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Lebesgue Integration Theory: Part I

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Lebesgue Integration Theory: Part I Part V 17 Introduction: What are measures and why “measurable” sets Lebesgue Integration Theory Definition 17.1 (Preliminary). A measure “on” a set [ is a function :2[ $ [0> 4] such that 1. (>)=0 Q 2. If Dl l=1 is a finite (Q?4) or countable (Q = 4) collection of subsets of {[ which} are pair-wise disjoint (i.e. D _ D = > if l = m) then l m 6 Q Q (^l=1Dl)= (Dl)= l=1 X Example 17.2. Suppose that [ is any set and { 5 [ is a point. For D [> let 1 if { 5 D (D)= { 0 if {@5 D= ½ Then = { is a measure on [ called the Dirac delta measure at {= Example 17.3. Suppose that is a measure on [ and A0> then is also a measure on [= Moreover, if are all measures on [> then· { }5M = 5M > i.e. P (D)= (D) for all D [ X5M is a measure on [= (See Section 2 for the meaning of this sum.) To prove 4 this we must show that is countably additive. Suppose that Dl l=1 is a collection of pair-wise disjoint subsets of [> then { } 4 4 4 (^l=1Dl)= (Dl)= (Dl) l=1 l=1 5M X 4 X X 4 = (Dl)= (^l=1Dl) 5M l=1 5M X 4X X = (^l=1Dl) 246 17 Introduction: What are measures and why “measurable” sets 17.1 The problem with Lebesgue “measure” 247 wherein the third equality we used Theorem 4.22 and in the fourth we used that is to say hlQ is Q rotated counter clockwise by angle = We now claim that fact that is a measure. that is invariant under these rotations, i.e. Example 17.4. Suppose that [ is a set : [ $ [0> 4] is a function. Then (}Q)=(Q) (17.2) 1 1 := ({){ for all } 5 V and Q V = To verify this, write Q = !(D) and } = !(w) for some w 5 [0> 1) and D [0> 1)= Then {X5[ is a measure, explicitly !(w)!(D)=!(w + D mod 1) (D)= ({) where for D [0> 1) and 5 [0> 1)> {X5D for all D [= w + D mod 1 := d + w mod 1 5 [0> 1) : d 5 Q { } =(d + D _ d?1 w ) ^ ((w 1) + D _ d 1 w ) = { } { } 17.1 The problem with Lebesgue “measure” Thus So far all of the examples of measures given above are “counting” type mea- (!(w)!(D)) = (w + D mod 1) sures, i.e. a weighted count of the number of points in a set. We certainly are = ((d + D _ d?1 w ) ^ ((w 1) + D _ d 1 w )) going to want other types of measures too. In particular, it will be of great { } { } = ((d + D _ d?1 w )) + (((w 1) + D _ d 1 w )) interest to have a measure on R (called Lebesgue measure) which measures { } { } the “length” of a subset of = Unfortunately as the next theorem shows, there = (D _ d?1 w )+ (D _ d 1 w ) R { } { } is no such reasonable measure of length if we insist on measuring all subsets = ((D _ d?1 w ) ^ (D _ d 1 w )) of { } { } R= = (D)=(!(D))= Theorem 17.5. There is no measure :2R$[0> 4] such that Therefore it su!ces to prove that no finite non-trivial measure on V1 such 1. ([d> e)) = (e d) for all d?eand that Eq. (17.2) holds. To do this we will “construct” a non-measurable set 2. is translation invariant, i.e. (D + {)=(D) for all { 5 R and D 5 2R> Q = !(D) for some D [0> 1)= Let where U := } = hl2w : w 5 = } = hl2w : w 5 [0> 1) _ D + { := | + { : | 5 D R= Q Q { } { } { } In fact the theorem is still true even if (1) is replaced by the weaker con- — a countable subgroup of V1= As above U acts on V1 by rotations and divides dition that 0 ?((0> 1]) ? 4= V1 up into equivalence classes, where }>z 5 V1 are equivalent if } = uz for some u 5 U= Choose (using the axiom of choice) one representative point q The counting measure (D)=#(D) is translation invariant. However from each of these equivalence classes and let Q V1 be the set of these ((0> 1]) = 4 in this case and so does not satisfy condition 1. representative points. Then every point } 5 V1 may be uniquely written as Proof. First proof. Let us identify [0> 1) with the unit circle V1 := } 5 { } = qu with q 5 Q and u 5 U= That is to say C : } =1 by the map | | } V1 = (uQ) (17.3) !(w)=hl2w =(cos2w + l sin 2w) 5 V1 ua5U for w 5 [0> 1)= Using this identification we may use to define a function on where D is used to denote the union of pair-wise disjoint sets D = By 1 2V by (!(D)) = (D) for all D [0> 1)= This new function is a measure on Eqs. (17.2) and (17.3), { } V1 with the property that 0 ?((0> 1]) ? 4= For } 5 V1 and Q V1 let ` (V1)= (uQ)= (Q)= }Q := }q 5 V1 : q 5 Q > (17.1) { } uX5U uX5U 248 17 Introduction: What are measures and why “measurable” sets 17.1 The problem with Lebesgue “measure” 249 The right member from this equation is either 0 or 4, 0 if (Q)=0and 4 if 1=([0> 1)) = (Q u)= (Q) 1 (Q) A 0= In either case it is not equal (V ) 5 (0> 1)= Thus we have reached u5U u5U X X the desired contradiction. 4 if (Q) A 0 = = Proof. Second proof of Theorem 17.5. For Q [0> 1) and 5 [0> 1)> 0 if (Q)=0 let ½ which is certainly inconsistent. Incidentally we have just produced an example Q = Q + mod 1 of so called “non — measurable” set. = d + mod 1 5 [0> 1) : d 5 Q Because of Theorem 17.5, it is necessary to modify Definition 17.1. Theo- { } =( + Q _ d?1 ) ^ (( 1) + Q _ d 1 ) = rem 17.5 points out that we will have to give up the idea of trying to measure { } { } all subsets of R but only measure some sub-collections of “measurable” sets. Then This leads us to the notion of — algebra discussed in the next chapter. Our revised notion of a measure will appear in Definition 19.1 of Chapter 19 below. (Q )= ( + Q _ d?1 )+ (( 1) + Q _ d 1 ) { } { } = (Q _ d?1 )+ (Q _ d 1 ) { } { } = (Q _ d?1 ^ (Q _ d 1 )) { } { } = (Q)= (17.4) We will now construct a bad set Q which coupled with Eq. (17.4) will lead to a contradiction. Set T{ := { + u 5 R : u5 Q ={ + Q= { } Notice that T{ _ T| = > implies that T{ = T|= Let = T{ : { 5 R —the 6 O { } orbit space of the Q action. For all D 5 choose i(D) 5 [0> 1@3) _ D1 and define Q = i( )= Then observe: O O 1. i(D)=i(E) implies that D _ E = > which implies that D = E so that i is injective. 6 2. = T : q 5 Q = O { q } Let U be the countable set, U := Q _ [0> 1)= We now claim that u v Q _ Q = > if u = v and (17.5) 6 u [0> 1) = ^u5UQ = (17.6) u v Indeed, if { 5 Q _ Q = > then { = u + q mod 1 and { = v + q0 mod 1> then 6 qq0 5 Q> i.e. Tq = Tq0 .Thatistosay,q = i(Tq)=i(Tq0 )=q0 and hence that v = u mod 1> but v> u 5 [0> 1) implies that v = u= Furthermore, if { 5 [0> 1) u mod 1 and q := i(T{)> then { q = u 5 Q and { 5 Q = Now that we have constructed Q> we are ready for the contradiction. By Equations (17.4—17.6) we find 1 We have used the Axiom of choice here, i.e. DM (D K [0> 1@3]) = F 6 Q 252 18 Measurability Proposition 18.4. Let be any collection of subsets of [= Then there exists a unique smallest algebraE ( ) and — algebra ( ) which contains = 18 A E E E Proof. The proof is the same as the analogous Proposition 10.6 for topolo- Measurability gies, i.e. ( ):= : is an algebra such that A E {A A E A} \ and ( ):= : is a — algebra such that = E {M M E M} \ Example 18.5. Suppose [ = 1> 2> 3 and = >>[> 1> 2 > 1> 3 > see Figure 18.1. { } E { { } { }} 18.1 Algebras and —Algebras Definition 18.1. Acollectionofsubsets of a set [ is an algebra if A 1. >>[ 5 2. D 5 Aimplies that Df 5 A A 3. is closed under finite unions, i.e. if D1>===>Dq 5 then D1 ^ ^Dq 5 A= A ··· InA view of conditions 1. and 2., 3. is equivalent to 30. is closed under finite intersections. A Definition 18.2. Acollectionofsubsets of [ is a — algebra (or some- times called a — field) if is an algebraM which also closed under countable unions, i.e. if 4 Mthen 4 (Notice that since is also Dl l=1 > ^l=1Dl 5 = Fig. 18.1. A collection of subsets. closed under taking{ } complements,M is also closedM under taking countableM in- tersections.) A pair ([> )> whereM[ is a set and is a — algebra on [> is called a measurableM space. M Then The reader should compare these definitions with that of a topology in Definition 10.1. Recall that the elements of a topology are called open sets. ( )= >>[> 1 > 1> 2 > 1> 3 Analogously, elements of and algebra or a —algebra will be called E { { } { } { }} A M ( )=( )=2[ = measurable sets.
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