Outer Measures, Measures and the Lebesgue Measure

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 : (X ) [0, ] satisfies ; P ! 1 (1) (normalization) µ( )=0; ; (2) (monotonicity) F, F 0 P(X ), F F 0 = µ(F) µ(F 0); (3) (countable subadditivity)2 if Fn⇢is a countable)  collection of subsets of X then { } 1 1 µ Fn µ(Fn). ✓n=1 ◆  n=1 [ X 1.2 Outer measure from a premeasure. Let X be a set, E be a collection of subsets of X and : E [0, ] be any function. Define for F (X ) ! 1 2P 1 1 µ(F)=inf (En) : over E0 = En : n N E with F En ®n=0 { 2 } ⇢ ⇢ n=0 ´ X [ 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. n Let " > 0 be given. For each Fn, we may find a subcollection En = Ej : j N of E which { 2 } covers Fn and n j (Ej ) µ(Fn)+"2 . j N  X2 n Then E0 = Ej : j, n N covers F, so that { 2 } n µ(F) (Ej ) µ(Fn)+"  n N j N  n N X2 X2 X2 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.

d 1.3 Lebesgue outer measure via dyadic cubes. A cube I R is the cartesian product ⇢ of bounded intervals (open, closed or half-open unless otherwise speciﬁed) I j, j = 1, . . . , d. d We now consider the problem of computing the volume of an arbitrary subset of R . Our d measuring tools will be dyadic cubes. Let Q0 =[0, 1) be the unit cube. The collection of 3 4 1. OUTER MEASURES, MEASURES AND THE LEBESGUE MEASURE

dilates and translates of Q0

1 n n n n = n, n = [k12 , (k1 + 1)2 ) [kd 2 , (kd + 1)2 ), k1,...,kd Z D n= D D ⇥ ··· 2 1 [ d n is termed the standard dyadic grid on R . We denote by `(Q)=2 the sidelength of Q if (k) Q n. We denote by Q the unique dyadic cube (k-th ancestor of Q) with sidelength k 2 `2D(Q) that contains Q. We record the following property for further use d (1.1) x R , n Z, !Q n with x Q. 8 2 8 2 9 2D 2 A consequence of this is that forms a grid in the sense that whenever Q,Q0 , either D 2D Q Q0 have trivial intersection or one of the inclusions Q Q0,Q0 Q holds. In symbols \ ⇢ ⇢ (1.2) Q,Q0 = Q Q0 Q,Q0, . 2D d) \ 2 { ;} We consider the outer measure µ : (R ) [0, ] generated by the volume premeasure P d !nd 1 (Q)=`(Q) = 2 , Q n, n Z. 2D 2 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, half- open intervals. A preliminary but fundamental lemma is the following.

d 1.3.1 LEMMA. Let I = I1 Id be an axis-parallel cube in R with inf I j = aj, sup I j = bj. ⇥ ···⇥ Assume I is the disjoint union of dyadic cubes Q j : j N . Then n { 2 }

I = (bj aj)= (Q j). | | j=1 j N Y X2 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 N is a collection of pairwise disjoint dyadic intervals whose union is I, we have{ 2 }

sup I inf I = (Q j). j N X2 Let a = inf I. Write Q j =[uj, vj) for definiteness. Let " > 0 be given and set x j = uj j j "2 , yj = vj +"2 , R j =(x j, yj). 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, yj), j = 1, . . . , N with x j increasingly { } ordered and such that if R j Rk = , then neither R j Rk nor the opposite inclusion holds. \ 6 ; ⇢ It must be that x1 < a, yN > b ", and x j+1 < yj so that N 1 N b " a yN x1 = yN xN + x j+1 x j + yj x j (Q j)+2"  j=1  j=1  j N X X X2 which by taking " 0 yields Q Q . To prove the reverse inequality, by taking ( ) j N ( j) limits, it suffices to! show that  2 N P (Q j) (Q)=1 j=1  X 1. OUTER MEASURES, MEASURES AND THE LEBESGUE MEASURE 5

We can rearrange Q j : j = 1, . . . , N so that supQ j = bj minQ j+1 = aj+1. Then { } N 1 N  N 1 bN a1 = bN aN + aj+1 aj bj aj = (Q j) j=1 j=1 j=1 X X X as claimed. ⇤ d 1.3.2 PROPOSITION (CONSISTENCY). Let µ denote the Lebesgue outer measure on R above deﬁned. Then d µ(Q)=`(Q) for all dyadic cubes Q .

2D d PROOF. First of all, since Q is a covering of Q, we have µ(Q) `(Q) = (Q). Assume now that Q j is a cover of Q{by} dyadic cubes such that  { } 1 (1.3) (Q j) (Q). j=1  X 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 Q Q, k ` Q Q jk = j = = = jk j` = k j 6 ) \ ; [ [ 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 Q were nontrivial, one of the inclusions would hold, contradicting maximality. jk j` But then \ 1 1 Q Q Q ( j) ( jk )= ( ) j=1 k=1 using Lemma 1.3.1 for the secondX equality.X 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.

d 1.3.3 WHITNEY COVERING LEMMA. Let O R be an open set. Then there exists a subcollection I of dyadic cubes such that ⇢ (1) O = I. I I (2) the cubes[2 of I are pairwise disjoint (3) the distance of the center of I, cI 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 c (1.4) I O, `(I) 10 dist(cI , O ). ⇢  These are pairwise disjoint by maximality. Let now x O. Since O is open, the collection of 1 2 c those dyadic cubes containing x with `(I) 10 dist(cI , O ) 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 I. Now, (1) 1 1 2 c we must have that I the parent of I violates (1.4). In either case, `(I ) 10 dist(cI(1) , O ), so that c c dist(cI , O ) 2`(I)+dist(cI(1) , O ) 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. |·| d (1) if I R is any axis parallel cube then µ(I)= I , the Euclidean volume of I; d (2) for all⇢ A R , | | ⇢ µ(A)=inf I I I | | X2 d where I ranges over all countable collections of cubes of R whose union covers A; (3) the Lebesgue outer measure is translation invariant and respects dilations, in the sense d d that, for all t R , s > 0, A R 2 ⇢ d µ(A + t)=µ(A), µ(sA)=s 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.

d 1.3.5 LEMMA. Let F R be any set. Then ⇢ µ F inf µ O ( )= O F ( ) O open PROOF. It sufﬁces to assume that µ(F) < , otherwise there is nothing to prove, and show that for all " > 0 there exists an open1 set O with µ(O) µ(F)+2". To do so, let  E0 = Q j : j N be a cover of F by dyadic cubes such that { 2 } (1.5) µ(Q j) µ(F)+". j N  X2 j Let Q j be the open cube with the same center as Q j and volume µ(Q j)+"2 . Then O = jQ j is an open set containing F. Furthermore, in view of the preceding corollary and of the above[ display,f f 1 j µ(O) µ(Q j)= (µ(Q j)+"2 ) µ(F)+2"  j=1 j N  X X2 as claimed. f ⇤ 1.4 -algebras and measures. A subcollection (X ) such that (1) (nontrivial) , F⇢P c (2) (closed under;2F complement) F = F , (3f) (closed under ﬁnite union) F, G2F )= F2FG hold is said to be an algebra of subsets of2FX . If (1),) (2)[ and2F

(3) (closed under countable union) Fn : n N = F = Fn { 2 } ⇢F ) n N 2F [2 1. OUTER MEASURES, MEASURES AND THE LEBESGUE MEASURE 7 hold instead, is called a -algebra of subsets of X . We refer to a pair (X , ) as a measurable space. F F

1.4.1 REMARK. Firstly, it follows from the deﬁnition that if ↵ : ↵ A is any collection of {F 2 } -algebras on X , then = ↵ ↵ is also a -algebra. Given any collection of sets E (X ) we denote by (E) theF intersection\ F of all the -algebras containing E. Such an intersection⇢P is nonvoid, since (X ) is a -algebra containing E. It is useful to observe that, by deﬁnition P E0 (E)= (E0) (E). ⇢ ) ⇢ Let now (X , ) be a measurable space. A positive measure is a set function F µ : [0, ], F! 1 with the countable additivity property

(1.6) µ(F)= µ(Fn) n N X2 whenever Fn : n N are pairwise disjoint sets whose union is F. Then (X , , µ) is called measure{ space2 and} ⇢F the elements of are called measurable (or -measurable)F sets. F F 1.4.2 PROPOSITION. (basic properties of measure) Let (X , , µ) be a measure space and throughout Ej be measurable sets. There holds F (P1) µ( )=0; ; (P2) (monotonicity) if E1 E2 then µ(E1) µ(E2); (P3) µ is countably subadditive;⇢ 

(P4) If Ej Ej+1 for all j, then µ Ej = lim µ(Ej) (P5) If µ E⇢ < and E E for all j, then µ E lim µ E , and the assumption ( 1) j j+1 [ j = ( j) 1 µ(E1) < is necessary; 1 \ (P6) (lower semicontinuity) µ lim inf Ej lim inf µ(Ej);  (P7) (upper semicontinuity) lim sup µ(Ej) µ lim sup Ej provided µ Ej < .  1 1.4.3 REMARK. In fact, for properties (P1) and (P2), the finite additivity property S (1.7) F, G , F G = = µ(F)+µ(G)=µ(F G) 2F \ ; ) [ would suffice. PROOF OF PROPOSITION 1.4.2. Exercise. ⇤ The following lemma provides a partial converse to (P3) 1.4.4 LEMMA. Let be a -algebra and µ : [0, ] be a countably subadditive set function satisfying theF finite additivity property (1.7)F!. Then1µ is a measure on . F PROOF. We need to show that µ is countably additive. Let Fn : n N be pairwise disjoint sets whose union is F. We need to prove (1.6). Relying{ on countable2 } ⇢F 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

µ(Fn)=µ Fn µ (F) . n=1 ✓n=1 ◆  X [ 8 1. OUTER MEASURES, MEASURES AND THE LEBESGUE MEASURE and the in (1.6) follows by taking the limit as N in the above left hand side. ⇤ !1 1.5 Complete measures. Carathéodory’s theorem. For a measure space (X , , µ), define F := F : µ(F)=0 , 0 := N X : F with N F . N { 2F } N { ⇢ 9 2N ⇢ } Note that and 0 are both closed under countable unions. The measure space (X , , µ) N N F is said to be complete if 0 . N ⇢F 1.5.1 LEMMA. Let (X , , µ) be a measure space. Then F := E N : E , N 0 F { [ 2F 2N } is a -algebra and there exists a unique measure µ on such that (X , , µ) is a complete measure space, with µ(F)=µ(F) for all F . F F 2F PROOF. Since and 0 are both closed under countable unions, so is . We show closure under complement.F N Let A = E N , with E, F and N F. ReplacingF possibly c c c c c F by F E , N by N E , we can assume[ E2 FF = . This2F way (E N⇢) =(E F) (F N ). c c But (F \ N ) F and\ (E F) . This proves\ the; closure under[ complement.[ [ \ \ ⇢ [ 2F Now define µ(E N) := µ(E). This is well defined, since if E N = F M, E, E0 , [ [ [ 2F N, N 0 0 subsets respectively of F, F 0 one has E E0 F 0, whence 2N 2N ⇢ [ µ(E) µ(E0 F 0) µ(E0)+µ(F 0) µ(E0)  [   and by symmetry, equality must hold. Clearly this extends µ on . The verification that µ is a measure is easy and we omit it. To show that it is unique, let ⌫Fbe another measure with the same properties, E , N 0, F such that µ(F)=0 and N F. Then 2F 2N 2N ⇢ ⌫(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 , N 0, µ(E N)=0 and F E N. By a similar argument as above we can assume2FE N2N= . [ ⇢ [ \ ; Then µ(E)=0, and F E E, so that F E 0. We also have N F 0. So actually \ ⇢ \ 2N \ 2N F =(F E) (F N) 0, and in particular F . Therefore µ is complete. ⇤ \ [ \ 2N 2 F Now, let µ be an outer measure on X . We say that F is a µ-measurable set if c (1.8) µ(G)=µ(F G)+µ(F G) G (X ). \ \ 8 2P 1.5.2 REMARK. Suppose µ is generated by a premeasure (, E). Then (1.8) holds for F if and only if c (1.9) µ(E)=µ(F E)+µ(F E) E E \ \ 8 2 We leave this proof as an exercise.

1.5.3 CARATHÉODORY’S THEOREM. Let µ be an outer measure on X . Then the subcollection F of (X ) such that (1.8) holds is a -algebra • the restriction of µ 2Fto isP a complete measure. • F The proof is divided into steps. The ﬁrst, trivial one is to observe that contains and is closed under complement by virtue of the symmetry of (1.8). We beginF the actual work.; 1. OUTER MEASURES, MEASURES AND THE LEBESGUE MEASURE 9

STEP 1. is closed under finite union and µ is finitely additive on . Let A, B X be such that (1.8)F holds for F = A and F = B. Let G X be arbitrary. Then,F using (1.8)⇢ twice and monotonicity ⇢ c c c µ(G)=µ(G A B)+µ(G A B)+µ(G A B )+µ(G (A B) ) \ \ \ \ c \ \ \ [ µ(G A B)+µ(G (A B) ) \ [ \ [ which is the nontrivial half of (1.8) for A B. Therefore A B . Now, if A and B are pairwise disjoint and belong to , the fact[ that µ(A B)=µ[(A)+2Fµ(B) is a triviality. F [ STEP 2. We show that is closed under countable disjoint union, which, together with closure with respect to complementF and finite union, suffices by the usual argument. Let Fn : n N be pairwise disjoint, A be their union, and AN = F1 FN . By the previous{ 2 step,} ⇢FAN belongs to . We then have for an arbitrary G, [ ···[ F N c c c µ(G)=µ(G AN )+µ(G (AN ) ) µ(G AN )+µ(G A )= µ(G Fn)+µ(G A ) \ \ \ \ n=1 \ \ X and letting N and using subadditivity of outer measure !1 1 c c c µ(G) µ(G Fn)+µ(G A ) µ G Fn + µ(G A )=µ(G A)+µ(G A ), n=1 \ \ ✓n N \ ◆ \ \ \ X [2 which is the nontrivial part of (1.8) for A. So is a -algebra, and using the finite additivity and Lemma 1.4.4, we conclude µ restricted toF is a measure. F STEP 3. We show that µ is complete. Let F be a set with µ(F)=0 and N F. Then µ(N)=0. We have using subadditivity and2F monotonicity that ⇢ c µ(G) µ(G N)+µ(G N ) µ(N)+µ(G) µ(G)  \ \   1.6 The Lebesgue and Borel -algebra. We keep denoting the Lebesgue outer measure d d on R by µ. Let be the -algebra of subsets of R such that (1.8) holds. We call the Lebesgue -algebra.L We already know that µ restricted to is a complete translationL invariant, dilation-respecting measure. As we have seen, givenL any topological space X , we denote the -algebra generated by the collection of open subsets of X by (X ) and call it the Borel -algebra. By the Whitney covering lemma,O each open set can beB written as a d countable union of dyadic cubes. Therefore (R )=( ). Let us explore the relationship d between (R ), or simply , and . B D B B L 1.6.1 LEMMA. The following statements hold: (1) . (2) D⇢L. 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, ( ), therefore ( ). On the other hand, it is easy to write the unit half closedO⇢ cube asD the countableB⇢ intersectionD of open rectangles. Thus ( ) as well. Hence, the first statement also implies .To disprove equality,B there areD two possibilities which are explored in later exercises.B⇢L 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 it sufﬁces to show that

2L c µ(Q0) µ(Q0 Q)+µ(Q0 Q ) Q0 . \ \ 8 2D 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 { } ( (R j) µ(Q0)+". j  X Clearly, there holds `(R j) > `(Q) for ﬁnitely many j. By breaking each such R j with into subcubes of sidelength `(Q) we can construct another countable disjoint cover of Q0, say Sj , with j (R j)= j (Sj) and such that either Sj Q = or Sj Q. Then, Sj : Sj Q { } c \ ; ⇢ { ⇢ } covers Q and Sj : Sj Q = covers Q0 Q whence P { \P ;} \ c µ(Q)+µ(Q0 Q ) (Sj)+ (Sj)= (R j) µ(Q0)+", \  Sj Q sj Q= j  X⇢ X\ ; X and we are done. ⇤

1.6.2 PROPOSITION. (continuities) Denote by µ the Lebesgue (outer) measure. Then (1) if A then µ(A)=inf µ(O) : A O, O open (2) if A 2L then µ(A)=sup{ µ(K) : K⇢ A, K compact} (3) A 2Lif and only if for all{ " > 0 there⇢ exists an open} set O such that 2L A O, µ(O A) < "; ⇢ \ (4) A if and only if for all " > 0 there exists an open set O and a closed set C such that2L C A O, µ(O C) < ", ⇢ ⇢ \ and C can be chosen to be compact if µ(A) < ; d 1 (5) if A then there exist a set G (R ) (countable intersection of open sets) and 2L d 2G a set F (R ) (countable union of closed sets) such that 2F 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. ⇤ d 1.6.3 COROLLARY. The measure space (R , , µ) is the completion of the measure space d (R , , µ) L B d PROOF. The measure space (R , , µ) is complete, it thus suffices to show that it is con- d tained in the abstract completion ofL(R , , µ). Let A . We can use (4) of Proposition 1.6.2 to find a Borel set F with M := A FBof null Lebesgue2L 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 [ . which finishes the proof. ⇤ B 1. OUTER MEASURES, MEASURES AND THE LEBESGUE MEASURE 11

1.7 Non-Lebesgue measurable sets. We present hereby an extended version of the classi- cal 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 2L 1.7.1 PROPOSITION. Let A be a Lebesgue measurable subset such that B (X ) : B A . Then µ(A)=0. { 2P ⇢ } ⇢ L PROOF. The equivalence relation x y x y Q ⇠ () 2 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 = n), En = R. Define A A E . Let K be any\ compact; subset6 of A . By assumption A , K are measurable. n = n n[ n \ 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 q Q: q <1 2 [| | is bounded, therefore µ(H) < . However, by the ﬁrst of the above, the sets K +q are pairwise disjoint and measurable (being1 translates of a measurable set), so that using translation invariance, >µ(H)= µ(K + q)= µ(K) 1 q Q: q <1 q Q: q <1 2X| | 2X| | and the claim follows. ⇤ 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 ⇢ c µ(G)=µ(F G)+µ(F G) G (X ) \ \ 8 2P if and only if c µ(E)=µ(F E)+µ(F E) E E. \ \ 8 2 1.2. Let be the dyadic grid on R and µ denote the Lebesgue outer measure on R, namely for A RD ⇢ 1 1 µ(A)=inf `(I j) : A I j, I j j . ® j=1 ⇢ j=1 2D8 ´ X [ n #1 Let n Z. Prove that µ does not change if we restrict to intervals with `(I j) 2 , namely2 

(n) (n) 1 1 n µ(A)=µ (A), µ (A) := inf `(I j) : A I j, I j j, `(I j) 2 . ® j=1 ⇢ j=1 2D8  ´ X [ #2 Let t be of the form t N k 2 n for suitable integers N, k ,...,k . R = n= N n N N Prove2 that µ is invariant under translations by t, namely P µ(A)=µ(A + t) A R. n 8 ⇢ Hints. For #2, reduce to the case t = 2 for some n. Then use #1 and that m = I : `(I)= m n D { 2D 2 is invariant under translation by 2 whenever m n. } d d 1.3. Let be the collection of all open cubes I R and deﬁne the outer measure on R given by O ⇢

1 1 ⌫(A)=inf Rn : A R j, R j j ®n=0 | | ⇢ n=0 2O8 ´ where R is the Euclidean volume ofX 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 d d respects dilations, in the sense that, for all t R , s > 0, A R 2 d ⇢ µ(A + t)=µ(A), µ(sA)=s µ(A). Hint. For #1 there are two inequalities. One is just the "/2j 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 of subsets of X is called a Dynkin system if (1) (nontrivial) , F c (2) (closed under;2F complement) F = F , 2F ) 2F 1. OUTER MEASURES, MEASURES AND THE LEBESGUE MEASURE 13

(3) (closed under countable disjoint union) if Fn : n N are pairwise disjoint { 2 } ⇢F sets, namely n = k = Fn Fk = , then 6 ) \ ; F := Fn . n N 2F [2 Given (X ), we denote by ( ) and by ⌃( ) respectively the smallest Dynkin system and theG⇢P smallest ⌃-algebra containingG . G G #1 Prove that ( ) is well deﬁned and that ( ) ⌃( ). #2 Let be a DynkinG system with the furtherG⇢ propertyG ⇢ thatG F (⇡) F, G = F G 2F ) \ 2F (stable under ﬁnite intersection). Prove that is a -algebra. #3 Prove that if has property (⇡) so does ( F). G G d 1.6 Uniqueness of Lebesgue measure. Let Q0 =[0, 1) . In this problem, we prove that d the Lebesgue measure on R is the unique complete measure µ such that µ(Q0)=1 which is translation invariant. Its solution is divided into three parts. For simplicity we work in d = 1. For a dyadic cube Q in R we indicate by (Q)= 2D D Q0 : Q0 Q and by (Q) the -algebra of Borel subsets of Q. Recall that (Q)= ⌃{ ( 2D(Q)) for instance.⇢ } B B D #1 Let µ, ⌫ be two measures on (Q0, (Q0)) with the property that µ(Q)=⌫(Q) for B all Q (Q0), and µ(Q0)=1. Prove that µ = ⌫. #2 Let µ,2D⌫ be two measures on (R, ) with the property that µ(Q)=⌫(Q) < for all Q . Prove that µ = ⌫. B 1 2D #3 Let ⌫ be a measure on (R, ) such that ⌫(Q0)=1 and B n A , n Z = ⌫(A + 2 )=⌫(A) 2B 2 ) Prove that the completion of ⌫ is the Lebesgue measure.

Hints. For 1. show that A (Q0) : µ(A)=⌫(A) is a Dynkin system and then a -algebra via Exercise 1.5. For 2. use 1.{ and2B exhaustion. For part} 3. verify the assumptions of part 2. by using dyadic translation invariance.

d d 1.7. #1 Let µ denote Lebesgue outer measure on R . Show that for all A R there exists a Lebesgue measurable set L with A L and µ(A)=µ(L). ⇢ ⇢ 1.8. Prove Proposition 1.6.2.

d d 1.9 (R ) = (R ), I. Deﬁne the n-th truncated Cantor set as B 6 L n n n n n Cn = an3 ,3 + an3 . n (a1,...,an) 0,2 ñ j=1 j=1 ô [2{ } X X Then C = n 0 Cn is called the Cantor middle thirds set. We have seen in class that C is compact and uncountable 1. T #1 Show that C contains no intervals: thus C is called nowhere dense or meager.

1Moreover, 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 1 still 2 1 satisfying #1 but whose Lebesgue measure is 2 . #3 Use point 1. and a cardinality argument to prove that (R) = (R). You can use without proof that (R) has the same cardinality as RB. See Folland,6 L Prop 1.23, p. 43 for a proof. B

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.

d d 1.11 (R ) = (R ), II (d 2). #1 Let E R be a non Lebesgue measurable set, E˜ = 2 2 2 E 0B R .6 ExplainL why E˜ (R ) (R )⇢. You can use without proof that ⇥ { } ⇢ 2 2L \B A (R ) x R : (x, y) A (R) y R. 2B () { 2 2 } 2B 8 2