m e,f µ a - unimi.it · 22 m. m. peloso closed under countable unions of disjoint sets – see the...
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22 M. M. PELOSO
closed under countable unions of disjoint sets – see the comment regarding formula (1). To thisend, we first show that M is closed under finite unions. Let E,F 2 M and let A ✓ X . Then
µ⇤(A) = µ⇤(A \ E) + µ⇤(A \ cE)
= µ⇤�(A \ E) \ F�+ µ⇤�(A \ E) \ cF
�+ µ⇤�(A \ cE) \ F
�+ µ⇤�(A \ cE) \ cF
�.
We observe that
E [ F = (E \ F ) [ (E \ cF ) [ ( cE \ F )
so that, using subadditivity
µ⇤�A \ E) \ F�+ µ⇤�(A \ E) \ cF
�+ µ⇤�(A \ cE) \ F
� � µ⇤�A \ (E [ F )�
Since A \ cE) \ cF = A \ c(E [ F we then have
µ⇤(A) � µ⇤�A \ (E [ F )�+ µ⇤�A \ c(E [ F )
�.
By (10) this implies that E [ F 2 M, and M is an algebra.Suppose that E,F 2 M and E \ F = ;. Then
µ⇤(E [ F ) = µ⇤�(E [ F ) \ E�+ µ⇤�(E [ F ) \ cE
�= µ⇤(E) + µ⇤(F ) ,
so that µ⇤ is finitely addivitive in M. In order to show that M is a �-algebra, it su�ces to showthat it is closed under countable union of disjoint subsets. Then, let {E
j
} ✓ M be a sequenceof disjoint sets. Setting F = [+1
j=1Ej
, we wish to show that F is µ⇤-measurable.To this end, set F
n
= [n
j=1Ej
, and let A ✓ X . Then
µ⇤(A \ Fn
) = µ⇤(A \ Fn
\ En
) + µ⇤(A \ Fn
\ cEn
)
= µ⇤(A \ En
) + µ⇤(A \ Fn�1) ,
as it is easy to check. Arguing by induction we then obtain that
µ⇤(A \ Fn
) =nX
j=1
µ⇤(A \ Ej
) .
It follows that
µ⇤(A) = µ⇤(A \ Fn
) + µ⇤(A \ cFn
) �nX
j=1
µ⇤(A \ Ej
) + µ⇤(A \ cF ) ,
for every n � 1. Letting n ! +1 we obtain
µ⇤(A) �+1Xj=1
µ⇤(A \ Ej
) + µ⇤(A \ cF ) � µ⇤⇣+1[
j=1
(A \ Ej
) + µ⇤(A \ cF )⌘
= µ⇤(A \ F ) + µ⇤(A \ cF ) = µ⇤(A) .
Thus, the above ones are all equalities and F is µ⇤-measurable, as we wanted to show. Moreover,taking A = F in the above equalities, we obtain
µ⇤⇣+1[
j=1
Ej
⌘= µ⇤(F ) =
+1Xj=1
µ⇤(Ej
) ,
that is, µ⇤ is countable additive on M.
MEASURE THEORY AND LEBESGUE INTEGRAL 23
Finally, we need to show that M is complete. Suppose that µ⇤(N) = 0 and let A ✓ X be anyset. Then,
µ⇤(A) µ⇤(A \N) + µ⇤(A \ cN) = µ⇤(A \ cN) µ⇤(A) ,
so that N 2 M and M is complete. ⇤
The application of Caratheodory’s theorem we have in mind deals with the notion of premea-sure.
Definition 3.4. Given a set X and an algebra A of subsets of X , a set function ⇢ : A ! [0,+1]is called a premeasure if
(i) ⇢(;) = 0;
(ii) if {Aj
} is a sequence of disjoint sets in A such that [+1j=1Aj
2 A, then
⇢⇣+1[
j=1
Aj
⌘=
+1Xj=1
⇢(Aj
) .
Notice that a premeasure satisfies the monotonicity condition: if E,F 2 A, E ✓ F , then⇢(E) ⇢(F ). Indeed, writing F = E [ (F \ cE), the conclusion follows easily.
Lemma 3.5. Let X be a set, A an algebra of subsets of X and ⇢ : A ! [0,+1] a premeasure,and let µ⇤ be given by
µ⇤(A) = infn +1X
j=1
⇢(Ej
) : A ✓+1[j=1
Ej
, Ej
2 Eo. (11)
Then µ⇤ is an outer measure such that the following properties hold true:
(i) µ⇤|A = ⇢;
(ii) every set E 2 A is µ⇤-measurable.
Proof. By Prop. 3.1 we know that µ⇤ is an outer measure. (i) Let E 2 A. Then E ✓ [+1j=1Ej
,where E1 = E and E
j
= ; for j = 2, 3, . . . . Then
µ⇤(E) +1Xj=1
⇢(Ej
) = ⇢(E) .
Thus, it su�ces to prove the reverse inequality. Let E 2 A and let {Ej
} be any covering of Ewith sets in A, i.e. E
j
2 A, j = 1, 2, . . . , and E ✓ [+1j=1Ej
. Set An
= En
\ [n�1j=1Ej
. Then theA
n
’s are dijoint elements of A, as it is easy to check, and their union is E. Therefore, by thedefinition of premeasure,
⇢(E) =+1Xn=1
⇢(An
) +1Xn=1
⇢(En
) .
Taking the infimum on the right hand side over the coverings of E by sets in A, it follows that⇢(E) µ⇤(E) for E 2 A. This proves (i).
24 M. M. PELOSO
(ii) Let E 2 A. In order to show that E is µ⇤-measurable, it su�ces to prove (10). Let B ✓ X .Given " > 0, there exists a collection {A
j
} ✓ A such that B ✓ [+1j=1Aj
andP+1
j=1 ⇢(Aj
) µ⇤(B) + ". Then, by the additivity of ⇢ on A,
µ⇤(B) + " �+1Xj=1
⇣⇢(A
j
\ E) + ⇢(Aj
\ cE)⌘� µ⇤(B \ E) + µ⇤(B \ cE) ,
since {Aj
\E} and {Aj
\ cE} are coverings of B\E and B\ cE, resp., by elements of A. Since," > 0 was arbitraty, this proves (ii). ⇤
Lemma 3.6. Let X ,A, ⇢ and µ⇤ be as in Lemma 3.5. Then
(i) Given any B ✓ X and " > 0 there exists a collection {Aj
} ✓ A be such that B ✓ [+1j=1Aj
and µ⇤� [+1j=1 Aj
� µ⇤(B) + ".
Assume further that ⇢ is �-finite and let M(A) be the �-algebra generated by A. The followingproperties hold true.
(ii) A set E ✓ X is µ⇤-measurable if and only if there exists A 2 M(A) such that E ✓ Aand µ⇤(A \ E) = 0.
(iii) A set E ✓ X is µ⇤-measurable if and only if there exists B 2 M(A) such that B ✓ Eand µ⇤(E \B) = 0.
Proof. (i) If µ⇤(B) we have nothing to prove. Suppose that µ⇤(B) is finite. Then, given " > 0, bydefinition of infimum, there exists {A
j
} ✓ A such that B ✓ [+1j=1Aj
andP+1
j=1 ⇢(Aj
) µ⇤(B)+".Using Lemma 3.5 (i) we then have
µ⇤⇣+1[
j=1
Aj
⌘
+1Xj=1
µ⇤(Aj
) =+1Xj=1
⇢(Aj
) µ⇤(B) + " ,
as we wished to prove.(ii) One direction is obvious. If E ✓ X and there exists B 2 M(A) such that B ✓ E and
µ⇤(E \B) = 0, then E = B [ (E \B). It follows that E is µ⇤-measurable since B 2 M(A) andall sets of µ⇤-measure 0 are µ⇤-measurable.
Conversely, suppose E ✓ X is µ⇤-measurable. Assume first that µ⇤(E) < +1. For each
k = 1, 2, . . . , we apply part (i) to E, with " = 1/k. We find a collection of sets {A(k)j
} such that
A(k) := [+1j=1A
(k)j
◆ E and µ⇤(A(k)) µ⇤(E) + 1k
. Set
A =+1\k=1
A(k) .
Then A ◆ E, so that µ⇤(E) µ⇤(A). Moreover, for each k,
µ⇤(A) = µ⇤⇣ +1\
k=1
A(k)⌘ µ⇤(A(k)) µ⇤(E) +
1
k,
so that µ⇤(A) = µ⇤(E). Since E is µ⇤-measurable,
µ⇤(A) = µ⇤(A \ E) + µ⇤(A \ cE) = µ⇤(E) + µ⇤(A \ E) .
MEASURE THEORY AND LEBESGUE INTEGRAL 25
Since µ⇤(A) = µ⇤(E) < +1, we can substract it from both sides and obtain that µ⇤(A\E) = 0.This proves (ii) when µ⇤(E) < +1.
Next, suppose E is µ⇤-measurable and µ⇤(E) = +1. Here we use the assumption that ⇢ is�-finite, that is that X = [+1
j=1Xj
with ⇢(Xj
) < +1. We may assume that the Xj
’s are disjoint.Let E
j
= E \ Xj
. By Lemma 3.5 (ii) we know that the Xj
’s are µ⇤-measurable, so is Ej
foreach j.
Then, for each j there exists Aj
2 M(A) such that Ej
✓ Aj
and µ⇤(Aj
\ Ej
) = 0. SettingA = [+1
j=1Aj
we have that E ✓ A and
µ⇤(A \ E) = µ⇤⇣+1[
j=1
(Aj
\ Ej
)⌘
+1Xj=1
µ⇤�(Aj
\ Ej
)�= 0 .
This proves (ii).
Finally we prove (iii). We first assume that µ⇤(E) < +1. We apply (ii) ( which is now validfor all µ⇤-measurable sets) to F = cE and find A 2 A, A ◆ cE and µ⇤(A \ cE) = 0. SettingB = cA, we have B 2 M(A), B ✓ E and since E is µ⇤-measurable,
µ⇤(B) = µ⇤(B \ E) + µ⇤(B \ cE) = µ⇤(B) + µ⇤(B \ cE) .
Now, µ⇤(B) µ⇤(E) < +1, so we can subtract it on both sides of the line of equations above.We then obtain
µ⇤(B) + µ⇤(B \ cE) = 0 .
The proof of the case µ⇤(E) is similar to the analogous case in (ii). This completes the proof. ⇤We are finally ready to prove the result about the construction of a complete measure, that
will be used in the next session to construct the Lebesgue measure on R as a particular instance.
Theorem 3.7. Let X ,A, ⇢ and µ⇤ be as in Lemma 3.5. Further, assume that ⇢ is �-finite. LetM be the �-algebra of the µ⇤-measurable sets and let µ = µ⇤
|M be the complete measure given
by Thm. 3.3.Let M(A) be the �-algebra generated by A and let ⌫ be any measure on M(A) whose restric-
tion to A coincides with ⇢. Then µ is the completion of ⌫ and
M = M(A) [N , (12)
where N =�N ✓ X : there exists F 2 M(A), N ✓ F, ⌫(F ) = 0
.
Proof. In order to show that (µ,M) is the completion of (⌫,M(A)) we need to show thatµ|M(A)
= ⌫ and that (12) holds. We begin with the former one.
Let µ⇤ be the outer measure given by (11) in Lemma 3.5. Let (M, µ) be the completemeasure constructed in Thm. 3.3, starting from µ⇤. By Lemma 3.5 we know that A ✓ M.Hence M(A) ✓ M. Now, let E 2 M(A) and let {A
j
} ✓ A be such that E ✓ [+1j=1Aj
. Then
⌫(E) +1Xj=1
⌫(Aj
) =+1Xj=1
⇢(Aj
)
so that
⌫(E) infn +1X
j=1
⇢(Aj
) : E ✓+1[j=1
Aj
, Aj
2 Ao= µ⇤(E) = µ(E) ,
26 M. M. PELOSO
since E in particular is in M. Thus, ⌫(E) µ(E) for all E 2 M(A). Moreover, settingA = [+1
j=1Aj
, since ⌫ and µ coincide on A, we have that
⌫(A) = limn!+1
⌫⇣ n[
j=1
Aj
⌘= lim
n!+1µ⇣ n[
j=1
Aj
⌘= µ(A) .
Conversely, assume first that µ(E) < +1. Then, given " > 0, there exists a collection{A
j
} ✓ A such that E ✓ [+1j=1Aj
and µ(A) µ(E) + ", so that µ(A \ E) < ". Then, using thefirst part too, we have that
µ(E) µ(A) = ⌫(A) = ⌫(E) + ⌫(A \ E) ⌫(E) + µ(A \ E) ⌫(E) + " .
Since " > 0 was arbitrary, we have µ(E) ⌫(E), hence µ(E) = ⌫(E), if µ(E) < +1 andE 2 M(A). Finally, suppose E 2 M(A) and µ(E) = +1. Here we assume that ⇢ is �-finite,that is that X = [+1
j=1Aj
with ⇢(Aj
) < +1. We may assume that the Aj
’s are disjoint. Then
µ(E) =+1Xj=1
µ(E \Aj
) =+1Xj=1
⌫(E \Aj
) ⌫(E) .
Hence, ⌫ = µ on M(A).The fact that (12) holds follows from Lemma 3.6 (ii). Indeed, Lemma 3.6 (ii) says that E ✓ X
is in M if and only if there exists B 2 M(A) such that
E = B [ (E \B)
with µ(E \B) = 0, where clearly (E \B) 2 M. This proves (12) and therefore the theorem. ⇤
3.2. The Lebesgue measure on R. We now are now ready to introduce the main object ofthis course.
We consider the collection A of finite unions of disjoint left-open/right-closed intervals in R,that is,
A =nE ✓ R : E =
n[j=1
Ij
, Ij
disjoint, Ij
= (aj
, bj
] or Ij
= (aj
,+1), �1 aj
< bj
< +1o.
Lemma 3.8. Set ⇢(;) = 0, ⇢(E) = +1 if E 2 A is unbounded, and if the intervals {(aj
, bj
]},j =, 1, 2, . . . , n are disjoint and E = [n
j=1(aj , bj ],
⇢(E) =nX
j=1
(bj
� aj
) .
Then ⇢ is a premeasure on the algebra A.
Proof. It is clear that E is well defined, that is, if E = [n
j=1(aj , bj ] = [m
k=1(ck, dk], then
nXj=1
(bj
� aj
) =mXk=1
(dk
� ck
) .
It is also easy to see that ⇢ is finitely additive. In fact, both assertions follow from the factthat an element of A can be written in a unique way as disjoint union of maximal disjointleft-open/right-closed intervals in R.
MEASURE THEORY AND LEBESGUE INTEGRAL 27
Thus, it remains to show that if E is countable union of disjoint left-open/right-closed intervalsE = [+1
j=1(aj , bj ], and E 2 A, then ⇢(E) =P+1
j=1 ⇢�(a
j
, bj
]�=P+1
j=1(bj � aj
). Since E 2 A, itcan be written as finite union of disjoint left-open/right-closed intervals I
j
j = 1, . . . , n. Usingthe finite subadditivity of ⇢, it su�ces to consider the case when E is itself a left-open/right-closed interval (a, b]. Suppose then that I = (a, b] = [+1
j=1Ij , where Ij
= (aj
, bj
]. Since for each
n, I =� [n
j=1 Ij� [ � [+1
j=n+1 Ij�=:
� [n
j=1 Ij� [ J , with J 2 A, we have
⇢(I) = ⇢⇣ n[
j=1
Ij
⌘+ ⇢(J) =
nXj=1
⇢(Ij
) + ⇢(J) �nX
j=1
⇢(Ij
)
for all n. Hence,
⇢(I) �+1Xj=1
⇢(Ij
) .
Conversely, ifP+1
j=1 ⇢(Ij) = +1 we have nothing to prove. Hence, assume thatP+1
j=1 ⇢(Ij) <+1. Given " > 0, there exists n such that
+1Xj=n+1
⇢(Ij
) " .
Writing I =� [n
j=1 Ij� [ � [+1
j=n+1 Ij�=:
� [n
j=1 Ij� [ J , as above, we have
⇢(J) +1X
j=n+1
⇢(Ij
) " .
Therefore,
⇢(I) = ⇢⇣ n[
j=1
Ij
⌘+ ⇢(J) =
nXj=1
⇢(Ij
) + " ,
which gives that
⇢(I)� " +1Xj=1
⇢(Ij
) ,
for every " > 0. Since " > 0 was arbitrary, this proves the inequality ⇢(I) P+1j=1 ⇢(Ij), hence
the lemma. ⇤
Definition 3.9. Let A and ⇢ be as in Lemma 3.8, µ⇤ the outer measure defined in (11) inLemma 3.5. We define the Lebesgue measure space on (R,L,m) where L is the �-algebra andas the complete measure m = µ constructed in Thm. 3.7. The measure m is called the Lebesguemeasure on R anf L the �-algebra of Lebesgue measurable sets.
Remark 3.10. We collect here some obvious but fundamental properties of the Lebesgue mea-sure in R.
(1) We begin by observing that the �-algebra of Lebesgue measurable sets L, by Thm. 3.3is the completion of the �-algebra generated by the algebra A of the finite unions of disjoint
28 M. M. PELOSO
left-open/right-closed intervals in R. Then L contains BR, the �-algebra of Borel sests in R.More precisely, using Thm. 3.7 we have that,
L = BR [N ,
where N =�N ✓ R : there exists F 2 BR, N ✓ F, m(F ) = 0
.
(2) For each interval I ✓ R, m(I) equals the length of I (possibly +1).
(3) Each open set A ✓ R has positive measure, or possibly = +1. Indeed, it su�ces tonotice that A is at most countable union of disjoint open intervals I
n
, n = 1, 2, . . . , since thenm(A) � m(I) > 0. Then, given x 2 A, there exists an open interval I
x
✓ A and containingx. Let I1 be the union of all such intervals. If I1 = A we are done. Otherwise, there exixtsy 2 A \ I1, and an open interval I
y
contained in A and disjoint from I1. Let z 2 Iy
\Q. DefineI2 be the union of all open intervals containing z and contained in A \ I1. This process is atmost of countably many steps, and the conclusion follows.
(4) Each point {x} has measure 0, so countable sets all have measure 0. In particular, the func-tion �[0,1]\Q is integrable and
R�[0,1]\Q dm = 0, and the function �[0,1]\(R\Q) is also integrable
andR�[0,1]\(R\Q) dm =
R ��[0,1] � �[0,1]\(R\Q)
�dm = 1.
We now see other fundamental properties of the Lebesgue measure.
Proposition 3.11. Let E 2 L. Then
(i) m(E) = inf�m(U) : E ✓ U, U open
,
and also
(ii) m(E) = sup�m(K) : K ✓ E, K compact
.
Proof. We preliminary observe that, if E 2 L, then
m(E) = infn +1X
j=1
(bj
� aj
) : E ✓+1[j=1
(aj
, bj
)o. (13)
Call ⌫(E) the quantity on the right hand side above. By construction we have that
m(E) = infn +1X
j=1
(bj
� aj
) : E ✓+1[j=1
(aj
, bj
]o,
so that clearly m(E) ⌫(E) (since in the case of m we take the infimum over a larger numericalset). On the other hand, given " > 0, let {(a
j
, bj
]} be a countable collection of left-open/right-closed intervals whose union covers E and such that m(E)+ " � P+1
j=1(bj �aj
). Then {(aj
, bj
+
"2�j)} is a collection of open intervals whose union covers E and such that
+1Xj=1
(bj
+ "2�j � aj
) =+1Xj=1
(bj
� aj
) + " m(E) + 2" .
Since " > 0 was arbitrary, ⌫(E) m(E) and (13) follows.(i) Since, if E ✓ U we have m(E) m(U), it is clear that
m(E) inf{m(U) : U open, U ◆ E} .
MEASURE THEORY AND LEBESGUE INTEGRAL 29
From (13) we know that given " > 0 there exists a collection of open intervals {Ij
} whose unioncovers E and such that
m(E) + " �+1Xj=1
m(Ij
) � m⇣+1[
j=1
Ij
⌘= m(U) .
This shows that m(E) � inf{m(U) : U open, U ◆ E}, and proves (i).
(ii) Suppose first that E is bounded. Notice that this implies that E is contained in somebounded interval (a, b], so it has finite measure. If E is closed, then it is compact and theconclusion is obvious. Assume now that E is not closed. Then E \E is not empty, and part (i)gives that given " > 0 there exists an open set U ◆ (E \ E) such that m(U) m(E \ E) + ".Let K = E \U = E \U . Then K ✓ E and it is compact, as they are both easy to check. Then,
m(K) = m(E)�m(E \ U) = m(E)� ⇥m(U)�m(U \ E)
⇤� m(E)�m(U) +m(E \ E)
� m(E)� " .
Thus (ii) holds in this case. Suppose now E is unbounded and define Ej
= E\(j, j+1]. We applythe argument above and, given " > 0, there exist compact sets K
j
, j 2 Z such that Kj
✓ Ej
andm(E
j
) m(Kj
) + "2�|j|. Let Hn
= [n
j=�n
Kj
, so that Hn
is compact and Hn
✓ [n
j=�n
Ej
✓ E.Thus,
m(Hn
) =nX
j=�n
m(Kj
) �nX
j=�n
�m(E
j
)� "2�|j|� � nXj=�n
m(Ej
)� 2" .
SinceP
n
j=�n
m(Ej
) ! m(E) as n ! +1, we have that
m(Hn
) � m(E)� 3" ,
for n large enough. This gives (ii) also for unbounded sets, and we are done. ⇤
Example 3.12. Given the sets D = R\Q and D1 = D\ (0, 1]. Given M, " > 0 determine H,Kcompacts, H ✓ D, K ✓ D1 such that m(H) > M and m(K) > 1� ".
Notice that the statement about D follows easily from the case of D1 by setting H =[n
j=�n
(K + j) and n large enough so that m(H) =P
n
j=�n
m(K) > (2n+ 1)(1� ") > M .In the case of D1, we observe that the proof of Thm. 3.11 (ii) provides the construction of
such a set. Let E = D1, so that E = [0, 1] and E \ E = Q \ [0, 1]. Let U be an open setsuch that U ◆ (Q \ [0, 1]) and m(U) m
�(Q \ [0, 1])
�+ " = ". In fact, given " > 0, we
may choose U = [+1j=1Iqj , where {q
j
} is a denumeration of the rational numbers in [0, 1] and
Iqj = (q
j
� "2�(j+1), qj
+ "2�(j+1)). Then, we set K = E \ U = E \ U , that is,
K = [0, 1] \+1[j=1
Iqj ,
which is clearly closed, bounded, hence compact, and
m(K) � 1�+1Xj=1
m(Iqj ) = 1�
+1Xj=1
"2�j = 1� " . ⇤
30 M. M. PELOSO
Proposition 3.13. Let E 2 L. Then E + y =�x + y : x 2 E
2 L for every y 2 R and
m(E + y) = m(E). If r 2 R and rE =�x+ y : x 2 E
, then rE 2 L and m(rE) = |r|m(E).
Proof. Since the collection of left-open/right-closed intervals in R is translation invariant, so isBR. Given y, r 2 R, we then define m
y
(E) = m(E + y) and mr(E) = m(rE). We consider thecase of m
y
first. It is easy to see that my
is a measure on BR. Next, since the premeasure ⇢ isinvariant by translation, if A 2 A then
my
(A) = m(A+ y) = ⇢(A+ y) = ⇢(A) ,
so that my |A = ⇢. Thus, we can apply the second part of Thm. 3.7 and see that (m,L) is the
completion of (my
,BR) and that m = my
on BR. In particular, if E 2 BR and m(E) = 0, thenm(E + y) = m
y
(E) = 0 which implies that the set of Lebesgue measure 0 are preserved bytranslation. Thus, also m
y
is complete and my
= m, that is, m is translation invariant.The same proof shows that mr = rm when r > 0 is positive, since A is preserved by positive
dilations, and mr = r⇢ on A. In this case we apply the second part of Thm. 3.7 to mr, thepremeasure r⇢ and the complete measure rm. We leave the elementary details to the reader.
To treat the case r < 0, it su�ces to consider the case r = �1, that is, to reflection aboutthe origin. In this case we need to start with the algebra A generated by left-closed/right-openintervals instead. Again, we leave the details to the reader. ⇤
The next result now follows easily from the previous result. We leave the proof as an exercise.
Proposition 3.14. Let E ✓ R. Then the following properties are equivalent.
(i) E 2 L;(ii) there exists a G
�
-set V and N 2 L with m(N) = 0 such that E = V \N ;
(iii) there exists an F�
-set W and N 0 2 L with m(N 0) = 0 such that E = W [N 0.
We now compare the Lebesgue integral with the Riemann integral. We write R�[a, b]� todenote the space of Riemann integrable functions on the compact interval [a, b] and we write
R Rb
a
f dt to denote the Riemann integral of f 2 R�[a, b]�. We recall that the definition off 2 R�[a, b]� requires f to be bounded on [a, b].
Proposition 3.15. The following properties hold true.
(i) If f : [a, b] ! R and f 2 R�[a, b]�, then f is Lebesgue integrable on [a, b] andZ[a,b]
f dm = RZ
b
a
f(t) dt .
(ii) If f : [a,+1] ! [0,+1), f 2 R�[a, b]�, for all b > a and there exists the improper
Riemann integral R R +1a
f(t) dt, then f is Lebesgue integrable on [a,+1) andZ[a,+1)
f dm = RZ +1
a
f(t) dt .
Analogously, if f : [a, b) ! [0,+1), f 2 R�[a, c]�, for all a < c < b and there exists
the improper Riemann integral R Rb
a
f(t) dt, then f is Lebesgue integrable on [a, b) andZ[a,b)
f dm = RZ
b
a
f(t) dt .
MEASURE THEORY AND LEBESGUE INTEGRAL 31
(iii) If f : [a, b] ! R is bounded, then f 2 R�[a, b]� if and only if the set F of points ofdiscountinuity of f is such that m(F ) = 0.
Proof. (i) Given a partition P = {a = x0 < x1 < · · · < xn
= b} of [a, b], let
s(f,P) =nX
i=1
mi
(xi
� xi�1), S(f,P) =
nXi=1
Mi
(xi
� xi�1),
be the inferior and superior sums of f , resp., w.r.t. the partition P, wheremi
= infx2(xi�1,xi) f(x)
and Mi
= supx2(xi�1,xi) f(x). Since f 2 R�[a, b]�, given n = 1, 2, . . . , there exist partitions P
n
of [a, b] such that Pn+1 is a refinement of P
n
and 0 S(f,Pn
)� s(f,Pn
) 1/n.For each n = 1, 2, . . . fixed, set I
i
(xi�1, xi), and define the simple functions
sn
=nX
i=1
mi
�Ii , S
n
=nX
i=1
Mi
�Ii .
Then,0 s1 s2 · · · f · · · S2 S1 .
Since the sequences {sn
} and {Sn
} are monotone, let L(x) = limn!+1 s
n
(x) and U(x) =lim
n!+1 Sn
(x). Then, 0 L(x) f(x) U(x) in [a, b]. Notice that L,U are bounded andmeasurable. Now, by the MCT and since f 2 R�[a, b]�,Z
Ldm = limn!+1
Zsn
dm = limn!+1
nXi=1
mi
(xi
� xi�1) = lim
n!+1s(f, P
n
) = RZ
b
a
f(t) dt ,
and also,ZU dm = lim
n!+1
ZSn
dm = limn!+1
nXi=1
Mi
(xi
� xi�1) = lim
n!+1S(f, P
n
) = RZ
b
a
f(t) dt .
Therefore, 0 L U andR[a,b](U � L) dm =
R[a,b] U dm � R
[a,b] Ldm = 0. This gives thatU = L = f a.e. and Z
[a,b]f dm =
Zb
a
f(t) dt ,
as we wished to prove.
Next we prove (iii). Let P = [+1n=1Pn
. Clearly P is countable, hence of measure 0. Forx 2 [a, b] \ P , the following are equivalent:
• f is continuous at x;
• U(x) = L(x).
Since x 2 [a, b] \ P , there exists a sequence of intervals In
such that In
◆ In+1, \+1
n=1In = {x}.Indeed, given n, x belongs to a unique interval I
n
determined by the partition Pn
, Pn+1 is a
refinement of Pn
. Hence, In
◆ In+1, and \+1
n=1In is non-empty, but its diameter is 0, sinceeach partition P
n
contains intervals of at most length 1/n. Then, L(x) = limn!+1 inf
x2In f(x),U(x) = lim
n!+1 supx2In f(x). It is now easy to convince oneself that the two conditions above
are equivalent.Next we observe that the following are equivalent:
(a) f is continuous a.e..;
32 M. M. PELOSO
(b) L = U a.e.;
(c) f 2 R�[a, b]�.Clearly (a) is equivalent to (b), and notice that in (i) we have proved that (c) implies (b).Suppose (b) does not hold, then 0 R
Ldm <RU dm, that is, U � L > � > 0 on a set of
positive measure. This implies that di↵erence infP S(f,P) � supP s(f,P) > 0, that is (c) doesnot hold. Hence, (b) implies (c), and we are done.
(ii) We only prove the first part, the proof of the second one being completely analogous.Since f 2 R�[a, b]� for all b > a, by (i) we have thatZ
[a,n]f dm =
Zn
a
f(t) dt
for all n su�ciently large. Set fn
= �[a,n]f . Then 0 f1 f2 · · · f and fn
! f pointwisein [a,+1). By the MCTZ
[a,+1]f dm = lim
n!+1
Z[a,n]
f dm = limn!+1
RZ
n
a
f(t) dt = RZ +1
a
f(t) dt .
This proves (ii), hence the theorem. ⇤
3.3. Examples.Here and in what follows, dx = dm(x) denotes the Lebesgue measure on the real line. We
also “import” the notationRb
a
f(x) dx to denote the Lebesgue integral on the set [a, b], that is,we write Z
[a,b]f dm =
Zb
a
f(x) dx ,
since, by the privious result the two expressions are equal for f when both integral exist.
(1) Consider the functions fn
: [0,+1) ! [0,+1), fn
= 1n
�[n,n+1). Show that fn
! 0
uniformly in [0,+1), butR +10 f
n
dx 6! R +10 f dx.
On the other hand, show that if fn
! f uniformly on a compact interval [a, b], fn
2 L1([a, b]),
thenRb
a
fn
dx ! Rb
a
f dx.
(2) We evaluate the following limits, if they exist:
limk!+1
Zk
0
✓1� x
k
◆k
ex/2 dx , limk!+1
Zk
0
�1 +
x
k
�k
e�2x dx .
(3) Let fn
: R ! [0,+1) be given by fn
(x) = (1 + 4nx2)�1. Setting f =P+1
n=1 fn, evaluateRf dx.We observe that for each n, f
n
is continuous on R, hence measurable. Since fn
� 0, f =P+1n=1 fn is well defined, non-negative, and measurable. We may apply Cor. 2.17 (2) to see that
+1Xn=1
Zfn
dx =
Z +1Xn=1
fn
dx =
Zf dx .
MEASURE THEORY AND LEBESGUE INTEGRAL 33
We observe that each fn
is integrable and thenZRf(x) dx =
+1Xn=1
ZRfn
(x) dx
= 2+1Xn=1
Z +1
0
1
(1 + (2nx)2)dx
=+1Xn=1
1
2n�1arctan(2nx)
���+1
0
= ⇡ .
(4) Let fn
: (0,+1) ! [0,+1), n = 1, 2, . . . be given by
fn
(x) =1
xen/x log2(1 + nx).
Show that fn
2 L1�(0,+1)
�for all n and that lim
n!+1R +10 f
n
dx = 0.
(5) Given the functions fn
: [0,+1) ! R,
fn
(x) =n4/3x
1 + n5/2x3,
evaluate the limits
(a) limn!+1
Z +1
1fn
(x) dx ,
(b) limn!+1
Z 1
0fn
(x) dx ,
justifying your answers.
It is easy to see that the fn
are measurable and non-negative. (a) When x � 1, fn
! 0pointwise and
0 fn
(x) 1
n7/6x2 1
x22 L1(1,+1) .
By the DCT it follows thatR +11 f
n
dx ! 0 as n ! +1. (b) With the change of variables nx = twe obtain that Z 1
0fn
(x) dx =
Z +1
0gn
(t) dt ,
where gn
(t) = t
n
2/3+n
1/6t
�(0,n)(t). Then, gn
! 0 pointwise on (0,+1) and |gn
t)| t
1+t
3 2L1�(0,+1)
�. The conclusion now follows from the DCT.
(5) Let fk
, gk
: [0,+1) ! R be given respectively, by
fk
(x) =1pk�(0,k)(x)
1pxe(x�
1k )
2and g
k
(x) =1
k2�[ 1k ,+1)(x)
1
x3e�(x�k)2 .
Evaluate, if they exist, the limits,
(i) limk!+1
Z +1
0fk
(x) dx , and (ii) limk!+1
Z +1
0gk
(x) dx .