The reference [Ho17] refers to the course lecture notes by Ilkka Holopainen.

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1 Department of Mathematics and Statistics Real Analysis I, Fall 207 Solutions to Exercise 6 (6 pages) riikka.schroderus at helsinki.fi Note. The course can be passed by an exam. The first possible exam is on November. You need to register for the exam through WebOodi, the registration ends 0 days before the exam. The reference [Ho7] refers to the course lecture notes by Ilkka Holopainen.. Let E R be Lebesgue-measurable and m(e) > 0. Show that contains an interval. E E = {a b : a, b E} Proof. It is enough to show that there exists δ > 0 such that E (E x) for every x ( δ, δ). Indeed, if this is true, then for every x ( δ, δ) there exist a, b E such that a = b x. That is, x = b a E E and hence ( δ, δ) would be an interval contained in E E. We can assume that m(e) < ; if m(e) =, we can show the claim for the set E ( k, k), where k N is so big that m(e ( k, k)) > 0, since then (E ( k, k)) E and m(e ( k, k)) 2k <. In particular, we can assume that χ E L (R). So, let us prove that the desired δ > 0 exists. Define a function f : R R by setting f(x) = χ E (x + t)χ E (t) dt = χ E x (t) dt = m(e (E x)). R By [Thm. 2.29, Ho7] and the assumption χ E L (R), it holds that lim f(0) f(h) = lim χ E (0 + t)χ E (t) χ E (h + t)χ E (t) dt h 0 h 0 Thm.2.29 = 0, R E lim χ E (t) χ E (t) χ E (h + t) dt h 0 R lim χ E (t) χ E (h + t) dt h 0 R and hence f is continuous at 0. Since, moreover f(0) = m(e) > 0, there exists δ > 0 such that f(x) = m(e (E x)) > 0 for all x ( δ, δ).

2 2 2. (a) Let G R n be an open set. Show that every point in G is a density point of G. (b) Demonstrate with the help of an example that a point in the complement of an open set G can be a density point of G. (c) Construct a subset A of R, such that A is not open, even though every point x A is a density point of A. Proof. (a) Let x G. Since the set G is open, there exists r x > 0 such that B(x, r x ) G. In particular, for all r (0, r x ), it holds that B(x, r) G = B(x, r) and hence m(b(x, r) G) lim r 0 + m(b(x, r)) = lim r 0 + =. We have that every point x G is a density point. (b) Let us consider the open set R \ {0}. Now, for all r > 0, it holds that m(b(0, r) (R\{0})) = m(b(0, r)\{0}) = m(b(0, r)). Hence we have that m(b(0, r) R \ {0}) lim r 0 + m(b(0, r)) = lim r 0 + =, and so the point 0 R \ {0} is a density point of the set R \ {0}. (c) Let A = R \ Q which is not open. Now, for every x R and r > 0, it holds that m(b(x, r) A) = m(b(x, r)\q) = m(b(x, r)). In particular, we have, for all x R, that m(b(x, r) A) lim r 0 + m(b(x, r)) = lim r 0 + =, so that every point x R is a density point of the set A. 3. Let Q = {q k : k N} be the set of all rational numbers. Define f(x) = 2 k, x R. Prove: (a) f : R (0, ) is strictly increasing. (b) f is continuous at x x R \ Q. Proof. (a) We need to show that f(y) > f(x) for all y > x. Recall that the rational numbers are dense in R. Hence, if x, y R are such that

3 x < y, then there exists a rational number q n such that x < q n < y. It follows that f(x) = 2 k < 2 k + 2 n q k y 2 k = f(y). (b) We will show first that the function f is not continuous at any rational point. Let x Q, so that x = q n for some n N. It follows that, for all y < x, f(y) + 2 n = q k y 2 k x=qn n k = f(x). But now, for all y < x, it holds that f(y) f(x) 2 n and hence lim y x f(y) f(x). Therefore, f is not continuous at any rational point. Let us then prove that the function f is continuous at every irrational point, i.e. for all ε > 0 there exists δ > 0 such that f(x) f(y) < ε when x y < δ and x R \ Q. So, let x R\Q and ε > 0. Since the series that defines f is convergent, we can choose k 0 N such that k=k k < ε. We will choose δ to be δ = min{ q k x : k =, 2,..., k 0 }. Now, q k x δ for all k =, 2,..., k 0 and so q, q 2,... q k0 R \ (x δ, x + δ). Let y (x δ, x). We have that f(y) f(x) = q k y x δ< 2 k 2 k 2 k = k=k 0 + y< 2 k < ε. The case y (x, x + δ) can be shown similarly and so we see that f is continuous at every irrational point. 4. (a) Exhibit a (simple) example of a function f : R R, which has bounded variation, but is not continuous. (b) Let f : [0, ] R, { x sin, if 0 < x, f(x) = x 0, if x = 0. 2 k 3

4 4 Show that f is continuous, but that f does not have bounded variation. Proof. (a) Consider the function f : R R, f = χ [0,]. Let a < b and let a = x 0 < x < x k = b be a finite division of the interval [a, b]. It holds that f(x i ) f(x i ) =, if x i (, 0) and x i [0, ];, if x i [0, ] and x i (, ); 0, otherwise. In particular, there exist at most two indeces i such that f(x i ) f(x i ) = and hence f(x i ) f(x i ) 2. By taking the supremum over all finite divisions, we see that V f (a, b) 2. Hence also V f (R) = sup a<b V f (a, b) 2 < so that f BV (R). (b) On the interval (0, ], the function f is defined as the product of continuous functions x x and x sin, so we need to show that x f is also continuous at 0. This is seen by noting that for all h > 0 it holds that h sin h 0 h sin h 0, as h 0 +. h To show that the function f does not have bounded variation on [0, ], consider the division a = x 0 < x < < x k = b, where x 0 = 0 and x n = {, π/2+nπ, nπ if n is odd if n is even. Now f(x n ) = /(π/2 + nπ), when n is odd, and f(x n ) = 0, when n is even. In particular, { π/2+nπ f(x n ) f(x n ) =, if n is odd, if n is even. π/2+(n )π π/2 + nπ. Hence, for all k N, it holds that f(x n ) f(x n ) n= n= π/2 + nπ 2nπ n=, as k, and by taking the supremum over all finite divisions of the interval [0, ] we see that V f (0, ) =. That is, f BV ([0, ]).

5 5. Let f : [a, b] R and g : [a, b] R be functions of bounded variation. Prove that V fg (a, b) M f V g (a, b) + M g V f (a, b), where M f = sup{ f(x) : x [a, b]} and M g = sup{ g(x) : x [a, b]}. Proof. Let a = x 0 < x < < x n = b be an arbitrary division of the interval [a, b]. By the triangle inequality, we have for all i =,..., n, fg(x i ) fg(x i ) = f(x i )g(x i ) f(x i )g(x i ) + f(x i )g(x i ) f(x i )g(x i ) Hence, it holds that n fg(x i ) fg(x i ) M f f(x i ) g(x i ) g(x i ) + f(x i ) f(x i ) g(x i ) M f g(x i ) g(x i ) + M g f(x i ) f(x i ). n g(x i ) g(x i ) + M g and by taking the supremum over all divisions, we have V fg (a, b) M f V g (a, b) + M g V f (a, b). n f(x i ) f(x i ) 6. Let f : [a, b] R be of bounded variation. Denote V (x) = V f (a, x). Prove that f is continuous V is continuous. Proof. Assume first that f is continuous. Let ξ (a, b] be arbitrary and let ε > 0. By [Lemma 3.58, Ho7], we have, for any points x, y [a, b] such that x < y, that V (y) V (x) = V f (a, y) V f (a, x) = ( V f (a, x) + V f (x, y) ) V f (a, x) = V f (x, y) 0. Hence V is increasing and so it has the left and right limits at every point on the interval [a, b]. Therefore, we need only to show that V is continuous on the left at the point ξ since the continuity from right at the point ξ follows by symmetry and, further, since ξ is arbitrary, it follows that V is continuous on the whole interval. To show that V is left continuous at the point ξ, it is enough to find a point x < ξ such that V (ξ) V (x) = V f (a, ξ) V f (a, x) L3.58 = V f (x, ξ) < ε. Let ξ < ξ a point in the interval [a, b]. Hence, by the definition of bounded variation, there exists a division ξ = x 0 < < x k = ξ of the interval [ξ, ξ] such that () V f (ξ, ξ) < f(x i ) f(x i ) + ε/2 =: H + ε/2. 5

6 6 Since f is continuous, there exists x (x k, x k ) = (x k, ξ) such that (2) f(x) f(ξ) < ε/2. Let us fine the division ξ = x 0 < < x k = ξ by adding the point x in it. By the triangle inequality the variation over this new division is bigger than the variation over the original division. By [Lemma 3.58, Ho7], it holds that V f (ξ, ξ) = V f (ξ, x) + V f (x, ξ) and hence it follows that V f (x, ξ) = V f (ξ, ξ) V f (ξ, x) H + ε/2 (H ( f(x) f(ξ) )) (2) < ε/2 + ε/2 = ε. In the first inequality above we use Inequality () as well as the fact that H ( f(x) f(ξ) ) is some variation over the interval [ξ, x] so that V f (ξ, x) H f(x) f(ξ). We have shown that V is continuous. Assume then that V is continuous. Let x, y [a, b], x < y. Since the points x, y form a division of the interval [x, y], it holds that By [Lemma 3.58, Ho7], we get f(x) f(y) V f (x, y). f(x) f(y) V f (x, y) = V f (a, y) V f (a, x) = V (y) V (x) = V (y) V (x). Since V is continuous by the assumption, we also get that f is continuous.