实变函数总复习题new

1、总复习题1. 证明: . (P.3) Proof. Let then for any there exists a such that , thus cannot be finite, conversely, if is infinite then for any there exists a such that , thus and , showing the first formula. Again let then there exists a such that , thus for any , thereby is finite, conversely if is finite, t

2、hen let , for any , so , and , showing the second formula. 2. 证明: . (P.11) Proof. is finite . is infinite . 3. 证明Cantor集是可测的,并且其测度为0.（P. 66）Proof. By the construction of Cantors set , first step take away one open interval of length , second step take away two open intervals and of length , third st

3、ep take away open intervals and of length , -th step take away open intervals of length , we see is measurable and . 4. (1) 设是中的开集, 是零测集. 证明: (提示: Let then since , there exists a such that , thus and ). (2) 写出Caratheodory条件. () (3) Let be an outer measure generated by the measure of ring spanned by

4、the left closed and right open intervals on straight line . Show for any and ,. (4) 设是直线上的Lebusgue测度, 证明对任意可测集, 有 . (P. 67) (1) Proof. Obviously, . Let then there exists a making . Since and for any there exists a and . Thus and . Since , and , showing . (3) Proof. Firstly we know the set of the dis

5、joint union of finite left closed and right open intervals. Let then . Obviously, . Let where then there exists a such that , thus for some ,and , showing . For any , since , thereby , showing . (4) Proof. Obviously, for and ,. Since for any . So is L-measurable and . 5. 设 , 证明. (P. 82) Proof. We mi

6、ght assume as well in and in where . In we have . Since , and a.e. 6. 设 =讨论: (1) 是否几乎处处收敛? (2) 是一致收敛? (3) 是否依测度收敛? (P. 82) Solution. (1) Let then in ,. (2) If is uniformly convergent to then naturally its also convergent to and so by (1). For and any , let be a rational number greater than then , a

7、contradiction. So is not uniformly convergent in . (3) If then there exists a sequence making . (4) Since and . But for any , contrary to . 7. 设在上可积并且一致连续, 证明. (P. 103) Proof. Since is L-measurable on , also is . If , then there exists a such that for any there exists a . Selecting we might assume .

8、 Let then there exists a such that as long so ,. Then , and , a contradiction. 8. 设在上可积. 对任意上的有界函数, 有, 证明 (a.e) (提示: Define , then ). (P. 103) Proof. Let then is bounded and . It follows that a.e and a.e. 9. 设在上可测, 问是否一定可积? (P. 103). Answer. Maybe not. For example, for , is obviously integrable on b

9、ut is not measurable on . Also let , then , obviously, but for . 10. 设,是可积函数, 证明: 若, 则. (P104) Proof. For any , let then . Since and . So . 11. 设都取为Lebusgue测度, 作证明: (1) 的两个累次积分存在且相等. (2) 在上不是Lebusgue可积. Proof. Since and are both continuous, they are Riemann integrable and naturally Lebesgue integrab

10、le. Since their integrand are both odd, their value can only equal to zero, thereby two successive integrations . But is not Lebesgue integrable in , if not, by integrability of in , we see also be integrable in , therefore there would exists the integral of second order. But its invalid, since for

11、, and for , if , which is not Lebesgue integrable in . 12. 设 证明: . (P. 113) Proof. Let , since , if then there existsthe limit where . Then there exists an and such that for . For , and the conclusion is obvious. For , let then for , and , a contradiction, so and , a contradiction, completing the verification. 13. Let , and are nonnegativ