Example 3.4.17 The integral
3.5 A MORE ADVANCED LOOK AT THE EXISTENCE OF THE PROPER RIEMANN INTEGRAL
In Section 3.2 we found necessary and sufficient conditions for existence of the proper Riemann integral, and in Section 3.3 we used them to study the properties of the integral. However, it is awkward to apply these conditions to a specific function and determine whether it is integrable, since they require computations of upper and lower sums and upper and lower integrals, which may be difficult. The main result of this section is an integrability criterion due to Lebesguethat does not require computation, but has to do with how badly discontinuous a function may be and still be integrable.
We emphasize that we are again considering proper integrals of bounded functions on finite intervals.
Definition 3.5.1Iff is bounded onŒa; b, theoscillation off onŒa; bis defined by WfŒa; bD sup
f .x/ f .x0/j; which can also be written as
WfŒa; bD sup axb
f .x/ inf
( Exercise3.5.1). Ifa < x < b, theoscillation off atxis defined by wf.x/D lim
h!0CWf.x h; xCh/:
The corresponding definitions forxDaandxDbare wf.a/D lim
h!0CWf.a; aCh/ and wf.b/Dh!lim0CWf.b h; b/:
For a fixedx in.a; b/,Wf.x h; xCh/is a nonnegative and nondecreasing function ofh for0 < h < min.x a; b x/; therefore,wf.x/ exists and is nonnegative, by Theorem2.1.9. Similar arguments apply towf.a/andwf.b/.
Theorem 3.5.2Letf be defined on Œa; b:Then f is continuous at x0 inŒa; b if and only ifwf.x0/ D 0: .Continuity at aor bmeans continuity from the right or left, respectively./
ProofSuppose thata < x0 < b. First, suppose thatwf.x0/D0and > 0. Then
WfŒx0 h; x0Ch < for someh > 0, so
jf .x/ f .x0/j< if x0 hx; x0x0Ch: Lettingx0Dx0, we conclude that
jf .x/ f .x0/j< if jx x0j< h: Therefore,f is continuous atx0.
Conversely, iff is continuous atx0and > 0, there is aı > 0such that
jf .x/ f .x0/j< 2 and jf .x 0/ f .x 0/j< 2 ifx0 ı x,x0x0Cı. From the triangle inequality,
jf .x/ f .x0/j jf .x/ f .x0/j C jf .x0/ f .x0/j< ; so
WfŒx0 h; x0Ch if h < ıI therefore,wf.x0/D0. Similar arguments apply ifx0Daorx0Db.
Lemma 3.5.3If wf.x/ < for a x b; then there is a ı > 0 such that
WfŒa1; b1;provided thatŒa1; b1Œa; bandb1 a1< ı:
ProofWe use the Heine–Borel theorem (Theorem1.3.7). Ifwf.x/ < , there is an
hx > 0such that
x 2hx < x0; x00< xC2hx and x0; x002Œa; b: (3.5.2) IfIx D.x hx; xChx/, then the collection
is an open covering ofŒa; b, so the Heine–Borel theorem implies that there are finitely many pointsx1,x2, . . . ,xninŒa; bsuch thatIx1,Ix2, . . . ,Ixn coverŒa; b. Let
and suppose thatŒa1; b1 Œa; bandb1 a1 < h. If x0 andx00 are inŒa1; b1, then
x02Ix rfor somer .1rn/, so jx0 xrj< hxr: Therefore, jx00 xrj jx00 x0j C jx0 xrj< b1 a1Chxr < hChxr 2hxr:
Thus, any two pointsx0andx00inŒa1; b1satisfy (3.5.2) withx Dxr, so they also satisfy (3.5.1). Therefore,is an upper bound for the set
jf .x0/ f .x00/jˇˇx0; x002Œa1; b1 ; which has the supremumWfŒa1; b1. Hence,WfŒa1; b1.
In the following,L.I /is the length of the intervalI.
Lemma 3.5.4Letf be bounded onŒa; band define
x2Œa; bˇˇwf.x/ :
ThenE is closed; andf is integrable onŒa; bif and only if for every pair of positive numbersandı; E can be covered by finitely many open intervalsI1; I2;. . .; Ip such that
L.Ij/ < ı: (3.5.3)
ProofWe first show thatEis closed. Suppose thatx0is a limit point ofE. Ifh > 0, there is anxfromEin.x0 h; x0Ch/. SinceŒx h1; xCh1Œx0 h; x0Chfor sufficiently smallh1andWfŒx h1; xCh1, it follows thatWfŒx0 h; x0Ch for allh > 0. This implies thatx02E, soEis closed (Corollary1.3.6).
Now we will show that the stated condition in necessary for integrability. Suppose that the condition is not satisfied; that is, there is a > 0and aı > 0such that
for every finite setfI1; I2; : : : ; Ipgof open intervals coveringE. IfP D fx0; x1; : : : ; xng is a partition ofŒa; b, then
S.P / s.P /D X j2A .Mj mj/.xj xj 1/C X j2B .Mj mj/.xj xj 1/; (3.5.4) where AD˚jˇˇŒxj 1; xj\E¤ ; and BD ˚ jˇˇŒxj 1; xj\ED ;:
SinceSj2A.xj 1; xj/contains all points ofE except any ofx0,x1, . . . ,xnthat may be inE, and each of these finitely many possible exceptions can be covered by an open interval of length as small as we please, our assumption onEimplies that
X j2A .xj xj 1/ı: Moreover, ifj 2A, then Mj mj ; so (3.5.4) implies that S.P / s.P /X j2A .xj xj 1/ı:
Since this holds for every partition ofŒa; b,f is not integrable onŒa; b, by Theorem3.2.7. This proves that the stated condition is necessary for integrability.
For sufficiency, letandıbe positive numbers and letI1,I2, . . . ,Ip be open intervals that coverEand satisfy (3.5.3). Let
eIj DŒa; b\Ij:
(Ij D closure ofI.) After combining any ofeI1,eI2, . . . ,eIp that overlap, we obtain a set of pairwise disjoint closed subintervals
Cj DŒ˛j; ˇj; 1j q .p/; ofŒa; bsuch that
a˛1< ˇ1< ˛2< ˇ2 < ˛q 1< ˇq 1< ˛q< ˇq b; (3.5.5) q X iD1 .ˇi ˛i/ < ı (3.5.6) and wf.x/ < ; ˇj x˛jC1; 1j q 1: Also,wf.x/ < forax˛1ifa < ˛1and forˇq xbifˇq < b.
LetP0be the partition ofŒa; bwith the partition points indicated in (3.5.5), and refine
P0 by partitioning each subintervalŒˇj; ˛jC1 (as well asŒa; ˛1if a < ˛1 andŒˇq; b ifˇq < b) into subintervals on which the oscillation off is not greater than. This is possible by Lemma 3.5.3. In this way, after renaming the entire collection of partition points, we obtain a partitionP D fx0; x1; : : : ; xngofŒa; bfor whichS.P / s.P /can be written as in (3.5.4), with X j2A .xj xj 1/D q X iD1 .ˇi ˛i/ < ı (see (3.5.6)) and Mj mj ; j 2B:
For this partition, X j2A .Mj mj/.xj xj 1/2K X j2A .xj xj 1/ < 2Kı;
whereKis an upper bound forjfjonŒa; band X
.Mj mj/.xj xj 1/.b a/:
We have now shown that ifandıare arbitrary positive numbers, there is a partitionP of Œa; bsuch that
S.P / s.P / < 2KıC.b a/: (3.5.7) If > 0, let ıD 4K and D 2.b a/: Then (3.5.7) yields S.P / s.P / < ; and Theorem3.2.7implies thatf is integrable onŒa; b.
We need the next definition to state Lebesgue’s integrability condition.
Definition 3.5.5A subsetS of the real line isof Lebesgue measure zeroif for every > 0there is a finite or infinite sequence of open intervalsI1,I2, . . . such that
S [ j Ij (3.5.8) and n X jD1 L.Ij/ < ; n1: (3.5.9)
Note that any subset of a set of Lebesgue measure zero is also of Lebesgue measure zero. (Why?)
Example 3.5.1The empty set is of Lebesgue measure zero, since it is contained in any open interval.
Example 3.5.2Any finite set S D fx1; x2; : : : ; xngis of Lebesgue measure zero, since we can choose open intervalsI1,I2, . . . ,Insuch thatxj 2 Ij andL.Ij/ < =n,
Example 3.5.3An infinite set isdenumerableif its members can be listed in a se- quence (that is, in a one-to-one correspondence with the positive integers); thus,
S D fx1; x2; : : : ; xn; : : :g: (3.5.10) An infinite set that does not have this property isnondenumerable. Any denumerable set (3.5.10) is of Lebesgue measure zero, since if > 0, it is possible to choose open intervals I1,I2, . . . , so thatxj 2Ij andL.Ij/ < 2 j,j 1. Then (3.5.9) holds because
1 2 C 1 22 C 1 23 C C 1 2n D1 1 2n < 1: (3.5.11)
There are also nondenumerable sets of Lebesgue measure zero, but it is beyond the scope of this book to discuss examples.
The next theorem is the main result of this section.
Theorem 3.5.6A bounded functionf is integrable on a finite intervalŒa; bif and only if the setSof discontinuities off inŒa; bis of Lebesgue measure zero:
S D˚x2Œa; bˇˇwf.x/ > 0 :
Sincewf.x/ > 0if and only ifwf.x/1= ifor some positive integeri, we can write
SD 1 [ iD1 Si; (3.5.12) where SiD ˚ x2Œa; bˇˇwf.x/1= i :
Now suppose thatf is integrable onŒa; band > 0. From Lemma3.5.4, eachSican be covered by a finite number of open intervalsIi1,Ii 2, . . . ,Ii n of total length less than
=2i. We simply renumber these intervals consecutively; thus,
I1; I2; DI11; : : : ; I1n1; I21; : : : ; I2n2; : : : ; Ii1; : : : ; Ii ni; : : : :
Now (3.5.8) and (3.5.9) hold because of (3.5.11) and (3.5.12), and we have shown that the stated condition is necessary for integrability.
For sufficiency, suppose that the stated condition holds and > 0. Then S can be covered by open intervalsI1; I2; : : : that satisfy (3.5.9). If > 0, then the set
E D˚x2Œa; b
of Lemma3.5.4is contained inS(Theorem3.5.2), and thereforeEis covered byI1; I2; : : :. SinceEis closed (Lemma3.5.4) and bounded, the Heine–Borel theorem implies thatE is covered by a finite number of intervals fromI1; I2; : : :. The sum of the lengths of the latter is less than, so Lemma3.5.4implies thatf is integrable onŒa; b.
1.In connection with Definition3.5.1, show that sup x;x02Œa;bj f .x/ f .x0/j D sup axb f .x/ inf axbf .x/:
2.Use Theorem3.5.6to show that iff is integrable onŒa; b, then so isjfjand, if f .x/ > 0 .axb/, so is1=f.
3.Prove: The union of two sets of Lebesgue measure zero is of Lebesgue measure zero.
4.Use Theorem 3.5.6and Exercise3.5.3to show that if f andg are integrable on Œa; b, then so aref Cgandfg.
5.Supposef is integrable onŒa; b,˛ D infaxbf .x/, andˇ D supaxbf .x/. Letgbe continuous onŒ˛; ˇ. Show that the compositionh Dgıf is integrable onŒa; b.
6.Letf be integrable onŒa; b, let˛Dinfaxbf .x/andˇDsupaxbf .x/, and suppose thatGis continuous onŒ˛; ˇ. For eachn1, let
aC .j 1/.b a/ n uj n; vj naC j.b a/ n ; 1j n: Show that lim n!1 1 n n X jD1 jG.f .uj n// G.f .vj n//j D0:
7.Leth.x/ D 0 for allx inŒa; bexcept for x in a set of Lebesgue measure zero. Show that ifRabh.x/ dxexists, it equals zero. HINT:Any subset of a set of measure zero is also of measure zero:
8.Suppose thatf andgare integrable onŒa; bandf .x/Dg.x/except forxin a set of Lebesgue measure zero. Show that
Z b a f .x/ dx D Z b a g.x/ dx: