# Thread: A question on Jordan content

1. ## 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!

2. 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 $g({\bf x})=\left\{ \begin{array}{ll}
f({\bf x}) & {\bf x}\in S \\
0 & {\bf x}\in I-S \\
\end{array} \right.$
where I is the closed interval containing S, then $\int_S f=\int_I g$. 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"