CHAPTER 1
Outer measures, measures and the Lebesgue measure
1.1 Outer measure. Let X be a set. An outer measure µ is a monotone, countably subadditive function on the power set of X with values in the nonnegative extended real numbers
and such that µ(;) = 0. In symbols µ : P (X ) ! [0, 1] satisfies
(1) (normalization) µ(;) = 0;
(2) (monotonicity) F, F 0 2 P(X ), F ⇢ F 0 =) µ(F )  µ(F 0 );
(3) (countable subadditivity) if {Fn } is a countable collection of subsets of X then
✓1 ◆ X
1
[
µ
Fn 
µ(Fn ).
n=1
n=1
1.2 Outer measure from a premeasure. Let X be a set, E be a collection of subsets of X
and : E ! [0, 1] be any function. Define for F 2 P (X )
®1
´
1
X
[
0
µ(F ) = inf
(En ) : over E = {En : n 2 N} ⇢ E with F ⇢
En
n=0
n=0
Then µ is an outer measure on X , generated by the premeasure . To see this, we need
to verify the three properties of Definition 1.1. First of all, the empty set is covered by the
empty subcollection of E. Thus µ(;) = 0. Secondly, if F ⇢ F 0 , then any countable E0 ⇢ E
covering F 0 covers F as well. It follows that µ(F )  µ(F 0 ) since for F we are taking an
infimum over a larger set. Finally, let Fn be a countable collection of sets and F be its union.
Let " > 0 be given. For each Fn , we may find a subcollection En = {E nj : j 2 N} of E which
covers Fn and
X
(E nj )  µ(Fn ) + "2 j .
j2N
Then E0 = {E nj : j, n 2 N} covers F , so that
XX
X
µ(F ) 
(E nj ) 
µ(Fn ) + "
n2N j2N
n2N
which yields the claim (3) due to " being arbitrary. We were able to exchange the summation
order because the terms in the summation are nonnegative.
1.3 Lebesgue outer measure via dyadic cubes. A cube I ⇢ Rd is the cartesian product
of bounded intervals (open, closed or half-open unless otherwise specified) I j , j = 1, . . . , d.
We now consider the problem of computing the volume of an arbitrary subset of Rd . Our
measuring tools will be dyadic cubes. Let Q 0 = [0, 1)d be the unit cube. The collection of
3
4
1. OUTER MEASURES, MEASURES AND THE LEBESGUE MEASURE
dilates and translates of Q 0
1
[
D=
Dn ,
Dn = [k1 2 n , (k1 + 1)2 n ) ⇥ · · · [kd 2 n , (kd + 1)2 n ), k1 , . . . , kd 2 Z
n= 1
is termed the standard dyadic grid on Rd . We denote by `(Q) = 2 n the sidelength of Q if
Q 2 Dn . We denote by Q(k) the unique dyadic cube (k-th ancestor of Q) with sidelength
2k `(Q) that contains Q. We record the following property for further use
8x 2 Rd ,
(1.1)
8n 2 Z,
9! Q 2 Dn with x 2 Q.
A consequence of this is that D forms a grid in the sense that whenever Q, Q0 2 D, either
Q \ Q0 have trivial intersection or one of the inclusions Q ⇢ Q0 , Q0 ⇢ Q holds. In symbols
Q, Q0 2 D =) Q \ Q0 2 {Q, Q0 , ;}.
(1.2)
We consider the outer measure µ : P (Rd ) ! [0, 1] generated by the volume premeasure
(Q) = `(Q)d = 2
nd
Q 2 Dn , n 2 Z.
,
We call this outer measure the Lebesgue outer measure. The remainder of this section will
be dedicated to the proof of the fact that the outer measure coincides with the Euclidean
volume on dyadic cubes, and, more generally, on cartesian products of open, closed, halfopen intervals. A preliminary but fundamental lemma is the following.
1.3.1 LEMMA. Let I = I1 ⇥ · · · ⇥ I d be an axis-parallel cube in Rd with inf I j = a j , sup I j = b j .
Assume I is the disjoint union of dyadic cubes {Q j : j 2 N}. Then
|I| =
n
Y
(b j
aj) =
X
j=1
(Q j ).
j2N
PROOF. We argue in the case d = 1 and leave the higher dimensions as an exercise. We
need to prove that whenever I is an interval and {Q j : j 2 N} is a collection of pairwise
disjoint dyadic intervals whose union is I, we have
X
sup I inf I =
(Q j ).
j2N
Let a = inf I. Write Q j = [u j , v j ) for definiteness. Let " > 0 be given and set x j = u j
"2 j , y j = v j +"2 j , R j = (x j , y j ). Then {R j } is an open cover of the compact set [a, a+ b "].
It thus admits a finite subcover, which we relabel {(x j , y j ), j = 1, . . . , N } with x j increasingly
ordered and such that if R j \ R k 6= ;, then neither R j ⇢ R k nor the opposite inclusion holds.
It must be that x 1 < a, yN > b ", and x j+1 < y j so that
b
"
a  yN
x 1 = yN
xN +
N 1
X
j=1
x j+1
xj  +
N
X
j=1
yj
xj 
X
j2N
(Q j ) + 2"
P
which by taking " ! 0 yields (Q)  j2N (Q j ). To prove the reverse inequality, by taking
limits, it suffices to show that
N
X
(Q j )  (Q) = 1
j=1
1. OUTER MEASURES, MEASURES AND THE LEBESGUE MEASURE
5
We can rearrange {Q j : j = 1, . . . , N } so that sup Q j = b j  min Q j+1 = a j+1 . Then
1
bN
a1 = bN
aN +
N 1
X
a j+1
j=1
aj
N
X
bj
j=1
aj =
N
X
(Q j )
j=1
⇤
as claimed.
1.3.2 PROPOSITION (CONSISTENCY). Let µ denote the Lebesgue outer measure on Rd above
defined. Then
µ(Q) = `(Q)d
for all dyadic cubes Q 2 D.
PROOF. First of all, since {Q} is a covering of Q, we have µ(Q)  `(Q)d = (Q). Assume
now that {Q j } is a cover of Q by dyadic cubes such that
1
X
(1.3)
j=1
(Q j )  (Q).
We prove that equality must hold in (1.3), whence µ(Q) = (Q). Notice that (1.3) entails
that Q j ✓ Q for all j (otherwise the opposite inequality would hold). Let jk be the indices
of those elements of {Q j } which are maximal with respect to the order relation induced by
inclusion. Then it must hold that
[
[
Q jk =
Q j = Q,
k 6= ` =) Q jk \ Q j` = ;
k
j
The first of these assertions follows by observing that every point of Q is contained in a
maximal Q j , since `(Q j ) is bounded above by `(Q). The second follows because if the intersection Q jk \Q j` were nontrivial, one of the inclusions would hold, contradicting maximality.
But then
1
1
X
X
(Q j )
(Q jk ) = (Q)
j=1
k=1
using Lemma 1.3.1 for the second equality. Hence equality must hold in (1.3), and the proof
is complete.
⇤
We now extend the above consistency to open intervals. The main tool (in fact, a more
general version which will be useful in the sequel) is the following covering lemma.
1.3.3 WHITNEY COVERING LEMMA. Let O ⇢ Rd be an open set. Then there exists a subcollection
I of dyadic cubes such that
[
(1) O =
I.
I2I
(2) the cubes of I are pairwise disjoint
(3) the distance of the center of I, c I from the complement of O is comparable to `(I).
PROOF (SKETCH). Let I be the collection of maximal dyadic cubes with the property that
(1.4)
I ⇢ O,
`(I) 
1
c
10 dist(c I , O ).
These are pairwise disjoint by maximality. Let now x 2 O. Since O is open, the collection of
1
those dyadic cubes containing x with `(I)  10
dist(c I , O c ) is nonempty. Thus there exists a
6
1. OUTER MEASURES, MEASURES AND THE LEBESGUE MEASURE
maximal one, which belongs to I. This proves that O is covered by the union of I 2 I. Now,
1
we must have that I (1) the parent of I violates (1.4). In either case, `(I 1 ) 10
dist(c I (1) , O c ),
so that
dist(c I , O c )  2`(I) + dist(c I (1) , O c )  30`(I),
and comparability follows.
⇤
1.3.4 COROLLARIES. Let µ denote the Lebesgue outer measure generated by the dyadic cubes
and | · | denote Euclidean volume.
(1) if I ⇢ Rd is any axis parallel cube then µ(I) = |I|, the Euclidean volume of I;
(2) for all A ⇢ Rd ,
X
µ(A) = inf
|I|
I2I
where I ranges over all countable collections of cubes of Rd whose union covers A;
(3) the Lebesgue outer measure is translation invariant and respects dilations, in the sense
that, for all t 2 Rd , s > 0, A ⇢ Rd
µ(A + t) = µ(A),
µ(sA) = s d µ(A).
⇤
PROOF. Exercise.
We now move onto the approximation properties of the Lebesgue outer measure µ. From
now on, in view of Proposition 1.3.2, we omit the distinction on µ and on dyadic cubes.
1.3.5 LEMMA. Let F ⇢ Rd be any set. Then
µ(F ) = inf µ(O)
O F
O open
PROOF. It suffices to assume that µ(F ) < 1, otherwise there is nothing to prove, and
show that for all " > 0 there exists an open set O with µ(O)  µ(F ) + 2". To do so, let
E0 = {Q j : j 2 N} be a cover of F by dyadic cubes such that
X
(1.5)
µ(Q j )  µ(F ) + ".
j2N
fj be the open cube with the same center as Q j and volume µ(Q j )+"2 j . Then O = [ j Q
fj
Let Q
is an open set containing F . Furthermore, in view of the preceding corollary and of the above
display,
1
X
X
fj ) =
µ(O) 
µ(Q
(µ(Q j ) + "2 j )  µ(F ) + 2"
j=1
j2N
⇤
as claimed.
1.4
-algebras and measures. A subcollection F ⇢ P (X ) such that
(1) (nontrivial) ; 2 F ,
(2) (closed under complement) F 2 F =) F c 2 F ,
(3f) (closed under finite union) F, G 2 F =) F [ G 2 F
hold is said to be an algebra of subsets of X . If (1), (2) and
[
(3) (closed under countable union) {Fn : n 2 N} ⇢ F =) F =
Fn 2 F
n2N
1. OUTER MEASURES, MEASURES AND THE LEBESGUE MEASURE
hold instead, F is called a
able space.
7
-algebra of subsets of X . We refer to a pair (X , F ) as a measur-
1.4.1 REMARK. Firstly, it follows from the definition that if {F↵ : ↵ 2 A} is any collection of
-algebras on X , then F = \↵ F↵ is also a -algebra. Given any collection of sets E ⇢ P (X )
we denote by (E) the intersection of all the -algebras containing E. Such an intersection
is nonvoid, since P (X ) is a -algebra containing E. It is useful to observe that, by definition
E0 ⇢ (E) =)
(E0 ) ⇢ (E).
Let now (X , F ) be a measurable space. A positive measure is a set function
µ : F ! [0, 1],
with the countable additivity property
(1.6)
µ(F ) =
X
n2N
µ(Fn )
whenever {Fn : n 2 N} ⇢ F are pairwise disjoint sets whose union is F . Then (X , F , µ) is
called measure space and the elements of F are called measurable (or F -measurable) sets.
1.4.2 PROPOSITION. (basic properties of measure) Let (X , F , µ) be a measure space and
throughout E j be measurable sets. There holds
(P1) µ(;) = 0;
(P2) (monotonicity) if E1 ⇢ E2 then µ(E1 )  µ(E2 );
(P3) µ is countably subadditive; [
(P4) If E j ⇢ E j+1 for all j, then µ
E j = lim µ(E j )
\
(P5) If µ(E1 ) < 1 and E j E j+1 for all j, then µ
E j = lim µ(E j ), and the assumption
µ(E1 ) < 1 is necessary;
(P6) (lower semicontinuity) µ lim inf E j  lim inf µ(E j );
S
(P7) (upper semicontinuity) lim sup µ(E j )  µ lim sup E j provided µ
E j < 1.
1.4.3 REMARK. In fact, for properties (P1) and (P2), the finite additivity property
(1.7)
would suffice.
PROOF
OF
F, G 2 F , F \ G = ; =) µ(F ) + µ(G) = µ(F [ G)
⇤
PROPOSITION 1.4.2. Exercise.
The following lemma provides a partial converse to (P3)
1.4.4 LEMMA. Let F be a -algebra and µ : F ! [0, 1] be a countably subadditive set
function satisfying the finite additivity property (1.7). Then µ is a measure on F .
PROOF. We need to show that µ is countably additive. Let {Fn : n 2 N} ⇢ F be pairwise
disjoint sets whose union is F . We need to prove (1.6). Relying on countable subadditivity,
we only need to prove the inequality in (1.6). To do so we may assume that the left hand
side µ(F ) is finite. This actually implies that the right hand side is finite as well. Indeed,
using finite additivity and monotonicity
✓N
◆
N
X
[
µ(Fn ) = µ
Fn  µ (F ) .
n=1
n=1
8
1. OUTER MEASURES, MEASURES AND THE LEBESGUE MEASURE
and the
in (1.6) follows by taking the limit as N ! 1 in the above left hand side.
⇤
1.5 Complete measures. Carathéodory’s theorem. For a measure space (X , F , µ), define
N := {F 2 F : µ(F ) = 0},
N 0 := {N ⇢ X : 9F 2 N with N ⇢ F }.
Note that N and N 0 are both closed under countable unions. The measure space (X , F , µ)
is said to be complete if N 0 ⇢ F .
1.5.1 LEMMA. Let (X , F , µ) be a measure space. Then
F := {E [ N : E 2 F , N 2 N 0 }
is a -algebra and there exists a unique measure µ on F such that (X , F , µ) is a complete
measure space, with µ(F ) = µ(F ) for all F 2 F .
PROOF. Since F and N 0 are both closed under countable unions, so is F . We show
closure under complement. Let A = E[N 2 F , with E, F 2 F and N ⇢ F . Replacing possibly
F by F \ E c , N by N \ E c , we can assume E \ F = ;. This way (E [ N )c = (E [ F )c [ (F \ N c ).
But (F \ N c ) ⇢ F and (E [ F )c 2 F . This proves the closure under complement.
Now define µ(E [ N ) := µ(E). This is well defined, since if E [ N = F [ M , E, E 0 2 F ,
N , N 0 2 N 0 subsets respectively of F, F 0 2 N one has E ⇢ E 0 [ F 0 , whence
µ(E)  µ(E 0 [ F 0 )  µ(E 0 ) + µ(F 0 )  µ(E 0 )
and by symmetry, equality must hold. Clearly this extends µ on F . The verification that µ
is a measure is easy and we omit it. To show that it is unique, let ⌫ be another measure with
the same properties, E 2 F , N 2 N 0 , F 2 N such that µ(F ) = 0 and N ⇢ F . Then
⌫(E [ N )  ⌫(E) + ⌫(N )  ⌫(E) + ⌫(F ) = µ(E) = µ(E [ N )
and by symmetry we conclude µ = ⌫ To show that it is complete, let E 2 F , N 2 N 0 ,
µ(E [ N ) = 0 and F ⇢ E [ N . By a similar argument as above we can assume E \ N = ;.
Then µ(E) = 0, and F \ E ⇢ E, so that F \ E 2 N 0 . We also have N \ F 2 N 0 . So actually
F = (F \ E) [ (F \ N ) 2 N 0 , and in particular F 2 F . Therefore µ is complete.
⇤
Now, let µ be an outer measure on X . We say that F is a µ-measurable set if
(1.8)
µ(G) = µ(F \ G) + µ(F c \ G)
8G 2 P (X ).
1.5.2 REMARK. Suppose µ is generated by a premeasure ( , E). Then (1.8) holds for F if
and only if
(1.9)
µ(E) = µ(F \ E) + µ(F c \ E)
We leave this proof as an exercise.
8E 2 E
1.5.3 CARATHÉODORY’S THEOREM. Let µ be an outer measure on X . Then
• the subcollection F 2 F of P (X ) such that (1.8) holds is a
• the restriction of µ to F is a complete measure.
-algebra
The proof is divided into steps. The first, trivial one is to observe that F contains ; and
is closed under complement by virtue of the symmetry of (1.8). We begin the actual work.
1. OUTER MEASURES, MEASURES AND THE LEBESGUE MEASURE
9
STEP 1. F is closed under finite union and µ is finitely additive on F . Let A, B ⇢ X be
such that (1.8) holds for F = A and F = B. Let G ⇢ X be arbitrary. Then, using (1.8) twice
and monotonicity
µ(G) = µ(G \ A \ B) + µ(G \ Ac \ B) + µ(G \ A \ B c ) + µ(G \ (A [ B)c )
µ(G \ A [ B) + µ(G \ (A [ B)c )
which is the nontrivial half of (1.8) for A [ B. Therefore A [ B 2 F . Now, if A and B are
pairwise disjoint and belong to F , the fact that µ(A [ B) = µ(A) + µ(B) is a triviality.
STEP 2. We show that F is closed under countable disjoint union, which, together with
closure with respect to complement and finite union, suffices by the usual argument. Let
{Fn : n 2 N} ⇢ F be pairwise disjoint, A be their union, and AN = F1 [ · · · [ FN . By the
previous step, AN belongs to F . We then have for an arbitrary G,
c
µ(G) = µ(G \ AN ) + µ(G \ (AN ) )
c
µ(G \ AN ) + µ(G \ A ) =
N
X
n=1
µ(G \ Fn ) + µ(G \ Ac )
and letting N ! 1 and using subadditivity of outer measure
✓
◆
1
X
[
c
µ(G)
µ(G \ Fn ) + µ(G \ A ) µ
G \ Fn + µ(G \ Ac ) = µ(G \ A) + µ(G \ Ac ),
n=1
n2N
which is the nontrivial part of (1.8) for A. So F is a -algebra, and using the finite additivity
and Lemma 1.4.4, we conclude µ restricted to F is a measure.
STEP 3. We show that µ is complete. Let F 2 F be a set with µ(F ) = 0 and N ⇢ F .
Then µ(N ) = 0. We have using subadditivity and monotonicity that
µ(G)  µ(G \ N ) + µ(G \ N c )  µ(N ) + µ(G)  µ(G)
1.6 The Lebesgue and Borel -algebra. We keep denoting the Lebesgue outer measure
on Rd by µ. Let L be the -algebra of subsets of Rd such that (1.8) holds. We call L
the Lebesgue -algebra. We already know that µ restricted to L is a complete translation
invariant, dilation-respecting measure. As we have seen, given any topological space X , we
denote the -algebra generated by the collection O of open subsets of X by B(X ) and call
it the Borel -algebra. By the Whitney covering lemma, each open set can be written as a
countable union of dyadic cubes. Therefore B(Rd ) = (D). Let us explore the relationship
between B(Rd ), or simply B, and L .
1.6.1 LEMMA. The following statements hold:
(1) D ⇢ L .
(2) B ( L .
PROOF. To derive the second statement from the first, Let us observe that, by virtue for
instance of Whitney’s covering Lemma 1.3.3, O ⇢ (D), therefore B ⇢ (D). On the
other hand, it is easy to write the unit half closed cube as the countable intersection of open
rectangles. Thus B
(D) as well. Hence, the first statement also implies B ⇢ L . To
disprove equality, there are two possibilities which are explored in later exercises.
10
1. OUTER MEASURES, MEASURES AND THE LEBESGUE MEASURE
We now move onto the proof of (1). Let Q be a dyadic cube. By Remark 1.5.2, to show
that Q 2 L it suffices to show that
µ(Q0 )
µ(Q0 \ Q) + µ(Q0 \ Q c )
8Q0 2 D.
If either Q0 \ Q = ;, or Q0 ⇢ Q, there is nothing to prove. The last remaining case is when
Q ( Q0 . Let then {R j } be a cover of Q0 by disjoint dyadic cubes such that
X
(R j )  µ(Q0 ) + ".
j
Clearly, there holds `(R j ) > `(Q) for finitely many j. By breaking each such R j with into
0
subcubes of
P sidelengthP`(Q) we can construct another countable disjoint cover of Q , say
{S j }, with j (R j ) = j (S j ) and such that either S j \Q = ; or S j ⇢ Q. Then, {S j : S j ⇢ Q}
covers Q and {S j : S j \ Q = ;} covers Q0 \ Q c whence
X
X
X
µ(Q) + µ(Q0 \ Q c ) 
(S j ) +
(S j ) =
(R j )  µ(Q0 ) + ",
S j ⇢Q
and we are done.
s j \Q=;
j
⇤
1.6.2 PROPOSITION. (continuities) Denote by µ the Lebesgue (outer) measure. Then
(1) if A 2 L then µ(A) = inf{µ(O) : A ⇢ O, O open}
(2) if A 2 L then µ(A) = sup{µ(K) : K ⇢ A, K compact}
(3) A 2 L if and only if for all " > 0 there exists an open set O such that
A ⇢ O,
µ(O\A) < ";
(4) A 2 L if and only if for all " > 0 there exists an open set O and a closed set C such
that
C ⇢ A ⇢ O,
µ(O\C) < ",
and C can be chosen to be compact if µ(A) < 1;
(5) if A 2 L then there exist a set G 2 G (Rd ) (countable intersection of open sets) and
a set F 2 F (Rd ) (countable union of closed sets) such that
F ⇢ A ⇢ G,
µ(G\A) = µ(A\F ) = 0.
PROOF. The property (1) is a simple consequence of Lemma 1.3.5. The properties (2)(5) are left as an exercise.
⇤
1.6.3 COROLLARY. The measure space (Rd , L , µ) is the completion of the measure space
(Rd , B, µ)
PROOF. The measure space (Rd , L , µ) is complete, it thus suffices to show that it is contained in the abstract completion of (Rd , B, µ). Let A 2 L . We can use (4) of Proposition
1.6.2 to find a Borel set F with M := A\F of null Lebesgue measure. Then applying again
the same property we find a Borel set N M with µ(N \M ) = 0. It follows that µ(N ) = 0
as well. So A = F [ M with F Borel and M subset of a Borel null set N . Thus A belongs to
the completion of B. which finishes the proof.
⇤
1. OUTER MEASURES, MEASURES AND THE LEBESGUE MEASURE
11
1.7 Non-Lebesgue measurable sets. We present hereby an extended version of the classical example of a subset of R which is not Lebesgue measurable. Coupled with completeness,
that is if A is Lebesgue measurable and µ(A) = 0, then all subsets of A are Lebesgue measurable, this yields a necessary and sufficient condition for the existence of nonmeasurable
subsets of a set A 2 L
1.7.1 PROPOSITION. Let A be a Lebesgue measurable subset such that {B 2 P (X ) : B ⇢ A} ⇢
L . Then µ(A) = 0.
PROOF. The equivalence relation
x ⇠ y () x
y 2Q
partitions R into (uncountably many) equivalence classes. Using the axiom of choice, we
may select from each a representative. Let E ⇢ R be the set whose members are the selected
representatives.
Let {qn } be an enumeration of the rationals and En = E + qn . The definition leads to the
identities
[
En \ Em = ; (m 6= n),
En = R.
Define An = A\ En . Let K be any compact subset of An . By assumption An , K are measurable.
We claim that µ(K) = 0, which implies µ(An ) = 0 for all n. Since A = [An by the second of
the above relations, µ(A) = 0 as well, which was to be proved.
We turn to the proof of the claim The set
[
H=
K +q
q2Q:|q|<1
is bounded, therefore µ(H) < 1. However, by the first of the above, the sets K + q are pairwise disjoint and measurable (being translates of a measurable set), so that using translation
invariance,
X
X
1 > µ(H) =
µ(K + q) =
µ(K)
q2Q:|q|<1
and the claim follows.
q2Q:|q|<1
⇤
1.7.2 REMARK. The two properties of the Lebesgue measure we have used in the above
construction are translation invariance and the fact that bounded sets have finite measure.
12
1. OUTER MEASURES, MEASURES AND THE LEBESGUE MEASURE
Chapter 1 Exercises
1.1. Let µ be an outer measure on X generated by the premeasure ( , E). Let F ⇢ X . Prove
that F is µ-measurable, namely
if and only if
µ(G) = µ(F \ G) + µ(F c \ G)
8G 2 P (X )
µ(E) = µ(F \ E) + µ(F c \ E)
8E 2 E.
1.2. Let D be the dyadic grid on R and µ denote the Lebesgue outer measure on R, namely
for A ⇢ R
®1
´
1
X
[
µ(A) = inf
`(I j ) : A ⇢
I j, I j 2 D 8 j .
j=1
j=1
#1 Let n 2 Z. Prove that µ does not change if we restrict to intervals with `(I j )  2 n ,
namely
®1
´
1
X
[
(n)
(n)
n
µ(A) = µ (A),
µ (A) := inf
`(I j ) : A ⇢
I j , I j 2 D 8 j, `(I j )  2
.
j=1
j=1
PN
#2 Let t 2 R be of the form t = n= N kn 2 n for suitable integers N , k
Prove that µ is invariant under translations by t, namely
µ(A) = µ(A + t)
Hints. For #2, reduce to the case t = 2
2 m } is invariant under translation by 2
n
n
N , . . . , kN .
8A ⇢ R.
for some n. Then use #1 and that Dm = {I 2 D : `(I) =
whenever m n.
1.3. Let O be the collection of all open cubes I ⇢ Rd and define the outer measure on Rd
given by
®1
´
1
X
[
⌫(A) = inf
|R n | : A ⇢
R j, R j 2 O 8 j
n=0
n=0
where |R| is the Euclidean volume of the cube R.
#1 Show that ⌫ coincides with the Lebesgue outer measure µ generated by dyadic
cubes. You can use the fact, proved in class, that µ(R) = |R| for all cubes R.
#2 Use part (1) to obtain that the Lebesgue outer measure is translation invariant and
respects dilations, in the sense that, for all t 2 Rd , s > 0, A ⇢ Rd
µ(A + t) = µ(A),
µ(sA) = s d µ(A).
Hint. For #1 there are two inequalities. One is just the "/2 j enlargement trick. For the other one,
assume ⌫(A) is finite. You can use the Whitney Lemma 1.3.3 from the notes to reduce a cover of A
by open cubes to a disjoint dyadic cover with lesser total volume.
1.4. Prove Proposition 1.4.2.
1.5 Dynkin systems. A collection F of subsets of X is called a Dynkin system if
(1) (nontrivial) ; 2 F ,
(2) (closed under complement) F 2 F =) F c 2 F ,
1. OUTER MEASURES, MEASURES AND THE LEBESGUE MEASURE
13
(3) (closed under countable disjoint union) if {Fn : n 2 N} ⇢ F are pairwise disjoint
sets, namely n 6= k =) Fn \ Fk = ;, then
[
F :=
Fn 2 F .
n2N
Given G ⇢ P (X ), we denote by (G ) and by ⌃(G ) respectively the smallest Dynkin system
and the smallest ⌃-algebra containing G .
#1 Prove that (G ) is well defined and that G ⇢ (G ) ⇢ ⌃(G ).
#2 Let F be a Dynkin system with the further property that
F, G 2 F =) F \ G 2 F
(⇡)
(stable under finite intersection). Prove that F is a
#3 Prove that if G has property (⇡) so does (G ).
-algebra.
1.6 Uniqueness of Lebesgue measure. Let Q 0 = [0, 1)d . In this problem, we prove that
the Lebesgue measure on Rd is the unique complete measure µ such that µ(Q 0 ) = 1 which
is translation invariant. Its solution is divided into three parts.
For simplicity we work in d = 1. For a dyadic cube Q 2 D in R we indicate by D(Q) =
0
{Q 2 D : Q0 ⇢ Q} and by B(Q) the -algebra of Borel subsets of Q. Recall that B(Q) =
⌃(D(Q)) for instance.
#1 Let µ, ⌫ be two measures on (Q 0 , B(Q 0 )) with the property that µ(Q) = ⌫(Q) for
all Q 2 D(Q 0 ), and µ(Q 0 ) = 1. Prove that µ = ⌫.
#2 Let µ, ⌫ be two measures on (R, B) with the property that µ(Q) = ⌫(Q) < 1 for
all Q 2 D. Prove that µ = ⌫.
#3 Let ⌫ be a measure on (R, B) such that ⌫(Q 0 ) = 1 and
A 2 B, n 2 Z =) ⌫(A + 2n ) = ⌫(A)
Prove that the completion of ⌫ is the Lebesgue measure.
Hints. For 1. show that {A 2 B(Q 0 ) : µ(A) = ⌫(A)} is a Dynkin system and then a -algebra via
Exercise 1.5. For 2. use 1. and exhaustion. For part 3. verify the assumptions of part 2. by using
dyadic translation invariance.
1.7. #1 Let µ denote Lebesgue outer measure on Rd . Show that for all A ⇢ Rd there exists
a Lebesgue measurable set L with A ⇢ L and µ(A) = µ(L).
1.8. Prove Proposition 1.6.2.
1.9 B(Rd ) 6= L (Rd ), I. Define the n-th truncated Cantor set as
ñ n
ô
n
X
X
[
Cn =
an 3 n , 3 n +
an 3 n .
T
(a1 ,...,an )2{0,2}n
j=1
j=1
Then C = n 0 Cn is called the Cantor middle thirds set. We have seen in class that C is
compact and uncountable1.
#1 Show that C contains no intervals: thus C is called nowhere dense or meager.
1
Moreover, every point of C is an accumulation point for C, but you don’t need to write the proof of this
fact here.
14
1. OUTER MEASURES, MEASURES AND THE LEBESGUE MEASURE
#2 Show that C has zero Lebesgue measure. Sketch the construction a set C 12 still
satisfying #1 but whose Lebesgue measure is 12 .
#3 Use point 1. and a cardinality argument to prove that B(R) 6= L (R). You can use
without proof that B(R) has the same cardinality as R. See Folland, Prop 1.23, p.
43 for a proof.
1.10. Let µ be the Lebesgue outer measure on R. Show that
µ(F ) = sup µ(O)
O⇢F
O open
fails in general, even if F is Borel.
Hint. Modify the construction of the Cantor set to obtain a set of positive Lebesgue measure containing no intervals.
1.11 B(Rd ) 6= L (Rd ), II (d 2). #1 Let E ⇢ R be a non Lebesgue measurable set, Ẽ =
E ⇥ {0} ⇢ R2 . Explain why Ẽ 2 L (R2 )\B(R2 ). You can use without proof that
A 2 B(R2 ) () {x 2 R : (x, y) 2 A} 2 B(R)
8 y 2 R.