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1、? chapter one 3.Prove that if f is a real function on a measurable space X such that x : f(x) r is measurable for every rational r, then f is measurable. proof:For each real number , there exists an descending sequence rn of rational numbers such that lim n rn= . Moreover,we have (,+) = n=1 rn,+). H
2、ence, f1(,+) = n=1 f1(rn,+). Since sets f1(rn,+) are measurable for each n, the set f1(,+) is also measurable. Then f is measurable. 4. Let an and bn be sequences in ,+, and prove the following assertions: (a) limsup n (an) = liminf n an. (b) limsup n (an+ bn) limsup n an+ limsup n bn. provided none
3、 of the sums is of the form . (c) If an bnfor all n, then liminf n an liminf n bn. Show by an example that strict inequality can hold in (b). proof: (a) Since sup kn(a k) = inf kn ak,n = 1,2,. Therefore, let n , it obtains lim n sup kn(a k) = lim n inf kn ak. 1 By the defi nations of the upper and t
4、he lower limits, that is limsup n (an) = liminf n an. (b) Since sup kn(a k+ bk) sup kn ak+ sup kn bk,n = 1,2,. Hence lim n sup kn(a k+ bk) lim nsup kn ak+ sup kn bk = lim n sup kn ak+ lim n sup kn bk. By the defi nations of the upper and the lower limits, that is limsup n (an+ bn) limsup n an+ limsu
5、p n bn. example: we defi ne an= (1)n,bn= (1)n+1,n = 1,2,. Then we have an+ bn= 0,n = 1,2,. But limsup n an= limsup n bn= 1. (c)Because an bnfor all n, then we have inf kn(ak) infkn bk,n = 1,2,. By the defi nations of the lower limits, it follows liminf n an liminf n bn. 5. (a)Supose f : X ,+ and g :
6、 X ,+ are measurable. Prove that the sets x : f(x) 0, then we can prove that the strict inequality in the Fatous lemma can hold. 10.Suppose that (X) 0 such that |fn(x)| M,|f(x)| M on X,n = 1,2,. By the Lebesgues dominated convergence theorem, we have lim n Z X fnd = Z X fd. counterexample: Let X = (
7、,+), and we defi ne fn= 1 n on X,n = 1,2,. 12. Suppose f L1(). Proved that to each 0 there exists a 0 such that R E |f|d 0, satisfying 0 s(x) M on X, and let = 2M, we have Z E |f|d Z E sd + 2 Mm(E) + 2 r x : f2(x) r and f1and f2are upper semicontinuous, therefore f1+ f2is upper semicon- tinuous. (c)
8、 This conclusion is false. counterexample: for each n N, defi ne fn= ( 1;x = rn 0;others 1 which rn are the all rationals on R1. Then fn is a sequence of real non- negative upper semicontinuous functions on R1, moreover we have f(x) = X 1 fn(x) = Q. But f(x) is not upper semicontinuous. (d) Accordin
9、g to the conclusion of (b), it is easy to obtain. If the word ”nonnegative” is omitted, (a) and (b) are still true but (c) and (d) is not. The truth of the statements is not aff ected if R1is replaced by a general topological space. 2. Let f be an arbitrary complex function on R1 , and defi ne (x,)
10、= sup|f(s) f(t)| : s,t (x ,x + ) (x) = inf(x,) : 0 Prove that is upper semicontinuous, that f is continuous at a point x if and only if (x) = 0, and hence that the set of points of continuity of an arbitrary complex function is a G. proof: It is obvious that (x) = inf(x, 1 k ) : k N 0 and (x, 1 k )
11、(x, 1 k + 1), k N For any 0, we will prove that the set x : (x) 0 such that (x,) N,xn= x;and|yn y| 0. 8 and for any 0 0. counterexample: (x) = x2is convex on (0,) but ln(x) = 2lnx is not convex. Moreover, for 0, it can show that the convexity of logc(c 1) implies the convexity of . 3. Assume that is
12、 a continuous real function on (a,b) such that (x + y 2 ) 1 2(x) + 1 2(y) for all x and y (a,b). Prove that is convex. Proof: According to the defi nition of convex function,we only need to prove the case 0 0 for any 0 0. For any p r, 0 2) Z X |f|pd = kfkp p. Therefore, lim n kfkp= +. Thirdly, assum
13、e that kfk= 1 and kfkr 0. Without generality, assume that |f(x)| 1,x X. Thus For any p r, it shows that kfkp p kfk r r. That is kfkp kfk r p r. And so lim p kfkp= 1 = kfk. The general case can reduce to the case kfk= 1 and kfkr 0. This is completed the proof. 5. Assume, in addition to the hypotheses
14、 of Exercise 4, that (X) = 1. (a) Prove that kfkr kfks, if 0 0 such that kfn fkp N. It shows by the Jensens Inequality that |Fn(x) F(x)| 1 x Rx 0 |fn(t) f(t)|dt) 1 x( Rx 0 |fn(t) f(t)|pdt) 1 p x 1 q x 1 pkfn fkp and thus Fn(x) F(x), n , x (0,). 10 Since kFnkp p p 1kfnkp, n = 1,2, it shows that Fn Lp
15、and Fn Lpis Cauchy sequence. By the completion of Lp, it exists G(x) Lpsatisfying kFn Gkp 0,n and so kFnkp kGkp G 2 when x N. Thus Z 0 F(x)dx Z N 1 x Z x 0 f(t)dt G 2 Z N 1 x dx = + 11 and therefore F / L1. 18.Let be a positive measure on X.A sequence fn of complex measurable functions on X is said to converge in measure to the measurable function f if to every 0 there corresponds an N su
