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Math 318 HW #4 Solutions 1. Abbott Exercise 7.4.6. Review the discussion immediately preceding Theorem 7.4.4. R1 (a) Produce an example of a sequence fn → 0 pointwise on [0, 1] where limn→∞ 0 fn does not exist. Answer. Consider the function fn whose graph is given in Figure 1. n2 1/n 2/n Figure 1: The function fn For any x ∈ [0, 1], either x = 0 and fn (x) = 0 for all n, or fn (x) = 0 for all n > 2/x; either way, fn (x) → 0, so we see that fn → 0 pointwise. On the other hand, Z 1 1 2 2 fn = · n = n, 2 n 0 R1 so limn→∞ 0 fn diverges to +∞. R1 (b) Produce another example (if necessary) where fn → 0 and the sequence 0 fn is unbounded. Answer. The same example from (a) still works. (c) Is it possible to construct each fn to be continuous in the examples of parts (a) and (b)? Answer. The example in (a) is already continuous. (d) Does it seem possible to construct the sequence (fn ) to be uniformly bounded? R1 Answer. Certainly we can’t get 0 fn to be unbounded if the fn are uniformly bounded: if R 1there’s a uniform bound M such that |fn (x)| ≤ M for all n and all x ∈ [0, 1], then 0 fn ≤ M (1 − 0) = M for any n. R1 Also, problem 2 below implies it is not possible to have 0 fn diverge at all when the convergence is uniform on any set of the form [δ, 1] (as it would be for, say, a power series centered at x = 1). 1 Of course, it’s easy to come up with non-integrable functions fn such that fn → 0 (even uniformly); for example, let fn (x) = n1 g(x) where g is the Dirichlet function. In this R1 case 0 fn won’t be well-defined. It may R 1be possible to get a sequence of integrable functions fn that are uniformly bounded with 0 fn diverging, but it seems hard. 2. Abbott Exercise 7.4.7. Assume that gn and g are uniformly bounded and integrable with gn → g on [0, 1]. The convergence is not uniform; however,Rthe convergence is uniform on any R1 1 set of the form [δ, 1] where 0 < δ < 1. Show that limn→∞ 0 gn = 0 g. Proof. Let > 0. Since (gn ) is uniformly bounded, there exists M > 0 such that |gn (x)| ≤ M for all n ∈ N and all x ∈ [0, 1]. Note that this implies that |g(x)| ≤ M for all x ∈ [0, 1] as well (by the Order Limit Theorem). Pick δ > 0 such that δ < /4M . R1 R1 Since gn → g uniformly on [δ, 1], we know, from Theorem 7.4.4, that limn→∞ δ gn = δ g. Therefore, there exists N ∈ N such that n ≥ N implies that Z 1 Z 1 gn − g < /2. δ δ On the other hand, Theorem 7.4.2 implies that δ Z −M δ ≤ gn ≤ M δ 0 and Z δ g ≤ M δ, −M δ ≤ 0 so Z δ Z gn − 0 0 δ Z g ≤ δ 0 Z gn + 0 δ g ≤ 2M δ < /2 for any n ∈ N. Therefore, if n ≥ N , we have Z 1 Z g − n 0 0 1 Z δ Z δ Z 1 Z 1 g = gn + gn − g− g 0 δ 0 δ Z δ Z δ Z 1 Z 1 gn − g + gn − g = 0 0 δ δ Z δ Z δ Z 1 Z 1 gn − g ≤ gn − g + 0 0 δ δ < /2 + /2 = . Since the choice of > 0 was arbitrary, we see that, indeed, limn→∞ 2 R1 0 gn = R1 0 g. 3. Abbott Exercise 7.6.11. Show that there exists a finite collection of disjoint open intervals {G1 , . . . , GN } whose union contains Dα and that satisfies N X |Gn | < n=1 . 4M Proof. Since Dα ⊆ D has measure zero, we know that there exists a countable collection U1 , U2 , . . . of open intervals such that Dα ⊆ ∞ [ Un and n=1 ∞ X |Un | < n=1 . 4M However, we proved last semester (cf. Exercise 7.6.8) that Dα is compact; since {Un : n = 1, 2, . . .} is an open cover for Dα , compactness implies that there is a finite subcover {G1 , . . . , GN }, where each Gi is one of the Uj . Then clearly Dα ⊆ N [ n=1 Gn and N X |Gn | ≤ n=1 ∞ X n=1 |Un | < . 4M The only concern is that the Gn may not be disjoint. Suppose Gi ∩ Gj 6= ∅. Since both Gi and Gj are open intervals, the union Gi ∪ Gj is also an open interval; removing both Gi and Gj from the collection and inserting Gi ∪ Gj eliminates the nontrivial intersection. Iterate for all nontrivial intersections until none are left: the resulting collection will have the same properties as above while also consisting of disjoint open intervals. 4. Abbott Exercise 7.6.13. Finish the proof in this direction by explaining how to construct a partition P of [a, b] such that U (f, P ) − L(f, P ) ≤ . It will be helpful to break the sum U (f, P ) − L(f, P ) = n X (Mk − mk )∆xk k=1 into two parts, one over those subintervals that contain points of Dα and the other over subintervals that do not. S Proof. Letting K = [a, b]\ N n=1 Gn , we know that f is uniformly α-continuous on K (by Exercise 7.6.12). In other words, there exists δ > 0 such that |x − y| < δ implies that |f (x) − f (y)| < α. Break K up into intervals of length < δ and construct P from the endpoints of these intervals 3 and from the endpoints of the intervals Gn . Then U (f, P ) − L(f, P ) = n X (Mk − mk )∆xk k=1 = X (Mk − mk )∆xk + (Mk − k s.t. [xk−1 ,xk ]∩Dα 6=∅ k s.t. [xk−1 ,xk ]⊆K < X X α∆xk + N X mk )∆xk 2M |Gn | n=1 k s.t. [xk−1 ,xk ]⊆K ≤ α(b − a) + 2M 4M = /2 + /2 = . Therefore, since the choice of > 0 was arbitrary, we see that Theorem 7.2.8 implies that f is Riemann integrable. 5. Wilcox & Myers Exercise 9.3. Prove that if µ is a measure on S and {x} ∈ S for every x ∈ [a, b], and µ({x}) = µ({y}) for all x, y ∈ [a, b], then µ(Q ∩ [a, b]) = 0. Proof. First, note that Q ∩ [a, b] ∈ S: if r1 , r2 , . . . is an enumeration of Q ∩ [a, b], then, by hypothesis, {rn } ∈ S, so, since S is closed under countable unions, Q ∩ [a, b] = ∞ [ {rn } ∈ S. n=1 Now, let M = µ{x}, which is the same fixed number for all x ∈ [a, b]. Then, since µ is a measure, it satisfies countable additivity, so µ(Q ∩ [a, b]) = ∞ X n=1 µ({rn }) = ∞ X M. n=1 Since M ≥ 0, the sum on the right hand side is either 0 or +∞. The latter is impossible (since, e.g., +∞ > b − a), so the sum, and hence µ(Q ∩ [a, b]), must be zero. 4