) lim f(t) dtA--+ A

exists for some # O, and to is fixed such that f(t) is integrable over each finiteinterval to <= <- B < , then

1(6.22) lim /B-..* Jto

exists; moreover the equality

(6.23) liraA--*oo A

f(t) dt limB’-O

f(t) dt

holds for each for which the left member exists.

Assuming first that h is a positive number for which (6.21) exists, we prove(6.23). On denoting the limit by hL and setting fl(t) f(t) L, we obtain

lim f(t) dt= O.

Suppose e > 0 and choose A0 > to such that

aa+f(t)

dt < /2, A > Ao.

Corresponding to ech B > A0, let ,(B) and N(B) be numbers such thatAo <= 9(B) < Ao + , N(B) is an integer, and

B (B) + N(B),.

Then, when B > Ao,

fl(t) dt f(t) dt + f(t) dtn=l o (B) +(n--1))

so that

(6.24) if" M N(B)kef(t) dt < +--ff 2B


where M denotes the maximum over the interval A0 _-< u =< A0 -[- X of the con-tinuous function

fl(t) dt

Hence the left member of (6.24) converges to 0 as B -- ; and, using the factthat f(t) f(t) L, we obtain

lim1 f(t) dt L.

B--*o

Since the limit in (6.21) is ,L, we obtain (6.23). The case in which =< 0 nowfollows easily and Theorem 6.2 is proved.

Certain functions of the form

f(t) L -+- t" sin

show that the hypothesis that (6.01) holds for each does not imply that f(t)is bounded. Other examples of functions f(t) for which (6.01) holds for each ),

have the form f(t) nz, when n is a positive integer and n -< < n -[- n-andf(t) 0 otherwise, the numbers , z, forming a real bounded sequence.It is easy to determine the sequence z in such a way that the functions f(t) and

B

fo f(t) dt

each have, as their arguments become infinite, inferior limit - and superiorlimit -- .When f(t) is not assumed bounded, the hypothesis that (6.01) holds for twovalues ), and . of for which /. is irrational does not imply that (6.01)holds for each real ),. This is established by the following example. Letn, n, n,.., be an increasing sequence of positive integers for whichn+/n as p . Suppose, to simplify typography,

d 2-, a 4n, b 4+1.

Suppose f(t) 0 except when is a point of one of the intervals a < < b.For each p 1, 2, 3, let f(x) be defined over the intervul a < < bs follows. For each integer/ for which 1 <= k <= (b a)/2d, suppose

f(x) (-1)2k/(b a), a -- (l 1)d < x -<_ a -b kdand let f(x) be defined over the remaining half of the interval by the formula

f(b x) -f(a + x), 0 < x < 1/2(b a).

For each j 1, 2, 3, the limit in (6.01) exists and is 0 when 2-; anexamination of the graph of f(t) indicates the manner in which proof proceeds.The limit in (6.01) also exists and is 0 when is the irrational number defined by

z E 21-n’;=1


proof of this depends upon the fact that t/d, is only a little greater than aneven integer when p is large, that is,

lim (/d 2[/2d]) 0,

where [x] denotes the greatest integer in x. That the conclusion of Theorem 3.1fails to hold for this function is an obvious consequence of the fact that, foreach p 1, 2, there is an interval of length d over which the integral off(t) is 1 and another of the same length over which the integral is -1. HenceTheorems 2.1 and 3.1 imply that the set of k for which the limit in (6.01) existsmust have measure 0. It can in fact be proved that, for this example, the limitin (6.01) exists for a given k if and only if k can be represented in the form

X ),o + 02-’,

where ),0 is an integer, each 0 is 0 or 1, and a sequence m of integers exists suchthat m -- and

0 Onv, np < j < n + mp

for each p 1, 2, 3, ....CORNELL UNIVERSITY.

limits of integrals

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