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Mathematics 4 Homework 2, solutions Prof. F. Brock 1. Let x0 ∈ R, and let µ be the point measure concentrated at x0 , that is, ( 1 if x0 ∈ A µ(A) := ∀A ∈ B1 . 0 if x0 6∈ A Find an increasing and right-continuous function α : R → R such that µ = µα . (Here, B1 is the Borel algebra on R and µα denotes the Lebesgue-Stieltjes measure on B1 induced by α.) Solution : Let α := χ[x0 ,+∞) (= characteristic function of the interval [x0 , +∞)). Since α(x) = 0 for x < x0 and α(x) = 1 for x ≥ 1, α is increasing and rightcontinuous, that is, α(x+) = α(x) for every x ∈ R. Defining µα as in the lecture, we find: If −∞ ≤ a ≤ b ≤ +∞, then µα ([a, b]) = α(b)−α(a−), µα ((a, b]) = α(b)−α(a), µα ((a, b)) = α(b−) − α(a) and µα ([a, b)) = α(b−) − α(a−). Let I be an interval, that is, one of the sets given above. Then an easy partition into cases shows that µα (I) = 1 if x0 ∈ I and µα (I) = 0 if x0 6∈ I. (As an example of that calculus, we have a look at the case a = x0 < b < ∞. Then µα ([x0 , b)) = α(b−) − α(x0 −) = 1 − 1 = 0, µα ((x0 , b)) = α(b−) − α(x0 ) = 1 − 0 = 1. ) In other words, we have µα = µ on the family of intervals. In view of the countable additivity of the set functions µ and µα , the same is true for Borel sets on R. 2. Let A be a σ-algebra over a set X. Prove χA is A-measurable if and only if A ∈ A. Solution : We have X {x ∈ X : χA (x) > t} = A ∅ Since X, A, ∅ ∈ A, the assertion follows. that the characteristic function if t < 0 if 0 ≤ t < 1 . if 1 ≤ t 3.∗ Let M be the set of all numbers in [0, 1] whose decimal representations contain the digit 5 infinitely many times. Show that M is a Borel set. Solution : To x ∈ [0, 1] we have the decimal expansion x = a0 + ∞ X ak 10−k , k=1 where a0 ∈ {0, 1} and ak ∈ {0, 1, . . . , 9}, (k ∈ N). To integers 0 < k1 < k2 < · · · < kn , (n ∈ N), we denote by A(k1 , . . . , kn ) the set of numbers x ∈ [0, 1] with ak1 = ak2 = · · · = akn = 5, and ak 6= 5 for k 6∈ {k1 , k2 , . . . , kn }. Then A(k1 , k2 , . . . , kn ) is closed. Let Bn be set of numbers x ∈ [0, 1] in whose decimal expansion the digit 5 occurs exactly n times, (n ∈ N ∪ {0}). Then B0 is closed. Further we have for n ∈ N, Bn = [ {A(k1 , k2 , . . . , kn ) : k1 < . . . < kn , k1 , . . . kn ∈ N} . Since Bn , (n ∈ N), is a countable union of closed sets, it is a Borel set. Finally, ! ∞ [ M = [0, 1] \ Bn , n=0 so that M is a Borel set, too. 4. Let f : R → R be an increasing function. Show that f is Borel measurable. Solution : Let t ∈ R. Since f is increasing, the set {x : f (x) > t} is either ∅, R, a half line of the form (a, +∞), or a half line of the form [a, +∞), for some number a ∈ R. Since each of those sets is a Borel set, the assertion follows. Remark: The situation {f > t} = [a, +∞), for some numbers t, a ∈ R, cannot occur if f was continuous (since the preimage of an open set is open for continuous functions f ). However, if f is discontinuous, this might not be the case, as the following example shows: Let f (x) = 1 for x ≥ 0 and f (x) = 0 for x < 0. Then {f > 0} = [0, +∞). 5. Consider Proposition 12.8, p.316, of A. Schüler ’Calculus’. Prove the implications (d) ⇒ (e) ⇒ (a). Solution : Proof of (d) ⇒ (e) : First observe that there hold the following properties for preimages of f , (A, Ak ⊂ R, n ∈ N ∪ {+∞}), f −1 f −1 f n [ ! n [ f −1 (Ak ), (1) f −1 (Ak ), (2) (R \ A) = X \ f −1 (A). (3) k=1 n \ k=1 −1 Ak = ! Ak = k=1 n \ k=1 Then, we leave it to the reader to show that the properties (a) through (d) are equivalent to each other. Using (2) this means that f −1 (I) ∈ A if I is an interval. Further, using (1), we find that f −1 (M ) ∈ A if M is an elementary set, M ∈ ε1 . Since every Borel set can be represented with sets from ε1 through a countable number of unions, intersections and complements, the properties (1)–(3) imply that f −1 (B) ∈ A for every Borel set B ∈ B1 . Proof of (e) ⇒ (a) : Assume that (e) holds, and let a ∈ R. Then (a, +∞] ∈ B1 , so that {x : f (x) > a} = f −1 ((a, +∞]) ∈ A.