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5.5.2), so that this definition is unambiguous.Lemma 2 a) A single point and a finite number of points are sets of measure zero.b) The union of a finite or countable number of sets of measure zero is a set ofmeasure zero.c) A subset of a set of measure zero is itself of measure zero.d) A closed interval [a, b] with a < b is not a set of measure zero.Proof a) A point can be covered by one interval of length less than any preassignednumber ε > 0; therefore a point is a set of measure zero.
The rest of a) then followsfrom b).b) Let E = n E n be an at most countable union of sets E n of measure zero.Given ε > 0, for each E n we construct a covering {Ikn } of E n such that k |Ikn | <ε2n .Since the union of an at most countable collection of at most countably manysets is itself at most countable, the intervals Ikn , k, n ∈ N, form an at most countablecovering of the set E, and εI n < + ε + · · · + ε + · · · = ε.k2 222nn,k6.1 Definition of the Integral345The order of summation n,k |Ikn | on the indices n and k is of no importance,since the series converges to the same sum for any order of summation if it convergesin even one ordering.
Such is the case here, since any partial sums of the series arebounded above by ε.Thus E is a set of measure zero in the sense of Lebesgue.c) This statement obviously follows immediately from the definition of a set ofmeasure zero and the definition of a covering.d) Since every covering of a closed interval by open intervals contains a finitecovering, the sum of the lengths of which obviously does not exceed the sum of thelengths of the intervals in the original covering, it suffices to verify that the sum ofthe lengths of open intervals forming a finite covering of a closed interval [a, b] isnot less than the length b − a of that closed interval.We shall carry out an induction on the number of intervals in the covering.For n = 1, that is, when the closed interval [a, b] is contained in one open interval(α, β), it is obvious that α < a < b < β and β − α > b − a.Suppose the statement is proved up to index k ∈ N inclusive.
Consider a covering consisting of k + 1 open intervals. We take an interval (α1 , α2 ) containing thepoint a. If α2 ≥ b, then α2 − α1 > b − a, and the result is proved. If a < α2 < b,the closed interval [α2 , b] is covered by a system of at most k intervals, the sum ofwhose lengths, by the induction assumption, must be at least b − α2 . Butb − a = (b − α2 ) + α2 − a < (b − α2 ) + (α2 − α1 ),and so the sum of the lengths of all the intervals of the original covering of theclosed interval [a, b] was greater than its length b − a.It is interesting to note that by a) and b) of Lemma 2 the set Q of rational pointson the real line R is a set of measure zero, which seems rather surprising at firstsight, upon comparison with part d) of the same lemma.Definition 8 If a property holds at all points of a set X except possibly the pointsof a set of measure zero, we say that this property holds almost everywhere on X orat almost every point of X.We now state Lebesgue’s criterion for integrability.Theorem A function defined on a closed interval is Riemann integrable on thatinterval if and only if it is bounded and continuous at almost every point.Thus,f ∈ R[a, b] ⇔ f is bounded on [a, b] ∧∧ f is continuous almost everywhere on [a, b] .It is obvious that one can easily derive Corollaries 1, 2, and 3 and Proposition 4from the Lebesgue criterion and the properties of sets of measure zero proved inLemma 2.3466IntegrationWe shall not prove the Lebesgue criterion here, since we do not need it to workwith the rather regular functions we shall be dealing with for the present.
However,the essential ideas involved in the Lebesgue criterion can be explained immediately.Propositionn 2 contained a criterion for integrability expressed by relation (6.10).The sum i=1 ω(f ; Δi )Δxi can be small on the one hand because of the factors ω(f ; Δi ), which are small in small neighborhoods of points of continuityof the function. But if some of the closed intervals Δi contain points of discontinuity of the function,then ω(f ; Δi ) does not tend to zero for these points, nomatter how fine we make the partition P of the closed interval [a, b]. However,ω(f ; Δi ) ≤ ω(f ; [a, b]) < ∞ since f is bounded on [a, b]. Hence the sum of theterms containing points of discontinuity will also be small if the sum of the lengthsof the intervals of the partition that cover the set of points of discontinuity is small;more precisely, if the increase in the oscillation of the function on some intervalsof the partition is compensated for by the smallness of the total lengths of theseintervals.A precise realization and formulation of these observations amounts to theLebesgue criterion.We now give two classical examples to clarify the property of Riemann integrability for a function.Example 1 The Dirichlet function!D(x) =1 for x ∈ Q,0 for x ∈ R\Q,on the interval [0, 1] is not integrable on that interval, since for any partition P of[0, 1] one can find in each interval Δi of the partition both a rational point ξi and anirrational point ξi .
Thenn σ f ;P,ξ =1 · Δxi = 1,i=1whilen 0 · Δxi = 0.σ f ; P , ξ =i=1Thus the Riemann sums of the function D(x) cannot have a limit as λ(P ) → 0.From the point of view of the Lebesgue criterion the nonintegrability of theDirichlet function is also obvious, since D(x) is discontinuous at every point of[0, 1] which, as was shown in Lemma 2, is not a set of measure zero.Example 2 Consider the Riemann function!1, if x ∈ Q and x =R(x) = n0, if x ∈ R\Q.mnis in lowest terms, n ∈ N,6.1 Definition of the Integral347We have already studied this function in Sect. 4.1.2, and we know that R(x) iscontinuous at all irrational points and discontinuous at all rational points except 0.Thus the set of points of discontinuity of R(x) is countable and hence has measure zero. By the Lebesgue criterion, R(x) is Riemann integrable on any interval[a, b] ⊂ R, despite there being a discontinuity of this function in every interval ofevery partition of the interval of integration.Example 3 Now let us consider a less classical problem and example.Let f : [a, b] → R be a function that is integrable on [a, b], assuming values inthe interval [c, d] on which a continuous function g : [c, d] → R is defined.
Thenthe composition g ◦ f : [a, b] → R is obviously defined and continuous at all thepoints of [a, b] where f is continuous. By the Lebesgue criterion, it follows that(g ◦ f ) ∈ R[a, b].We shall now show that the composition of two arbitrary integrable functions isnot always integrable.Consider the function g(x) = | sgn |(x). This function equals 1 for x = 0 and 0for x = 0. By inspection, we can verify that if we take, say the Riemann functionf on the closed interval [1, 2], then the composition (g ◦ f )(x) on that interval isprecisely the Dirichlet function D(x).
Thus the presence of even one discontinuityof the function g(x) has led to nonintegrability of the composition g ◦ f .6.1.4 Problems and Exercises1. The theorem of Darboux.a) Let s(f ; P ) and S(f ; P ) be the lower and upper Darboux sums of a realvalued function f defined and bounded on the closed interval [a, b] and corresponding to a partition P of that interval. Show thats(f ; P1 ) ≤ S(f ; P2 )for any two partitions P1 and P2 of [a, b].b) Suppose the partition P̃ is a refinement of the partition P of the interval[a, b], and let Δi1 , . . . , Δik be the intervals of the partition P that contain points of.
that do not belong to P . Show that the following estimates hold:the partition P.) ≤ ω f ; [a, b] · (Δxi1 + · · · + Δxik ),0 ≤ S(f ; P ) − S(f ; P.) − s(f ; P ) ≤ ω f ; [a, b] · (Δxi1 + · · · + Δxik ).0 ≤ s(f ; Pc) The quantities I = supP s(f ; P ) and I = infP S(f ; P ) are called respectivelythe lower Darboux integral and the upper Darboux integral of f on the closedinterval [a, b]. Show that I ≤ I .3486Integrationd) Prove the theorem of Darboux:I = lim s(f ; P ),λ(P )→0I = lim S(f ; P ).λ(P )→0e) Show that (f ∈ R[a, b]) ⇔ (I = I ).f) Show that f ∈ R[a, b] if and only if for every ε > 0 there exists a partition Pof [a, b] such that S(f ; P ) − s(f ; P ) < ε.2.
The Cantor set of Lebesgue measure zero.a) The Cantor set described in Problem 7 of Sect. 2.4 is uncountable. Verify thatit nevertheless is a set of measure 0 in the sense of Lebesgue. Show how to modifythe construction of the Cantor set in order to obtain an analogous set “full of holes”that is not a set of measure zero.
(Such a set is also called a Cantor set.)b) Show that the function on [0, 1] defined to be zero outside a Cantor set and1 on the Cantor set is Riemann integrable if and only if the Cantor set has measurezero.c) Construct a nondecreasing continuous and nonconstant function on [0, 1] thathas a derivative equal to zero everywhere except at the points of a Cantor set ofmeasure zero.3. The Lebesgue criterion.a) Verify directly (without using the Lebesgue criterion) that the Riemann function of Example 2 is integrable.b) Show that a bounded function f belongs to R[a, b] if and only if for any twonumbers ε > 0 and δ > 0 there is a partition P of [a, b] such that the sum of thelengths of the intervals of the partition on which the oscillation of the function islarger than ε is at most δ.c) Show that f ∈ R[a, b] if and only if f is bounded on [a, b] and for any ε > 0and δ > 0 the set of points in [a, b] at which f has oscillation larger than ε can becovered by a finite set of open intervals the sum of whose lengths is less than δ (thedu Bois-Reymond criterion).2d) Using the preceding problem, prove the Lebesgue criterion for Riemann integrability of a function.4.
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