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  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.
Th 14.11 of Apostol's "Mathematical Analysis"
Th 14.12 of Apostol's "Mathematical Analysis"
Th 14.13 of Apostol's "Mathematical Analysis"