If A is Jordan-measurable with Jordan content c(A), and is of content zero. Prove that A-B is also Jordan-measurable with the Jordan-content equal to c(A). Thanks!
I finally proved it. First we prove A-B is Jordan-measurable. This is equivalent to showing that has content zero and it suffices to show , here means boundary. Let , then . If , here Int means interior, then and we get what we want. If , we try to prove that . Supposing , this means there is a ball not intersecting B. Because , there is an another ball . Take , then for the ball , any element of it satisfies both and , that is, , so , but this contradicts the supposition , so . On the other hand, c(B)=0 implies B contains no interior point, otherwise, there would be a ball in square metric, or a subinterval, lying completely in Int B, then would >0 and this would in turn leads to c(B)>0, a contradiction of the hypothesis. Therefore , so , as desired.
Secondly let me prove c(A-B)=0. For this purpose, I introduce two lemmas:
1)If S is of Jordan-content zero, f is defined and bounded on S, then exists and equals 0.
Proof. By hypothesis, S is of measure 0, so discontinuities of f in S, as a subset of a content 0 set, is content 0 and therefore measure 0, so f is integrable on S . Since c(S)=0, there is a partition P some finite subintervals (denote it as A) of which cover and the sum of measures of these subintervals < any given . Let where I is the closed interval containing S, then . Assume , then and . So . So by arbitraryness of . Similar argument can be applied to get . So , that is, .
2)If S is Jordan-measurable, then exists and equals c(S). This is an extension of [2] where S is required to be compact.
Proof. It is easy to see exists since . Let be characteristic function of S and P any partition. Since those subintervas contributing to contain elements of S and therefore elements of , they must also contribute to . But the converse is not true, since it is likely that some contains elements of but no element of S. So . Since these two numbers are both indexed by P, we can impose inf on both sides and get . On the other hand, for those subintervals contributing to , since they are all contained in S, assumes 1 throughout and as a result. So these subintervals also contribute to . But the converse is not true since there may be which is contained completely in S but contains points not belonging to Int S, in this situation, is still 1, but is not counted in . So . Taking sup on both sides, we obtain . From above we have , but , so .
Now the work becomes easy. Since A is Jordan-measurable, by Lemma 2) we have exists. Note that A-B and B are both Jordan-measurable, by the additive property of the Riemann integral , , in which by Lemma 2), by Lemma 1) and . The conclusion is thus immediate.
If there is any mistakes or there is any other good proof, please kindly point out. PS: Just as it is greatly beneficial to study general topology before studying mathematical analysis, it is greatly beneficial to have some knowledge of measure theory before studying Integral Theory.
[1]Th 14.11 of Apostol's "Mathematical Analysis"
[2]Th 14.12 of Apostol's "Mathematical Analysis"
[3]Th 14.13 of Apostol's "Mathematical Analysis"