THE LEBESGUE INTEGRAL AS A RIEMANN INTEGRAL

The object of this paper is to develop a very direct theory of the Lebesgue integral that is easily accessible to any audience familiar with the Riemann integral.

per Riemann integrals as a prerequisite because, beyond the definition, all that is needed is the fact that such an integral is a linear and increasing function of the integrand).Such a theory is presented in Rey Pastor [5], but only for functions defined on a set of finite measure and his development makes use of the Riemann- Stieltjes integral.The main object of this paper is to reconstruct this theory from the beginning (with one minor modification} for functions defined over sets of arbitrary measure and, for additional simplicity, avoiding all use of the Riemann- Stieltjes integral.This has necessitated a complete set of new proofs and a collec- tion of new auxiliary results (Propositions 2, 3 and 4 and Lemmas 3, 4 and 5 below).
In addition, we have included the Levi monotone convergence theorem and have adopted an entirely different approach to the Lebesgue dominated convergence theorem.Instead of basing its proof on the additivity of the integral (not an easy fact to establish), our approach highlights the role of uniform convergence.Lemmas 3 and 4 show that the contribution of the dominated convergence is just to enable the uniform convergence to do the job via Egorov's theorem and Proposition 4. 2. THE LEBESGUE MEASURE AND MEASURABLE FUNCTIONS.
For easy reference, and to specify how much Lebesgue measure theory is necessary for our presentation, we collect in this section those results that will be used in the rest of the article.For proofs see [I] or [6].By a rectangle in n we shall mean the product space of n bounded intervals, open, closed, or neither.The volume of such a rectangle R, denoted by v(R), is the product of the lengths of its com- ponent intervals.
DEFINITION I.The outer measure of a set E C n is def m(E) : inf v(R.) (2.1) where the infimum is taken over all finite or countable collections of open rectan- gles {Ri such that EC R i.
It turns out that it is not always true that the outer measure of a finite union of disjoint sets is the um of their outer measures.But this will always be true if we restrict ourselves to sts of the following type.DEFINITION 2. A set E n is called measurable if and only if for any set S n m(S) m(SE} + m{S/hEC), where E c is the complement of E in n.If and zero respectively.It is also clear that a set is measurable if and only it its complement is measurable and that if E and F are measurable with F C E then m(F) < m(E).
THEOREM I. (I) If E is a finite or countable set then m(E) O.
(2) If {Ei} is a finite or countable collection of measurable sets then their union E is measurable and m{U E i) < m(E If, in addition, the E i are i i disjoint then equality holds. (3) If {El} is a finite or countable collection of measurable sets then their intersection E i is measurable.If, in addition, m(E I) < and Ei+1 C E i for all i then m(E i) -m(f]E i) as i --.
(4) If E and F are measurable then E F (5) Every half space in n is measurable. (6)Every rectangle in n is measurable and its Lebesgue measure is its volume.
(7) Open and closed sets are measurable.DEFINITION 3. A function f:E C Bn B is measurable if and only if E is measurable and for any y e the set {x e E f(x) > y} is measurable.
THEOREM 2. The following statements are equivalent: (I) For each y e the set {x e E f(x) > y} is measurable.
(2) For each y e the set {x e E f(x) y} is measurable.
(3) For each y the set {x e E f(x) < y} is measurable.
(4) For each y e the set {x e E f(x) y} is measurable.EXERCISE I.If E is measurable and f:E --B is continuous then f is measurable.
THEOREM 3. Let E cn be a measurable set.
(I) If f,g:E-- are measurable then f + g is measurable.
(2) If {fN} is a sequence of measurable functions on E and if fN--f on E then f is measurable.
THEOREM 4. (Egorov's theorem) Let E C n be a measurable set and let be a sequence of measurable functions on E such that fN-f on E. For any 0 there is a subset G of E with m(G) < d and such that fN -f uniformly on E Go 3. THE LEBESGUE INTEGRAL.
The Lebesgue approach to define the integral of a function f:E n _ _ is to partition its range, not its domain as in the Riemann theory.To be specific, consider the case of a measurable set E with m(E) < and a measurable function f:E-whose values are between zero and M > O.If 0 YO < Y < < Yk M is a partition of [0, M] and if we define S. {x e E f(x) > Yi} for i O, k, then f is said to be integrable on E if and only if where the supremum and infimum are taken over all partitions of [0, M], and this E.A. GONZALEZ-VELASCO common value will be denoted by IE f.EXERCISE 2. Define a set E.
{x e E yi_ that for each i.Prove k k " i m(Ei)Yi-1 i:I m(Si)(Yi-Yi-1 " i:I m(Ei)Yi Z {:I m(Si-1 i Now, if we define a function f:[O, M] 3 by f(y) m {x e E f(x) > y} then f is clearly bounded and decreasing and, thus, Riemann integrable on [0, M].
PROPOSITION I. Let E C ]R n be a measurable set with m(E)<=o.Every bounded, nonnegative measurable function f:E is integrable on E and IEf: RI M 0 f (3.4) where the R indicates Riemann integral.
PROOF.Notice that m(S i) f(yi and then, since f is decreasing, i=kl m(Sl)(Yi Yi-1 and i=kl m(Si_1)(y i Yi_1 (3.5) are, respectively, the lower and upper Riemann sums of f with respect to the parti- If f is nonpositlve instead of nonnegative we would like its integral on E to equal the opposite of the integral of -f, that is, we want This suggests that we define Mf(y) -M_f(-y) if y < O, and then the entire prece- ding discussion motivates the following general definitions.
DEFINITION 4. Let E n be a measurable set and let f:E be a measu- Notice that f 0 on (-, O) and f 0 on (O,oo), and that Yl Y2 f(yl f(y2 on each of these intervals.The measure function can have infinite oo, but, if it does not, it is Riemann integrable on every bounded subinterval of (-, O) or (0,o}, and then the improper integrals and R 0 'f .Let E C n be a measurable set, f:E--a measurable func- tion, and f the measure function of f on E. Then the (Lebesgue} integral of f on E is defined to be if the right-hand side exists {that is if the two improper Riemann integrals over (-4, O} and {0,} exist} If, in addition, the right-hand side is finite then f is said to be {Lebesgue} integrable on E nd we write f e [{E}. EXAMPLE I.If E {x 0 < x < I} and f(x} EXAMPLE 3. If E {x ]R 0 < x < I} and f(x) when x is rational and f(x) 0 when x is irrational, then f{y) 0 for all y 0 and f O. E EXERCISE 3. Show that if f > 0 and E f 0 then m {x e E f(x) # O} O.

BASIC PROPERTIES OF THE LEBESGUE INTEGRAL.
First we establish the basic properties of the measure function.We shall write E f instead of f when we need to specify the domain E of f.LEMMA I. Let E cn be a measurable set, let f,g:E be measurable functions and let c e Then (I} f<_ g f g.
(2) cf is measurable and cf(y} sgn(c}f(y/c} for all y O. PROOF.
(I} If y > O, {x e E f(x) > y} x e E g(x} > y} f(y) <_ g(y}.If y < O, {x e E g(x} < y} {x e E f(x} < y} -g(y} -f{y} f(y} g(y}.LEMMA 2. If {Ei} is a finite or countable collection of disjoint measurable sets in n is a measurable function, then f <_ f on (0, } and f >_ f on (-, 0).
THEOREM 5. Let E,F ]R n be measurable sets, let f,g:E ]R be measurable functions and let c e Then, if the integrals below exist, g |E f IE g and if, in addition, f 0 then g L(E) ( IEcf c E f and then f [(E) cf (E). ( (2) Assume that c < 0 (proceed analogously if c > 0).By Lemma I(2), and taking the limit as b--ee, 0 gcf cIO_f ( Similarly for the integral fromto O. Adding the two proves (2).
(3) By Corollary I, If the outer integrals are finite, so are the inner integrals.Thus, f e L(E) f [(E).The last assertion follows from the last two inequalities on the right.
fc   x e E, then am(E) E f bm(E) and m(E)< f e [(E).
EXERCISE 5. Let E c Bn be a measurable set and let f,g:E --B be measu- rable functions If f g on E except, possibly, on a subset of E of measure zero, then IE f =IEE THEOREM 7. Let E C n be a measurable set and let f:E B be a measurable function If Ill e L(E).
PROOF If y > 0 the identity {x e E f(x)l > Y) {x e E f(x) > y} U {x E f(x) <-y} shows that Ifl is measurable and that Ifl (y) f(Y)-f Now, R 0 /f(-Y) dy :-/ 0 .f(u)du R p.f(y) dy (4.10) The right-hand side is always well defined as a real number or oo since f _ 0 on (-, 0} and f >_ 0 on (0, ), and it is finite if and only if both integrals are finite, that is f e [(E} fl e i(E}.In either case, the inequality la + bl a b holds when a > 0 and b < O, when a > 0 and b -, and when a and b <_ O. Thus, (4.12) Q.E.D. EXERCISE 6.Let R be a rectangle in Bn If f:R is Riemann integrable then f e i(E) and IE f RIE f

THE ADDITIVITY OF THE LEBESGUE INTEGRAL.
This is one of the week points of the Lebesgue theory and it will not be straightforward to establish this fact.We start by considering two particular cases: those of piecewise continuous functions and nonnegative bounded functions.
PROPOSITION 2. Let E cn be a measurable set and let f,g:E B be measu- " i " j:1 (ci + Cj) m(Fif]Gj) N K . K Cj .i N Z i:I ci Z j:1 m(FiGj j:1 Q.E.D. PROPOSITION 3. Let E c n be a measurable set and let f,g :E--be non- negatlve, bounded, measurable functions.Then IE(f + g): IE f + IEg M be a partition of [0, M].Then define F.I {x e E f(x) > yi }_ and a func- tion sf } by sf(x} Yi-1 if x e Fi_ Fi, i 1, k. Noticing that YO mlF k) 0 and using Theorem 6 we have k and, as the norm of the partition approaches zero, If Sg is defined for g as sf was defined for f, applying Proposition 2 to sf and Sg, observing that sf + Sg <_ f + g, using Theorem 5(I}, and letting the norm of the partition approach zero, we obtain The first integral above is well defined because f + g is measurable, nonnegative and bounded.Similarly, if we define a function Sf f by Sf(x) Yi if x Fi F. and if S is defined for g as Sf is for f, a similar argument would 1 g yield IE (f + g) IE f + IEg (5.7) and the conclusion follows from the last two inequalities, Q.E.D. THEOREM 8. Let E c n be a measurable set and let f,g [(E}.Then f + g L(E) and IE(f + g): IEf + IEg (5.8) PROOF.Assume first that f and g are nonnegative and for M > 0 define fM M Since fM g f + g, Proposi- mln {f, M) and, similarly, g and {f g}M M tlon  and, since the right-hand side is finite, f g e [(E).

5.5) E-E E
Subtracting the last equation from the previous one and using Theorem 6 proves the desired result in this case.
The two cases proved above imply the truth of the assertion in the general case after decomposing E into the union of four disjoint subsets, on each of which f and g do not change sign, Q.E.D.

THE CONVERGENCE THEOREMS.
In this section we study the validity of the limit IE fN IE f when fN f" In the case of Riemann integration the uniform convergence of this sequence is a sufficient condition.We start by proving a similar result for the Lebesgue integral.
PROPOSITION 4. Let S n be a measurable set with m(S) < o and let {fN} be a sequence of measurable functions such that fN [(S) and fN f e [(S)   uniformly on S. Then ISfN--IS f (6.1) PROOF.Given e > 0 there is a K e + such that f 6 fN f ' and then f-fN f+e' for N > K.
For y > 0 and N > K, and then f{y + ) m {x S f(x) > y + } m {x e S f(x) > y} 0 f em(S) <_ i oo f oo0 gf (y + e) dy _ i For y > The following theorem is usually cited as one of the triumphs of the Lebesgue theory because it replaces the uniform convergence of {fN by the less restrictive condition that the fN be unifdmly bounded.It also allows the domain of these functions to have arbitrary Lebesgue measure.THEOREM 9. (The Lebesgue dominated convergence theorem) Let E C n be a measurable set and let {fN } be a sequence of integrable functions on E that con- verges to an integrable function f on E. If there is a function g [(E) such (6.8).+that fNl <_ g for all N e then IEfN 1E f (6.9} as N --oo, This theorem is sometimes stated with the fN and f measurable rather than integrable, but, in fact, these functions must be integrable.For the fN this is so because g 6 [(E) -g (E) and -g IN g -g fN g" For f because fN f and -g < fN < g -g < f < g" The mechanism through which the uniform boundedness by the integrable function g operates in this theorem can be better understood by means of the next two lemmas, whose proof is temporarily postponed.LEMMA 3. Let E n be a measurable set and let g e [(E).For any e > 0 there is a subset F of E with m(F) < oo such that IIE_Fg < (6.10) LEMMA 4. Let F c ]R n be a measurable set, let g e (F) and let G be a subset of F. For any > 0 there is a > 0 such that if m(G) < then IIGgl <e ( Because of Lemma 3, the integrals of all functions involved in Theorem 9 are negligeable outside a set F of finite measure and, because of Lemma 4, the same is true in a sufficiently small subset G of F. This is significant because, according to Egorov's theorem (Theorem 4), the set G can be chosen so that fN f uniformly on F G. And now, as in the case of Riemann integration, uniform convergence is seen to be the main driving force in the proof of Theorem 9 through Proposition 4. PROOF OF THEOREM 9. Given > 0 let F be as in Lemma 3, let be as in Lemma 4 and, for this choice of , let G be as in Egorov's theorem.Using the triangle inequality and Theorems 6 and 7 we obtain By Proposition 4, N can be chosen so large that the first term on the right is smaller than , and, since e is arbitrary, the result follows, Q.E.D.
PROOF OF LEMMA 3. In view of Theorem 7 we need only consider the case g > O.
By Definition 5, given > 0 there is an a > 0 such that la E 0 g < (6.13) E(a) < ee since g e L(E).
If we define F {x e E g(x) > a} we have m(F) gg PROOF OF LEMMA 4. In view of Theorem 7 we need only consider the case g O.
Given e > 0 there is an a > 0 such that ( In some applications the sequence {fN } may not be uniformly bounded by an inte- grable function.Such a case frequently occurs when the fN are the partial sums of a series of positive functions.In this type of situation Theorem 11 below is useful, but in order to prove it we need some preparation.First we observe that the Lebesgue integral of a nonnegative function can be characterized in terms of those of bounded integrable functions.
LEMMA 5. Let E C n be a measurable set and let f:E-- be a nonnegative measurable function.Then IE f suPlEg (6.177 where the supremum is taken over all bounded functions g e L(E7 such that 0 g f. PROOF.If f e {E), given e > 0 there is a b > 0 such that b 0 <_ IEf RIO f < e (6.18) If we define g min {f, b} then g is bounded, 0 g f, g(y) f(y7 if O< y < b and g(y} 0 if y >_ b. (2) m(E o for some a > O.Given M > 0 let R be a rectangle so large a that m(RE > M/a.If we define g(x)

Call for Papers
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( 3
.8) exist (they may be infinite).For convenience, if f (-oo, O) or on (0,=o) we define has infinite values on RI o f or R Of: oo- (3.9) LEBESGUE INTEGRAL AS A RIEMANN INTEGRAL 697DEFINITION 5

{2)
We shall consider the case c < O, y > O, and the others are similar cf(y} m {x e E cf(x} > y} m x e E f(x} < y/c} -f(y/c} sgn(c}f(y/c} (4.1} Q.E.D.
first assertion follows from Lemma 2 and Definition 5. Now, if c. > E.E i 0 then f(y) m(E i) when 0.< y < c. (here m(E.) can be ) and f (y) 0 otherwise.Then ci Ei m(E Elf R 0 f ci i Similarly if c. < 0, Q.E.D.

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The last two limits and Definition 5 establish the desired convergence, Q.E.D.
) define E {x E f(x) > a} for each a > 0 and consider two cases.{I}m(E < oo for all a > O.For any M > 0 there are a,(y} m(Ea} if 0 y < a, g{y) f(y} if a _ y < b and g{yl 0 if y b.It follows that g[{E} and 10. (Fatou's lena} Let E :21R n be a measurable set, let {fN) be a sequence of nonnegative measurable functions on E and let fN f" Then for any If we define gN min {g, fN} for each N then g is measurable because {x e E :gNIx} > y} {x e E :g{xl > y}h{x e E :fN(x} > y} for each y > O, and THEOREMas N oo, Q.E.D.

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