Carathéodory s extension of a measure on a semi-ring

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1 Carathéodory s extension of a measure on a semi-ring Reinhardt Messerschmidt 7 October Introduction This article presents Carathéodory s extension of a measure on a semi-ring, and the construction of the Lebesgue-Stieltjes σ-algebra and measure with respect to a real-valued function on R d satisfying certain nonnegativity and continuity conditions. It follows a combination of the approaches in [B86], [C80], [F99] and [S09]. 2 Some properties of sets of subsets Suppose X is an arbitrary set. Let PX be the power set of X, i.e. the set of all subsets of X. If A is a subset of X, let A c be the complement of A with respect to X, i.e. A c = X A. A subset S of PX is i closed under finite unions if A 1 A 2 S whenever A 1, A 2 S, ii closed under finite intersections if A 1 A 2 S whenever A 1, A 2 S, iii closed under differences if A 1 A 2 S whenever A 1, A 2 S, iv partially closed under differences if for every A 1, A 2 S there exist disjoint sets B 1, B 2,..., B n in S such that A 1 A 2 = n B j, v closed under complements if A c S whenever A S, vi closed under countable unions if S whenever is a sequence in S, vii closed under disjoint countable unions if S whenever is a sequence of disjoint sets in S. Proposition 2.1. If S is a subset of PX that is closed under complements, closed under finite unions, and closed under disjoint countable unions, then it is closed under countable unions. 1

2 Proof. Suppose is a sequence in S. Let B 1 = A 1, and for every j 2 let B j = j 1 A k, then B j = j 1 j 1 A k = A k c = j 1 c A c j A k, B j is a sequence in S. The sets B 1, B 2,... are also disjoint, B j S, but = B j, S. 3 Proto-rings, semi-rings and σ-algebras A subset S of PX is i a proto-ring if S, ii a semi-ring if it is a proto-ring, closed under finite intersections, and partially closed under differences, iii a σ-algebra if it is a proto-ring, closed under complements, and closed under countable unions. Proposition 3.1. If S is a semi-ring and A, B 1, B 2,..., B n S, then there exist disjoint sets C 1, C 2,..., C m S such that A n B j = m C j. Proof. We will use induction on n. The base case follows from the partial closure of S under differences. For the inductive case, note that n+1 A B j = A n B j B n+1. By the inductive hypothesis, there exist disjoint sets C 1, C 2,..., C m S such that A n B j = m C j, n+1 m m A B i = C j B n+1 = C j B n+1. For every j, there exist disjoint sets D j1, D j2,..., D jmj S such that It follows that C j B n+1 = m j D jk j =1,2,...,m,2,...,m j D jk. is a finite sequence of disjoint sets in S whose union is A n+1 B j. Proposition 3.2. If S is a σ-algebra then it is closed under finite unions, closed under finite intersections, closed under differences, and a semi-ring. 2

3 Proof. Closed under finite unions. Suppose A 1, A 2 S. Let B 1 = A 1, and for every j 2 let B j = A 2, then A 1 A 2 = B j and B j S, A 1 A 2 S. Closed under finite intersections. This follows from closure under finite unions and the identity A 1 A 2 = A c 1 A c 2 c. Closed under differences. This follows from closure under finite intersections and the identity A 1 A 2 = A 1 A c 2. Semi-ring. Closure under differences implies partial closure under differences. 4 Some properties of functions on sets of subsets If S is a subset of PX, then a function µ from S into [0, ] is i monotone if µa 1 µa 2 whenever A 1, A 2 S and A 1 A 2, ii finitely additive if n µ = µ whenever A 1, A 2,..., A n are disjoint sets in S such that n S, iii subtractive if µa 1 A 2 = µa 1 µa 2 whenever A 1, A 2 S are such that A 1 A 2 S and µa 2 <, iv countably additive if µ = µ whenever is a sequence of disjoint sets in S such that S, v countably subadditive if µ A i µ whenever is a sequence in S such that S, vi σ-finite if there exists a sequence in S such that µ < for every j and = X. Proposition 4.1. If S is a subset of PX and µ is a finitely additive function on S, then µ is subtractive. Proof. Suppose A 1, A 2 S are such that A 1 A 2 S and µa 2 <. We have µa 1 = µa 2 + µa 1 A 2, and since µa 2 < it can be subtracted from both sides. Proposition 4.2. Suppose S is a semi-ring and µ is a finitely additive function on S. 3

4 a If A S and B 1, B 2,..., B n are disjoint sets in S such that n B j A, then µb j µa. b If A S and B 1, B 2,..., B n are sets in S such that A n B j, then c µ is monotone. µa µb j. Proof. a There exist disjoint sets C 1, C 2,..., C m S such that A n B j = m C j. It follows that B 1, B 2..., B n, C 1, C 2..., C m are disjoint sets in S whose union is A, b Note that µa = m µb j + µc j A = A n B j = µb j. n A B j and A B j S for every j. Let C 1 = A B 1, and for every j 2 let j 1 C j = A B j A B k. For every j, there exist disjoint sets D j1, D j2,..., D jmj in S such that C j = mj D jk. It follows that D jk j =1,2,...,m,2,...,m j is a finite sequence of disjoint sets in S whose union is A, By part a, c Let n = 1 in part a. µa = m j m j µd jk. µd jk µb j, µa m µb j. Proposition 4.3. If S is a semi-ring and µ is a finitely additive and countably subadditive function on S, then µ is countably additive. 4

5 Proof. Suppose is a sequence of disjoint sets in S such that S. By proposition 4.2a, µ µ for every n, µ µ. The opposite inequality holds by countable subadditivity. Proposition 4.4. If S is a semi-ring and µ is a σ-finite and monotone function on S, then there exists a sequence C j of disjoint sets in S such that µc j < for every j and C j = X. Proof. There exists a sequence of sets in S, not necessarily disjoint, such that µ < for every j and = X. Let B 1 = A 1, and for every j 2 let B j = For every j, there exist disjoint sets C j1, C j2,..., C jmj S such that B j = mj C jk. It follows that C jk j =1,2,...,2,...,m j is a sequence of disjoint sets in S whose union is X, and µc jk µ < for every j, k. j 1 A k. 5 Proto-measures and measures If S is a proto-ring, then a function µ from S into [0, ] is i a proto-measure if µ = 0, ii a measure if it is a proto-measure and countably additive. Proposition 5.1. If S is a proto-ring and µ is a measure on S, then µ is finitely additive. Proof. Suppose A 1, A 2..., A n are disjoint sets in S such that n S. For every j = 1, 2,..., n let B j =, and for every j > n let B j =. It follows that B j is a sequence of disjoint sets in S and n µ B j = = µ n S, B j = µb j = µ. Proposition 5.2. If S is a semi-ring and µ is a measure on S, then µ is countably subadditive. 5

6 Proof. Suppose is a sequence in S such that S. Let B 1 = A 1, and for every j 2 let B j = For every j, there exist disjoint sets C j1, C j2,..., C jmj S such that B j = mj C jk. It follows that C jk j =1,2,...,2,...,m j j 1 A k. is a sequence of disjoint sets in S whose union is, µ = m j µc jk. By proposition 4.2a, In summary, m j µc jk µ, µ µ. a if S is a proto-ring and µ is a measure on S, then µ is finitely additive and subtractive propositions 5.1 and 4.1, b if S is a semi-ring and µ is a measure on S, then, in addition to being finitely additive and subtractive as in a, the measure µ is also monotone and countably subadditive propositions 4.2c and 5.2, c if S is a semi-ring and µ is a finitely additive and countably subadditive proto-measure on S, then µ is a measure proposition 4.3, d every σ-algebra is a semi-ring proposition 3.2, b and c also hold if S is a σ-algebra. 6 Outer measures A function µ from PX into [0, ] is an outer measure if it is a proto-measure, monotone, and countably subadditive. Suppose µ is an outer measure and A is a subset of X. The set A is µ- measurable if µb = µb A + µb A c for every subset B of X. Since µ is a proto-measure and countably subadditive, it always holds that µb µb A + µb A c. 6

7 The opposite inequality clearly holds if µb =. It follows that A is µ- measurable if and only if µb µb A + µb A c for every B such that µb <. Let M µ be the set of all µ-measurable subsets of X. We will show that M µ is a σ-algebra and µ Mµ the restriction of µ to M µ is a measure on M µ. Proposition 6.1. If µ is an outer measure, B is a subset of X, and A 1, A 2,..., A n are disjoint sets in M µ, then µb = µb + µ B Proof. We will use induction on n. The base case follows from the µ-measurability of A 1. Consider the inductive case. Since A n+1 is µ-measurable, µ B n A c j = µ B n Since A 1, A 2,..., A n, A n+1 are disjoint, µ B B n A c j n A c j A c j A n+1 + µ B n A c j A n+1 = B A n+1, It follows from the inductive hypothesis that µb = µb + µ B. n A c j A c n+1. n+1 = µb A n+1 + µ B n A c j A c j. n+1 n+1 = µb + µ B Theorem 6.2. If µ is an outer measure, then M µ is a σ-algebra. Proof. By proposition 2.1, it is sufficient to show that M µ is a proto-ring, closed under complements, closed under finite unions, and closed under disjoint countable unions. Proto-ring. For every subset B of X, µb + µb c = µ + µb = µb, is µ-measurable. Closed under complements. Suppose A is µ-measurable. For every subset B of X, µb A c + µb A c c = µb A c + µb A = µb. A c j. 7

8 Closed under finite unions. Suppose A 1, A 2 are µ-measurable. For every subset B of X, µ B A 1 A 2 + µ B A 1 A 2 c = µ B A 1 A 2 + µb A c 1 A c 2 = µ B A 1 A 2 A 1 + µ B A 1 A 2 A c 1 + µb A c 1 A c 2 = µb A 1 + µb A c 1 A 2 + µb A c 1 A c 2 = µb A 1 + µb A c 1 = µb. Closed under disjoint countable unions. Suppose is a sequence of disjoint sets in M µ, and B is a subset of X such that µb <. For every n, by proposition 6.1 and monotonicity, n µb = µb + µ B µb µb + µ B µb + µ B By countable subadditivity, µb µ B = µ B A c j A c j, c., µb µ B + µ B c. Theorem 6.3. If µ is an outer measure, then µ Mµ is a measure on M µ. Proof. Suppose is a sequence of disjoint sets in M µ. Since M µ is a σ- algebra, we have M µ. For every n, by proposition 6.1, n µ = µ A k + µ µa k, µ µ. The opposite inequality holds by countable subadditivity. A c k 8

9 7 Constructing an outer measure If µ is a function from a subset S of PX into [0, ], let µ be the function from PX into [0, ] defined by { µ A = inf µ : is a sequence in S such that A with the understanding that inf =. }, Theorem 7.1. If S is a proto-ring and µ is a proto-measure on S, then µ is an outer measure and µ µ on S. Proof. Proto-measure. Let be the sequence in S defined by = for every j, then, Monotone. If A B then 0 µ µa i = 0. { } µ : is a sequence in S such that B i=1 { µ : is a sequence in S such that A }, µ A µ B. Countably subadditive. Suppose is a sequence in PX. Countable subadditivity clearly holds if µ =, so suppose µ < and ɛ > 0. For every j, there exists a sequence k,2,... in S such that k and µk < µ + ɛ/2 j. Choose a bijection f from N onto N N and let B j be the sequence defined by B j = A fj, then B j is a sequence in S such that B j, µ µb j. By proposition 11.1, µb j = µk, µ µk µ + ɛ/2 j = µ + ɛ. 9

10 Inequality. Suppose A S. Let be the sequence defined by A 1 = A and = for every j 2. We then have that is a sequence in S such that A, µ A µ = µa. 8 Extending a measure on a semi-ring We have shown that if S is a proto-ring and µ is a proto-measure on S, then a µ is an outer measure theorem 7.1, b M µ is a σ-algebra theorem 6.2, c µ Mµ is a measure on M µ theorem 6.3, d µ µ on S theorem 7.1. In this section, we will show that if i S is a semi-ring, ii µ is a finitely additive, countably subadditive and monotone proto-measure on S, then, in addition to a-d, the measure µ Mµ is an extension of µ. In other words, S M µ and µ = µ on S. Note that S and µ satisfy i-ii if and only if S is a semi-ring and µ is a measure on S. Theorem 8.1. If S is a semi-ring and µ is a finitely additive proto-measure on S, then S M µ. Proof. Suppose that A S, that B is a subset of X such that µ B <, and that ɛ > 0. There exists a sequence B j in S such that B B j and We have µb j < µ B + ɛ. B A B j A = B j A and B j A S for every j, µ B A µb j A. We also have B A c B j A c = B j A. 10

11 For every j, there exist disjoint sets C j1, C j2,..., C jmj in S such that It follows that B j A = m j C jk. C jk j =1,2,...,2,...,m j is a sequence of sets in S whose union contains B A c, Combining these inequalities, µ B A c m j µc jk. m j µ B A + µ B A c µb j A + µc jk = m j µb j A + µc jk. For every j, we have that B j A, C j1, C j2,..., C jmj union is B j, are disjoint sets in S whose m j µb j A + µc jk = µb j, µ B A + µ B A c µb j < µ B + ɛ. Theorem 8.2. If S is a semi-ring and µ is a countably subadditive and monotone function on S, then µ µ on S. Proof. Suppose A S and is a sequence in S such that A. We have A = A = A and A S for every j, µa = µ A µa µ, µa µ A. 11

12 9 Uniqueness of the extension We will show that the extension of the previous section is unique if the measure is σ-finite. Theorem 9.1. Suppose S is a semi-ring and µ is a σ-finite measure on S. If ν is a measure on M µ such that ν = µ on S, then ν = µ on M µ. Proof. Suppose A M µ. We will show that νa = µ A. Througout this proof, remember that M µ is a σ-algebra, µ Mµ is a measure, S M µ and µ = µ = ν on S. Step 1. For every sequence in S such that A, we have νa ν ν = µ, νa µ A. Step 2. For the opposite inequality, first suppose that µ A < and ɛ > 0. There exists a sequence in S such that A and By monotonicity, µ < µ A + ɛ. Let B 1 = A 1, and for every j 2 let µ A µ. B j = For every j, there exist disjoint sets C j1, C j2,..., C jmj in S such that B j = mj C jk. It follows that C jk j =1,2,...,2,...,m j j 1 A k. is a sequence of disjoint sets in S whose union is, m j m j µ = µ C jk = νc jk = ν, µ A ν. We have shown in step 1 that ν µ on M µ, ν = νa + ν A νa + µ A, 12

13 µ A νa + µ A. Since µ A <, µ A = µ µ A µ µ A < ɛ, µ A < νa + ɛ. Step 3. Now suppose µ A =. There exists a sequence C j of disjoint sets in S such that µc j < for every j and C j = X. For every j, µ A C j µ C j = µc j <, µ A C j = νa C j by step 2. It follows that µ A = µ A C j = µ A C j = νa C j = νa. 10 Lebesgue-Stieltjes measures We will use the results of the previous sections to construct a σ-algebra and a measure from a real-valued function on R d satisfying certain nonnegativity and continuity conditions. A subset A of R d is a rectangle if A = or A = d a k, b k ] with a k < b k for every k. Let S d be the set of rectangles in R d. If a, b R, let a b and a b be the minimum and maximum of a, b respectively. Theorem S d is a semi-ring. Proof. Proto-ring. S d by definition. Closed under finite intersections. If A 1 = d a 1k, b 1k ] and A 2 = d a 2k, b 2k ] are nonempty rectangles, then A 1 A 2 = = a 1k, b 1k ] a 2k, b 2k ] a 1k a 2k, b 1k b 2k ]. Partially closed under differences. Suppose A 1 = d a 1k, b 1k ] and A 2 = d a 2k, b 2k ] are nonempty rectangles. Let Y be the set of all {0, 1}-valued sequences of length d, excluding 1, 1,..., 1, i.e. Y has 2 d 1 elements. For every y = y 1, y 2,..., y d Y and k {1, 2,..., d}, let { a1k, b 1k ] a 2k, b 2k ] if y k = 1 B k y = a 1k, b 1k ] a 2k, b 2k ] if y k = 0 { a1k a 2k, b 1k b 2k ] if y k = 1 = a 1k, b 1k a 2k ] a 1k b 2k, b 1k ] if y k = 0, 13

14 b 12 A 1 B0, 0 B1, 0 b 22 A 2 B0, 1 a 22 B0, 0 B1, 0 a 12 a 11 a 21 b 11 b 21 Figure 10.1: An example of the decomposition in the proof of theorem and let see figure We have By = B k y A 1 A 2 = By, y Y which is a disjoint union of rectangles. If A = d a k, b k ] is a nonempty rectangle, let V A = { x 1, x 2,..., x d R d : x k = a k or x k = b k for every k }, and for every x = x 1, x 2,..., x d V A let { 1 if xk = a k for an odd number of k sgn A x = +1 if x k = a k for an even number of k. Suppose F is a function from R d into R that is i nondecreasing, i.e. for every nonempty rectangle A we have sgn A xf x 0, x V A ii continuous from above, i.e. for every x R d and ɛ > 0 there exists δ > 0 such that if h d [0, δ then F x + h F x < ɛ. 14

15 Let µ F be the function from S d into [0, ] defined by sgn A xf x if A µ F A = x V A 0 if A =. For example, if d = 2 and A = a 1, b 1 ] a 2, b 2 ] is a nonempty rectangle, then Proposition If µ F A = F b 1, b 2 F a 1, b 2 F b 1, a 2 + F a 1, a 2. i A = d a k, b k ] is a nonempty rectangle, ii k 0 {1, 2,..., d}, iii c a k0, b k0, iv for every k, B k = { ak, b k ] if k k 0 a k, c] if k = k 0, C k = { ak, b k ] if k k 0 c, b k ] if k = k 0, v B = d B k and C = d C k, then µ F A = µ F B + µ F C. Proof. Let V B = {x 1, x 2,..., x d V B : x k0 = a k } V B = {x 1, x 2,..., x d V B : x k0 = c} V C = {x 1, x 2,..., x d V C : x k0 = b k } V C = {x 1, x 2,..., x d V C : x k0 = c}. The sets V B, V C are disjoint and their union is V A. If x V B then sgn Bx = sgn A x, and if x V C then sgn Cx = sgn A x. The sets V B, V C are in fact equal, so let V = V B = V C see figure If x V then sgn C x = sgn B x, because c is the right-endpoint of B k0 and the left-endpoint of C k0. It follows that µ F B = sgn A xf x + sgn B xf x x V x V B and µ F C = sgn A xf x x V C µ F B + µ F C = sgn B xf x, x V sgn A xf x = µ F A. x V B V C 15

16 b 2 V B V V C A B C a 2 V B V V C a 1 c b 1 Figure 10.2: An example of the decomposition in the proof of proposition Proposition Suppose A = d a k, b k ] is a nonempty rectangle and ɛ > 0. a There exists δ > 0 such that δ < b k a k for every k and µ F A < µ F k + δ, b k ] + ɛ. a b There exists δ > 0 such that µ F k, b k + δ] < µ F A + ɛ. a Proof. a Since F is continuous from above, for every x V A there exists δ x > 0 such that δ x < b k a k for every k and if h d [0, δ x then Let F x + h F x < ɛ/2 d. δ = 1 2 min{δ x : x V A}, B = a k + δ, b k ]. For every x = x 1, x 2,..., x d V A, let h x be the element of R d whose k-th entry is 0 if x k = b k and δ if x k = a k, h x d [0, δ x. The function x x + h x is a bijection between V A and V B, and sgn A x = sgn B x + h x, 16

17 µ F B = sgn B yf y y V B = sgn B x + h x F x + h x x V A = sgn A xf x + h x. x V A It follows that µ F A µ F B = b Similar to part a. x V A x V A < = ɛ. x V A sgn A xf x F x + h x F x + hx F x ɛ/2 d Theorem µ F is a σ-finite measure on S d. Proof. σ-finite. For every j, let = d j, j], then is a sequence in S d. Furthermore, µ F < for every j and = R d. Proto-measure. We have µ F = 0 by definition. Finitely additive. Suppose A = d a k, b k ] is a nonempty rectangle and A 1 = a 1k, b 1k ], A 2 = a 2k, b 2k ],..., A n = a nk, b nk ] are disjoint nonempty rectangles such that A = n. For every k {1, 2,..., d}, let c k1, c k2,..., c k,mk be the distinct elements in the set labelled such that {a 1k, b 1k, a 2k, b 2k,..., a nk, b nk }, c k1 < c k2 < < c k,mk. Since a 1k < b 1k we have m k 2, and since A = n we have a k = c 1k and b k = c k,mk. Let Y be the set of all sequences y = y 1, y 2,..., y d such that y k {1, 2,..., m k 1} for every k, Y has m 1 1m 2 1 m k 1 elements. For every y = y 1, y 2,..., y d Y, let By = c k,yk, c k,yk +1]. see figure After multiple applications of proposition 10.2, we find that both µ F A and n µ F are equal to y Y µ F By. 17

18 c 24 B1, 3 B2, 3 B3, 3 c 23 B1, 2 B2, 2 B3, 2 c 22 B1, 1 B2, 1 B3, 1 c 21 c 11 c 12 c 13 c 14 Figure 10.3: An example of the decomposition in the proof of theorem 10.4 Countably subadditive. Suppose that A = d a k, b k ] is a nonempty rectangle, that A 1 = a 1k, b 1k ], A 2 = a 2k, b 2k ],... are nonempty rectangles such that A =, and that ɛ > 0. By proposition 10.3, for every j there exists δ j > 0 such that µ F jk, b jk + δ j ] < µ F + ɛ/2 a j+1, and there exists δ > 0 such that δ < b k a k for every k and µ F A < µ F k + δ, b k ] + ɛ/2. a We have and [a k + δ, b k ] A = = a k, b k ] = A, a jk, b jk ] [a k + δ, b k ] a jk, b jk + δ j, a jk, b jk + δ j. Since the left-hand side is compact and the right-hand side is an open cover, there exists n such that n [a k + δ, b k ] a jk, b jk + δ j, 18

19 a k + δ, b k ] By proposition 4.2b, µ F k + δ, b k ] a n a jk, b jk + δ j ]. µ F a jk, b jk + δ j ]. It follows that µ F A < µ F k + δ, b k ] + ɛ/2 a a jk, b jk + δ j ] + ɛ/2 µ F < µ F + ɛ/2 j+1 + ɛ/2 = µ F + ɛ/2 j+1 + ɛ/2 µ F + ɛ. Since S d is a semi-ring and µ F is a σ-finite measure on S d, a µ F is an outer measure theorem 7.1, b M µ F is a σ-algebra theorem 6.2, c µ F M µ F is a measure on M µ F theorem 6.3, d µ F M µ F is an extension of µ F theorem 8.1 and 8.2, e this extension is unique theorem 9.1. The measure µ F M µ F is the Lebesgue-Stieltjes measure with respect to F. 11 Appendix: a result on double series The following result on double series was used in the proof of theorem 7.1. Proposition If a jk is a double sequence in [0, ] and f is a bijection from N onto N N, then a fj = a jk. 19

20 Proof. For every j, n, let s n j = t n = u n = v n = a jk s j = s n j = a fj. a jk a jk s j = lim n sn j = t = lim n tn = a jk a jk Note that, for every n, u n t n t. Step 1. We will first show that u n t. Step 1a. Suppose that s j0 = for some j 0 which implies that t =, and that x <. There exists n 0 such that if n n 0 then If n max{j 0, n 0 } then u n = x < s n j 0. j j 0 a jk + s n j 0 > x. Step 1b. Now suppose that s j < for every j, and that x < y < t. There exists m 0 such that y < t m0 t. For every j = 1, 2,..., m 0, there exists n j such that if n n j then If n max{m 0, n 1, n 2,..., n m0 } then u n = s n j m 0 s j y x 2 j < s n j s j. m 0 s n j > s j y x 2 j m 0 = t m0 y x 2 j > y y x 2 j = x. Step 2. We can now show that v n t. Suppose x < t. There exists m 0 such that x < u m0 t. Since f is a bijection, there exists n 0 such that {1, 2,..., m 0 } {1, 2,..., m 0 } f {1, 2,..., n 0 }. 20

21 Suppose n n 0. There exists m such that f {1, 2,..., n} {1, 2,..., m} {1, 2,..., m}, x < u m0 v n0 v n u m t. References [B86] P. Billingsley, Probability and Measure, 2nd ed., John Wiley & Sons, New York, Chichester, Brisbane, Toronto, Singapore, [C80] D.L. Cohn, Measure Theory, Birkhäuser, Boston, Basel, Stuttgart, [F99] G.B. Folland, Real Analysis: Modern Techniques and Their Applications, 2nd ed., John Wiley & Sons, New York, Chichester, Weinheim, Brisbane, Singapore, Toronto, [S09] S. Sawyer, Measures on semi-rings in R 1 and R k, wustl.edu/~sawyer/handouts/measuressemirings.pdf, 2009, accessed 7 October Copyright This work is licensed under a Creative Commons Attribution 4.0 International License. 21