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Lebesgue Integration Theory: Part I

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Lebesgue Integration Theory: Part I Part V Lebesgue Integration Theory 17 Introduction: What are measures and why “measurable” sets Definition 17.1 (Preliminary). A measure µ “on” a set X is a function µ :2X [0, ] such that → ∞ 1. µ( )=0 ∅ N 2. If Ai is a finite (N< ) or countable (N = ) collection of subsets { }i=1 ∞ ∞ of X which are pair-wise disjoint (i.e. Ai Aj = if i = j) then ∩ ∅ 6 N N µ( Ai)= µ(Ai). ∪i=1 i=1 X Example 17.2. Suppose that X is any set and x X is a point. For A X, let ∈ ⊂ 1 if x A δ (A)= x 0 if x/∈ A. ½ ∈ Then µ = δx is a measure on X called the Dirac delta measure at x. Example 17.3. Suppose that µ is a measure on X and λ>0, then λ µ · is also a measure on X. Moreover, if µα α J are all measures on X, then { } ∈ µ = α J µα, i.e. ∈ P µ(A)= µα(A) for all A X ⊂ α J X∈ is a measure on X. (See Section 2 for the meaning of this sum.) To prove this we must show that µ is countably additive. Suppose that Ai i∞=1 is a collection of pair-wise disjoint subsets of X, then { } ∞ ∞ µ( ∞ Ai)= µ(Ai)= µα(Ai) ∪i=1 i=1 i=1 α J X X X∈ ∞ = µα(Ai)= µα( ∞ Ai) ∪i=1 α J i=1 α J X∈ X X∈ = µ( ∞ Ai) ∪i=1 246 17 Introduction: What are measures and why “measurable” sets wherein the third equality we used Theorem 4.22 and in the fourth we used that fact that µα is a measure. Example 17.4. Suppose that X is a set λ : X [0, ] is a function. Then → ∞ µ := λ(x)δx x X X∈ is a measure, explicitly µ(A)= λ(x) x A X∈ for all A X. ⊂ 17.1 The problem with Lebesgue “measure” So far all of the examples of measures given above are “counting” type mea- sures, i.e. a weighted count of the number of points in a set. We certainly are going to want other types of measures too. In particular, it will be of great interest to have a measure on R (called Lebesgue measure) which measures the “length” of a subset of R. Unfortunately as the next theorem shows, there is no such reasonable measure of length if we insist on measuring all subsets of R. Theorem 17.5. There is no measure µ :2R [0, ] such that → ∞ 1. µ([a, b)) = (b a) for all a<band − 2. is translation invariant, i.e. µ(A + x)=µ(A) for all x R and A 2R, where ∈ ∈ A + x := y + x : y A R. { ∈ } ⊂ In fact the theorem is still true even if (1) is replaced by the weaker con- dition that 0 <µ((0, 1]) < . ∞ The counting measure µ (A)=#(A) is translation invariant. However µ((0, 1]) = in this case and so µ does not satisfy condition 1. Proof.∞ First proof. Let us identify [0, 1) with the unit circle S1 := z { ∈ C : z =1 by the map | | } φ(t)=ei2πt =(cos2πt + i sin 2πt) S1 ∈ for t [0, 1). Using this identification we may use µ to define a function ν on 1 ∈ 2S by ν(φ(A)) = µ(A) for all A [0, 1). This new function is a measure on S1 with the property that 0 <ν((0⊂, 1]) < . For z S1 and N S1 let ∞ ∈ ⊂ zN := zn S1 : n N , (17.1) { ∈ ∈ } 17.1 The problem with Lebesgue “measure” 247 that is to say eiθN is N rotated counter clockwise by angle θ. We now claim that ν is invariant under these rotations, i.e. ν(zN)=ν(N) (17.2) for all z S1 and N S1. To verify this, write N = φ(A) and z = φ(t) for some t ∈[0, 1) and A ⊂ [0, 1). Then ∈ ⊂ φ(t)φ(A)=φ(t + A mod 1) where for A [0, 1) and α [0, 1), ⊂ ∈ t + A mod 1 := a + t mod 1 [0, 1) : a N { ∈ ∈ } =(a + A a<1 t ) ((t 1) + A a 1 t ) . ∩ { − } ∪ − ∩ { ≥ − } Thus ν(φ(t)φ(A)) = µ(t + A mod 1) = µ ((a + A a<1 t ) ((t 1) + A a 1 t )) ∩ { − } ∪ − ∩ { ≥ − } = µ ((a + A a<1 t )) + µ (((t 1) + A a 1 t )) ∩ { − } − ∩ { ≥ − } = µ (A a<1 t )+µ (A a 1 t ) ∩ { − } ∩ { ≥ − } = µ ((A a<1 t ) (A a 1 t )) ∩ { − } ∪ ∩ { ≥ − } = µ(A)=ν(φ(A)). Therefore it suffices to prove that no finite non-trivial measure ν on S1 such that Eq. (17.2) holds. To do this we will “construct” a non-measurable set N = φ(A) for some A [0, 1). Let ⊂ i2πt i2πt R := z = e : t Q = z = e : t [0, 1) Q { ∈ } { ∈ ∩ } — a countable subgroup of S1. As above R acts on S1 by rotations and divides S1 up into equivalence classes, where z,w S1 are equivalent if z = rw for some r R. Choose (using the axiom of choice)∈ one representative point n from each∈ of these equivalence classes and let N S1 be the set of these representative points. Then every point z S1 may⊂ be uniquely written as z = nr with n N and r R. That is to say∈ ∈ ∈ S1 = (rN) (17.3) r R a∈ where α Aα is used to denote the union of pair-wise disjoint sets Aα . By Eqs. (17.2) and (17.3), { } ` ν(S1)= ν(rN)= ν(N). r R r R X∈ X∈ 248 17 Introduction: What are measures and why “measurable” sets The right member from this equation is either 0 or , 0 if ν(N)=0and if ν(N) > 0. In either case it is not equal ν(S1) (0,∞1). Thus we have reached∞ the desired contradiction. ∈ Proof. Second proof of Theorem 17.5. For N [0, 1) and α [0, 1), let ⊂ ∈ N α = N + α mod 1 = a + α mod 1 [0, 1) : a N { ∈ ∈ } =(α + N a<1 α ) ((α 1) + N a 1 α ) . ∩ { − } ∪ − ∩ { ≥ − } Then µ (N α)=µ (α + N a<1 α )+µ ((α 1) + N a 1 α ) ∩ { − } − ∩ { ≥ − } = µ (N a<1 α )+µ (N a 1 α ) ∩ { − } ∩ { ≥ − } = µ (N a<1 α (N a 1 α )) ∩ { − } ∪ ∩ { ≥ − } = µ(N). (17.4) We will now construct a bad set N which coupled with Eq. (17.4) will lead to a contradiction. Set Qx := x + r R : r Q =x + Q. { ∈ ∈ } Notice that Qx Qy = implies that Qx = Qy. Let = Qx : x R —the ∩ 6 ∅ O { ∈ } orbit space of the Q action. For all A choose f(A) [0, 1/3) A1 and define N = f( ). Then observe: ∈ O ∈ ∩ O 1. f(A)=f(B) implies that A B = which implies that A = B so that f is injective. ∩ 6 ∅ 2. = Qn : n N . O { ∈ } Let R be the countable set, R := Q [0, 1). ∩ We now claim that N r N s = if r = s and (17.5) ∩ ∅ 6 r [0, 1) = r RN . (17.6) ∪ ∈ r s Indeed, if x N N = then x = r + n mod 1 and x = s + n0 mod 1, then ∈ ∩ 6 ∅ n n0 Q, i.e. Qn = Qn0 .Thatistosay,n = f(Qn)=f(Qn0 )=n0 and hence that− s ∈= r mod 1, but s, r [0, 1) implies that s = r. Furthermore, if x [0, 1) ∈ r mod 1 ∈ and n := f(Qx), then x n = r Q and x N . Now that we have constructed N, 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. A (A [0, 1/3]) = ∈F ∩ 6 ∅ Q 17.1 The problem with Lebesgue “measure” 249 1=µ([0, 1)) = µ(N r)= µ(N) r R r R X∈ X∈ if µ(N) > 0 = . ∞0 if µ(N)=0 ½ which is certainly inconsistent. Incidentally we have just produced an example of so called “non — measurable” set. Because of Theorem 17.5, it is necessary to modify Definition 17.1. Theo- 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. This leads us to the notion of σ — algebra discussed in the next chapter. Our revised notion of a measure will appear in Definition 19.1 of Chapter 19 below. 18 Measurability 18.1 Algebras and σ —Algebras Definition 18.1. Acollectionofsubsets of a set X is an algebra if A 1. ,X 2. A∅ ∈ Aimplies that Ac ∈ A ∈ A 3. is closed under finite unions, i.e. if A1,...,An then A1 An 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 X is a σ — algebra (or some- times called a σ — field) if is an algebraM which also closed under countable M unions, i.e. if Ai i∞=1 , then i∞=1Ai . (Notice that since is also closed under taking{ } complements,⊂ M ∪ is also∈ closedM under taking countableM in- tersections.) A pair (X, ), whereMX is a set and is a σ — algebra on X, is called a measurableM space. M 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. Analogously, elements of and algebra or a σ —algebra will be called measurable sets.
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