Question 1. Question 3. Question 4. Graduate Analysis I Exercise 4

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1 Graduate Analysis I Exercise 4 Question 1 If f is easurable and λ is any real nuber, f + λ and λf are easurable. Proof. Since {f > a λ} is easurable, {f + λ > a} = {f > a λ} is easurable, then f + λ is easurable. If λ = 0, then λf = 0, of course is easurable. If λ > 0, then {λf > a} = {f > a }, if λ λ < 0, {λf > a} = {f < a }, these two sets are both easurable, so λf is easurable. λ Question 3 theore4.3 can be used to define easurability for vector-valued (e.g, coplexvalued) functions. Suppose, for exaple, that f and g are real-valued and defined in R n, and let F (x) = (f(x), g(x)). Then F is said to be easurable if F 1 (G) is easurable for every open G R 2. Prove that F is easurable if and only if both f and g are easurable in R n. Proof. ( = ) Since G is open, by Theore1.11 we have G = I k, I k = {y : a (k) i y i b (k) i, i = 1, 2} are nonoverlapping intervals. F 1 (I k ) = {a (k) 1 f b (k) 1 } {a (k) 2 g b (k) 2 }, since f and g are easurable, F 1 (I k ) is easurable. F 1 (G) = F 1 (I k ) is easurable, thus F is easurable. ( = ) For every finite a, since F is easurable, F 1 ((a, + ) R) = {f > a} is easurable, then f is easurable in R n. Siilarly we can get g is easurable in R n. Question 4 Let f be defined and easurable in R n. If T is a nonsingular linear transforation of R n, show that f(t x) is easurable. [ If E 1 = {x : f(x) > a} and E 2 = {x : f(t x) > a}, show that E 2 = T 1 E 1 ] Proof. E 2 = {x : f(t x) > a} = {x : T x f 1 ((a, + ))} = {x : T x E 1 } = {x : x T 1 E 1 } = T 1 E 1. Since T is nonsingular, T 1 is a Lipschitz transforation. By Theore3.33, E 2 = T 1 E 1 is easurable, thus f(t x) is easurable. 1

2 Question 5 Give an exaple to show that φ(f(x)) ay not be easurable if φ and f are easurable. Exaple: Let F be the Cantor-Lebesgue function. Define g : [0, 1] [0, 1] and g = (F (x) + x)/2, note that g is continuous and strictly increasing. Let C denote the Cantor set, then g(c) = 1. By Corollary3.39 we know there exists a noneasurable 2 set N g(c). Put f = g 1, then f(n) C and f(n) = 0 since C = 0. Put Z = f(n) and φ is the characteristic function of Z. Consider {φ(f(x)) > 0}, when f(x) Z, φ(f(x)) > 0, then we have {φ(f(x)) > 0} = N which is noneasurable, thus φ(f(x)) is not easurable. Lecturer s REMARK It ay not be clear that g(c) > 0. defining F 1 (x) = inf{y : F (y) = x} for x [0, 1] and let f = F 1. Thus I will prefer Question 6 Let f and g be easurable functions on E. (a) If f and g are finite a.e. in E, show that f + g is easurable no atter how we define it at the points when it has the for + + ( ) or +. (b) Show that fg is easurable without restriction on the finiteness of f and g. Show that f + g is easurable if it is defined to have the sae value at every point where it has the for + + ( ) or +. Proof. (a) Put Z 1 = {x E : f = ± } and Z 2 = {x E : g = ± }, Z 1 = 0, Z 2 = 0, then Z 1 Z 2 = 0. Put Z = {x E (f(x) = + and g(x) = ) or (f(x) = and g(x) = + )} Z 1 Z 2, then Z = 0. Let { f(x) x E Z f(x) = 0 x Z { g(x) x E Z g(x) = 0 x Z Since f = f a.e.and g = g a.e., f, g are easurable. By Theore4.9, f +g is easurable on E\Z and hence easurable on E. Since f + g = f + g a.e. no atter how we define the value of f + g on Z, we conclude that f + g is easurable on E. 2

3 (b) First, for any easurable set D, we will let M F (D) be the set of all easurable functions on D. Define E i, i = 1,..., 6 : E 1 E 2 E 3 E 4 E 5 E 6 f 0 0 ± 0 g ± fg β 1 β 2 β 3 β 4 ± ± Since f, g M F (E), we have f, g M F (E i ), i = 1,..., 6 and f, g M F (E\ E i ), then fg M F (E i ), i = 1,..., 6 and fg M F (E\ E i ). Thus, fg M F (E) without restriction on the finiteness of f and g. We now define E i, i = 1,..., 4: E 1 E 2 E 3 E 4 f g f+g β 1 β 2 Since f, g M F (E), we have f, g M F (E i ), i = 1,..., 4 and f, g M F (E\ E i ), then f + g M F (E i ), i = 1,..., 4 and f + g M F (E\ E i ). Hence, f + g M F (E). Question 7 Let f be usc and less than + on a copact set E. Show that f is bounded above on E. Show also that f assues its axiu on E, i.e., that there exists x 0 E such that f(x 0 ) f(x) for all x E. Proof. Since f is usc on E, then {x E : f(x) < k} is open for k = 1, 2,.... Since f is less than +, {x E : f(x) < k} is an open cover of E. Since E is copact, then there exist a finite subcover, that is N N such that N {x E : f(x) < k} E, this eans f(x) < N, so f is bounded above on E. Put M = sup f(x), then M < N, for each x E k = 1, 2,..., there exists x k E such that f(x k ) > M 1. Since E is copact, there k exists a subsequence x nk of x k which converges to a point x 0 of E. f(x nk ) > M 1 n k, let k, we have f(x 0 ) M, since f(x 0 ) M, then f(x 0 ) = M. Thus, f assues its axiu on E. 3

4 Question 8 (a) Let f and g be two functions which are usc at x 0. Show that f + g is usc at x 0. Is f g usc at x 0? When is fg usc at x 0? Proof. If f(x 0 ) = + or g(x 0 ) = +, then f(x 0 ) + g(x 0 ) = +, so f + g is usc at x 0. If f(x 0 ) < + and g(x 0 ) < +, since f is usc at x o, given M 1 > f(x 0 ), there exists δ 1 > 0 such that f(x) < M 1 for all x E which lie in the ball x x 0 < δ 1. Since g is also usc at x o, given M 2 > f(x 0 ), there exists δ 2 > 0 such that g(x) < M 2 for all x E which lie in the ball x x 0 < δ 2. Put δ = in(δ 1, δ 1 ), then f(x) + g(x) < M 1 + M 2 for all x E which lie in the ball x x 0 < δ, thus f + g is usc at x 0. f g is not usc at x 0. Since g is usc at x 0, g is lsc at x 0, put f = 0, then f g = g is lsc at x 0. When f 0 and g 0, fg is usc at x 0. If f(x 0 ) = + or g(x 0 ) = +, obviously fg is usc at x 0. Otherwise, by the above proof, put δ = in(δ 1, δ 1 ), then f(x)g(x) < M 1 M 2 for all x E which lie in the ball x x 0 < δ, thus fg is usc at x 0. at x 0. (b) If {f k } is a sequence of functions which are usc at x 0. Show that inf k f k (x) is usc Proof. Since f k is usc at x o, given ε > 0, f k (x 0 ) + ε > f k (x 0 ), then δ k > 0 such that f k (x) < f k (x 0 ) + ε for x E k = ({x : x x 0 < δ k } E). When x E k, inf k f k (x) < inf k f k (x 0 ) + ε, so inf k f k (x) is usc at x 0. (c) If {f k } is a sequence of functions which are usc at x 0 and which converge uniforly near x 0, show that li f k is usc at x 0. Proof. Since f k is usc at x o, given ε > 0, there exists δ k > 0 such that f k (x) < f k (x 0 )+ ε 2 for all x E which lie in the ball x x 0 < δ k. Since f k converge uniforly near x 0, there exists K, when k > K and x {x : x x 0 < δ k } E, li f k (x) f k (x) < ε 2, then li f k (x) < f k (x) + ε 2 < f k(x 0 ) + ε, li f k (x) < li f k (x 0 ) + ε, thus li f k is usc at x 0. Question 9 Show that the liit of a decreasing ( increasing ) sequence of functions usc (lsc) at x 0 is usc (lsc) at x 0. In particular, the liit of a decreasing ( increasing ) sequence of functions continuous at x 0 is usc (lsc) at x 0. Proof. Let {f k } be the decreasing sequences usc at x 0, since f k is decreasing, we have f = li k f k = inf k f k, by Question8(b) we know that inf k f k is usc at x o, then 4

5 f = li x f k is usc at x 0. Question 10 (a) If f is defined and continuous on E, show that {a < f < b} is relatively open, and that {a f b} and {f = a} are relatively closed. (b) If f is a finite function on R n. Show that f is continuous on R n if and only if f 1 (G) is open for every open G in R 1, or if and only if f 1 (F ) is closed for every closed F in R 1. Proof. (a) For each x k {a < f < b}, since a < f(x k ) < b and f is continuous on E, there exists δ 1, δ 2 > 0 such that f(x) < b when x B(x k, δ (k) 1 ) E and f(x) > a when x B(x k, δ (k) 2 ) E, put δ k = in(δ (k) 1, δ (k) 2 ), we have a < f(x) < b when x B(x k, δ k ) E. Then B(x k, δ k ) E = {a < f < b}, since B(x k, δ k ) is open, {a < f < b} is relatively open. We prove the copleent of {a f b} is relatively open, the copleent of {a f b} is {f > b} {f < a}. Fro the above proof we know {f > b} and {f < a} are relatively open, then {f > b} {f < a} is relatively open, so {a f b} is relatively closed. Put b = a, {f = a} is also relatively closed. (b) ( = ) If f is continuous, put f 1 (G). For x 0 f 1 (G), then f(x 0 ) G. Since G is open, there exists ε > 0 such that (f(x 0 ) ε, f(x 0 ) + ε) G. Since f is continuous, for the above ε, δ > 0 such that ε < f(x) f(x 0 ) < ε when x (x 0 δ, x 0 + δ), so when x (x 0 δ, x 0 + δ), f(x) G, then (x 0 δ, x 0 + δ) f 1 (G). This eans x 0 is the interior point of f 1 (G), so f 1 (G) is open. ( = ) Since f 1 (G) is open for every open G in R 1, for each x 0 and ε > 0, the pre-iage U of (f(x 0 ) ε, f(x 0 ) + ε) is open. Since x 0 U, then δ > 0 such that (x 0 δ, x 0 + δ) U. So when x (x 0 δ, x 0 + δ), we have f(x) (f(x 0 ) ε, f(x 0 ) + ε), this eans f is continuous at x 0. Thus f is continuous. Question 11 Let f be defined on R n and let B(x) denote the open ball {y : x y < r} with center x and fixed radius r. Show that the function g(x) = sup{f(y) : y B(x)} is lsc and that the function h(x) = inf{f(y) : y B(x)} is usc on R n. Is the sae true for the closed ball {y : x y r}? 5

6 Proof. For any x 0 R n, g(x 0 ) = sup{f(y) : y B(x 0 )}, then y B(x 0 ), f(y) g(x 0 ). Moreover, ε > 0, there exists x ε B(x 0, 1) such that f(x ε ) > g(x 0 ) ε. Choose r ε < 1 such that x 0 x ε < r ε. Since x ε B(z, r) if z B(x 0, r r ε ) and g(x) = sup{f(y) : y B(z)}, we have g(x) f(x ε ) > g(x 0 ) ε when x B(x 0, r r ε ), so g(x) is lsc at x 0. Thus, g(x) is lsc on R n. We can use the siilar way to prove h(x) = inf{f(y) : y B(x)} is usc on R n. It is not the sae for the closed ball {y : x y r}, for exaple, let f(0) = 1 and f(x) = 0 for x 0, it is clear that h(x) = sup{f(y) : y B(y, 1)} is not lsc. Question 12 If f(x), x R 1, is continuous at alost every point of an interval [a, b], show that f is easurable on [a, b]. Proof. Put Z = {f is not continuous}, then Z = 0. Put E = [a, b] Z, then f is continuous relative to E. {f > α} = {x E : f > α} {x Z : f > α}. Since f is continuous on E, {x E : f > α} is easurable. {x Z : f > α} z = 0, then {x Z : f > α} is easurable. Thus, f is easurable on [a, b]. Question 15 Let {f k } be a sequence of easurable functions defined on a easurable E with E < +. If f k (x) M x < + for all k for each x E, show that given ε > 0, there is a closed F E and a finite M such that E F < ε and f k (x) M for all k and all x F. Proof. Fix ε > 0, for each n, let E n = {x E : M x n for all k}. Thus, E n = {x : M x n}, so that E n is easurable. Clearly E n E n+1 and E n E. Hence, k E n E. Since E < +, it follows that E E n 0. Choose n 0 such that E E n0 < ε, and let F be a closed subset of E 2 n 0 with E n0 F < ε. Then E F < ε 2 and f k (x) n 0 for all k and all x F. Question 17 Suppose that f k if E < +, that f k g k f and g k g on E. Show that f k + g k f + g on E and, fg on E. If, in addition, g k g on E, g 0 a.e., and 6

7 E < +, show that f k /g k f/g on E. Proof. (1) {x E : f + g (f k + g k ) > ε} {x E : f f k > ε 2 } {x E : g g k > ε }, since li {x E : f f 2 k > ε } = 0 and li {x E : g g k 2 k > ε } = 0, k 2 then li {x E : f + g (f k + g k ) > ε} = 0, f k + g k f + g on E. k (2) Since f k g k fg = (f k f)(g k g)+f(g k g)+g(f k f), {x E : f k g k fg > ε} {x E : (f k f)(g k g) > ε 3 } {x E : f(g k g) > ε 3 } {x E : g(f k f) > ε 3 }. {x E : (f k f)(g k g) > ε 3 } {x E : f k f > ε 3 } {x E : g k g > ε 3 }. Since f is the liit (in easure) of a sequence of functions, it ust be finite on E. Since E < +, for any η > 0, there exists Z 1, Z 2 E, Z 1, Z 2 < η such that f is bounded on E Z 1 and g is bounded on E Z 2, and. Put f M 1 on E Z 1 and g M 2 on E Z 2. So {x E : f(g k g) > ε 3 } {x E : g k g > ε 3M 1 } Z 1, {x E : g(f k f) > ε} {x E : f 3 k f > ε 3M 2 } Z 2, since f k f and g k g on E and Z 1, Z 2 < η, we have {x E : f k g k fg > ε} < 5η if k is sufficiently large. Hence f k g k fg on E. (3) First, since g 0 a.e., we have 1/g is finite a.e.. Hence given any η > 0, since E <, there exists Z E such that Z < η and 1/g M on E\Z for soe constant M > 0. Since g k g, K such that E k = { g k g > 1/(2M)} < η when k K. Note that g k g g g k 1/(2M) on E\(Z E k ). Now observe that 1 g k 1 g = g k g g k g = 1 g k g g k g. Note that g k g 1/(2M 2 ) on E\(Z E k ). Since g k such that { g k g > ε/(2m 2 )} < η for k K. It is now clear that { 1 g k 1 > ε} < 2η when k ax{k, g K }. Thus 1 g k 1 on E. By the conclusion of (2), we have f g k/g k Question 18 g, for any ε > 0, there exists K f/g on E. If f is easurable on E, define ω f (a) = {f > a} for < a < +. If f k f, show that ω fk ω f. If f k f, show that ω fk ω f at each point of continuity of ω f. Proof. Put E k = {f k > a}, since f k f, we have E k E k+1 and E k E = {f > a}, by Theore3.26 we have E k E, then ω fk ω f. 7

8 Let a be a continuous point of ω f, then η > 0, δ > 0, when b a δ, ω f (b) ω f (a) < η. Let E k = {x : f f k > δ}. Then there exists N N such that E k < η for k N. Since f k = (f k f) + f, {f k > a} E k {f > a δ}, then ω fk η + ω f (a δ) < η + ω f (a) + η for k N. Next, since f = (f f k ) + f k, we have {f > a + δ} (E k {f k > a}), then ω fk ω f (a + δ) η, so ω fk (a) ω f (a) < 2η. Thus ω fk (a) ω f (a). Question 19 Let f(x, y) be a function defined on the unit square 0 x 1, 0 y 1 which is continuous in each variable separately. Show that f is a easurable function of (x, y). proof. Divide [0, 1] into n equal subintervals, put f n (x, y) = f(x, k ) when k y < k+1. n n n Clearly for any fixed (x, y), f n (x, y) f(x, y). {f n (x, y) > a} = n 1 {{x [0, 1] : f(x, k ) > a} [ k, k+1)}, since f is continuous in each variable separately, so {f n n n n(x, y) > a} is easurable. Since f n (x, y) f(x, y), f is a easurable function of (x, y). Question 20 If f is easurable and finite a.e. on [a, b], show that given ε > 0, there is a continuous g on [a, b] such that {x : f(x) g(x)} < ε. Proof. By Lusin stheore we have ε > 0, there exists a closed set F E such that E F < ε and f is continuous relative to F. Put G = F, then G is open, then G = (a k, b k ), (a k, b k ) are disjoint. When x F, Let g(x) = f(x). When x / F, define g(x) is a linear on each open interval (a k, b k ) and let g(x) be continuous at the endpoint of the interval. When x (a k, b k ) which is a finite interval, g(x) = f(a k ) b k x b k a k + f(b k ) x a k b k a k. When soe open interval is infinite, if x (c, + ), g(x) = f(c), when x (, d), g(x) = f(d). So g is well-defined on R, and obviously {f g} E F < ε. Fro the above construction of g, we have g is continuous at the endpoint of the open interval, so g is continuous on R, of course g is continuous on [a, b]. 8