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Math Help - A question on Jordan content

  1. #1
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    A question on Jordan content

    If A is Jordan-measurable with Jordan content c(A), B\subseteq A and is of content zero. Prove that A-B is also Jordan-measurable with the Jordan-content equal to c(A). Thanks!
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  2. #2
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    I finally proved it. First we prove A-B is Jordan-measurable. This is equivalent to showing that \partial(A-B) has content zero and it suffices to show \partial(A-B)\subseteq\partial A\cup\partial B, here \partial means boundary. Let {\bf x}\in\partial(A-B), then {\bf x}\in\overline{A-B}\subseteq\bar A. If {\bf x}\notin{\rm Int} A, here Int means interior, then {\bf x}\in\bar A-{\rm Int} A=\partial A and we get what we want. If {\bf x}\in {\rm Int}A, we try to prove that {\bf x}\in\partial B. Supposing {\bf x}\notin\bar B, this means there is a ball B({\bf x},\epsilon_1) not intersecting B. Because {\bf x}\in {\rm Int}A, there is an another ball B({\bf x},\epsilon_2)\subseteq A. Take \epsilon=\min(\epsilon_1,\epsilon_2), then for the ball B({\bf x},\epsilon), any element of it satisfies both \in A and \notin B, that is, \in A-B, so {\bf x}\in {\rm Int}(A-B), but this contradicts the supposition {\bf x}\in\partial(A-B), so {\bf x}\in\bar B. 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 \underline J(P,B) would >0 and this would in turn leads to c(B)>0, a contradiction of the hypothesis. Therefore {\bf x}\notin {\rm Int} B, so {\bf x}\in\bar B-{\rm Int} B=\partial B, 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 \int_S f 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 ^{[1]}. Since c(S)=0, there is a partition P some finite subintervals (denote it as A) of which cover \bar S and the sum of measures of these subintervals < any given \epsilon>0. Let g({\bf x})=\left\{ \begin{array}{ll}<br />
f({\bf x}) & {\bf x}\in S \\ <br />
0 & {\bf x}\in I-S \\ <br />
\end{array} \right. where I is the closed interval containing S, then \int_S f=\int_I g. Assume |f|\leq M, then M_k=\sup\{f({\bf x})|{\bf x}{\rm\;in\;subinterval\;}I_k\}\leq M and m_k=\inf\{f({\bf x})|{\bf x}{\rm\;in\;subinterval\;}I_k\}\geq -M. So U(P,g)=\sum\limits_A M_k\mu(I_k)\leq M\sum\limits_A\mu(I_k)\leq M\epsilon. So \bar\int\leq 0 by arbitraryness of \epsilon. Similar argument can be applied to get \underline\int\geq 0. So \underline\int=\bar\int=0, that is, \int_I g({\bf x})d{\bf x}=\int_S f({\bf x})d{\bf x}=0.
    2)If S is Jordan-measurable, then \int_S 1 exists and equals c(S). This is an extension of [2] where S is required to be compact.
    Proof. It is easy to see \int_S 1 exists since c(\partial S)=0. Let \chi_S be characteristic function of S and P any partition. Since those subintervas I_k contributing to U(P,\chi_S) contain elements of S and therefore elements of \bar S, they must also contribute to \bar J(P,S). But the converse is not true, since it is likely that some I_k contains elements of \bar S but no element of S. So U(P,\chi_S)\leq\bar J(P,S). Since these two numbers are both indexed by P, we can impose inf on both sides and get \bar\int_I\chi_S\leq\bar c(S). On the other hand, for those subintervals I_k contributing to \underline J(P,S), since they are all contained in S, \chi_S assumes 1 throughout and m_k=1 as a result. So these subintervals also contribute to L(P,\chi_S). But the converse is not true since there may be I_k which is contained completely in S but contains points not belonging to Int S, in this situation, m_k is still 1, but I_k is not counted in \underline J(P,S). So \underline J(P,S)\leq L(P,\chi_S). Taking sup on both sides, we obtain \underline c(S)\leq\underline\int_I\chi_S. From above we have \underline c(S)\leq\underline\int_I\chi_S\leq\bar\int_I\chi_S  \leq\bar c(S), but \underline c(S)=\bar c(S), so \int_I\chi_S=\int_S 1=c(S).
    Now the work becomes easy. Since A is Jordan-measurable, by Lemma 2) we have \int_A 1 exists. Note that A-B and B are both Jordan-measurable, by the additive property of the Riemann integral ^{[3]}, \int_A 1=\int_B 1+\int_{A-B} 1, in which \int_A 1=c(A) by Lemma 2), \int_B 1=0 by Lemma 1) and \int_{A-B} 1=c(A-B). 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"
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