Jordan Measure

Subjects: Vector Analysis
Links: Riemann Integral in Rn, Lebesgue Measure in Rn

Def: Let $A\subseteq {\mathbb{R}}^n$ and define ${\chi }_A:{\mathbb{R}}^n\to \mathbb{R}$, the function characteristic of $A$, as follows

Def: Let $A\subseteq {\mathbb{R}}^n$ bounded, we say that $A$ is Jordan-measurable if the characteristic function of $A$ is integrable over some rectangle $R$, that contains $A$. In this case we say that the Jordan measure of $A$ is (denoted by $J(A)$)

Def: In the context of Jordan measure we can define the Jordan outer-measure as

Similarly we can define the Jordan inner measure as

Th: Let $A\subseteq R^{\prime }\subseteq R$, where $R$ and $R^{\prime }$ be rectangles, if ${\chi }_A$ is integrable over $R$, then ${\chi }_A$ is integrable over $R^{\prime }$ and ${\int }_R{\chi }_A={\int }_{R^{\prime }}{\chi }_A$. If $A\subseteq R^{\prime },R$ and ${\chi }_A$ be integrable over $R$ and $R^{\prime }$, then ${\int }_R{\chi }_A={\int }_{R^{\prime }}{\chi }_A$.

This ensures that the Jordan measure of $A$ is well defined.

Th: Let $A\subseteq {\mathbb{R}}^n$ be bounded. Then all of the following are equivalent

• $A$ is Jordan measurable
• for any $\epsilon >0$, there’s $R_1,\dots ,R_k$ rectangles such that
  • $\partial A\subseteq \bigcup _{i=1}^k{R}_i$
  • $\sum _{i=1}^{k}m(R_i)<\epsilon$
• $\partial A$ is Jordan measurable and $J(\partial A)=0$

Cor: Let $A\subseteq {\mathbb{R}}^n$ be a bounded set. $A$ is Jordan-measurable and $J(A)=0$ iff for any $\epsilon >0$ there’s $R_1,\dots ,R_k$ rectangles such that

• $A\subseteq \bigcup _{i=1}^k{R}_i$
• $\sum _{i=1}^{k}m(R_i)<\epsilon$

Lemma: Some basic properties are if $A\subseteq {\mathbb{R}}^n$ be bounded and Jordan-measurable:

• If $B\subseteq {\mathbb{R}}^n$ is Jordan measurable with $J(A)=J(B)=0$, then $A\cup B$ is Jordan-measurable and $J(A\cup B)=0$.

• If $B\subseteq A$, and $J(A)=0$, then $B$ is Jordan-measurable and $J(B)=0$.

• This pair of results help us show that the set of Jordan-measurable sets is an algebra, although not a $\sigma$-algebra
  If $A,B\subseteq {\mathbb{R}}^n$ are Jordan-measurable then the following sets are Jordan measurable:

  • $A\cup B$
  • $A\cap B$
  • $A\setminus B$
    And we have the following additional properties:
  • $J(A\cup B)+J(A\cap B)=J(A)+J(B)$
  • $J(A\setminus B)=J(A)-J(A\cap B)$
  • If $B\subseteq A$, then $J(B)\le J(A)$
    If $A\subseteq B\subseteq C\subseteq {\mathbb{R}}^n$ such that $A,C$ are Jordan-measurable with $J(A)=J(C)$, then $B$ is Jordan-measurable and $J(A)=J(B)=J(C)$.

• Some topological properties
  If $A$ is Jordan-measurable then

  • $J(A)>0$ iff $\mathrm{int}(A)\ne \varnothing$
  • $J(A)=0$ iff $A\subseteq \partial A$
  • $\mathrm{int}(A)$ and $\mathrm{cl}(A)$ are Jordan-measurable and $J(\mathrm{int}(A))=J(A)=J(\mathrm{cl}(A))$
  • For any $\epsilon >0$, there’s a compact set $K\subseteq \mathrm{int}(A)$ such that $K$ is Jordan-measurable and $J(A)-J(K)<\epsilon$

Lemma: Let $A$ be a Jordan-measurable. If $f:A\subseteq {\mathbb{R}}^n\to \mathbb{R}$ is continuous, then the set $G_f=\{(x,f(x))\in {\mathbb{R}}^{n+1}\mid x\in A\}$, is Jordan-measurable and $J(G_f)=0$

Th: Let $A$ be a Jordan-measurable. If $f,g:A\subseteq {\mathbb{R}}^n\to \mathbb{R}$ be continuous, such that for any $x\in A$, $f(x)\le g(x)$, then the set $D=\{(x,y)\in {\mathbb{R}}^{n+1}\mid x\in A\wedge f(x)\le y\le g(x)\}$ is Jordan-measurable.

Th: Let $f:R\subset {\mathbb{R}}^n\to \mathbb{R}$ bounded over the rectangle $R$. If the set of $f$ discontinuities over $R$, $D(f,R)$ is Jordan-measurable. Then $J(D(f,R))=0$, iff $f$ is integrable over $R$.