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