$\newcommand{\cA}{ {\cal A} }$ $\newcommand{\bR}{ {\mathbb{R}} }$ $\newcommand{\cM}{ {\cal M} }$ $\newcommand{\cP}{ {\cal P} }$

Given the following definition, I would like to prove the theorem below.

Definition 1. Let $\cA$ be a set of subsets of $\bR^n$.
We say that $\cA$ is a {\em field of subsets of $\bR^n$} when it has the following properties:

(FS-1) The empty set $\emptyset$ belongs to $\cA$.

(FS-2) If $A,B \in \cA$, then the set-difference $A \setminus B$ still belongs to $\cA$.

(FS-3) If $A_1, \ldots , A_k \in \cA$, then the union $A_1 \cup \cdots \cup A_k$ and the intersection $A_1 \cap \cdots \cap A_k$ still belong to $\cA$.

Definition 2. A set $A \subseteq \bR^n$ is Jordan measurable when it satisfies the following two conditions (i) and (ii):

(i) $A$ is a bounded set;

(ii) the boundary bd$(A)$ is a zero-content set.

The collection of all Jordan measurable subsets of $\bR^n$ will be denoted by $\cM_n$.

Definition 3 The collection of all half-open rectangles in $\bR^n$ will be denoted by $\cP_n$.

A volume function $\mbox{vol}_o : \cP_n \to [0, \infty )$, where for $P = ( a_1, b_1 ] \times \cdots \times ( a_n, b_n ] \in \cP_n$ we put $\mbox{vol}_o (P) := (b_1 -a_1) \cdots (b_n -a_n)$.

Theorem. $\cM_n $, the collection containing all Jordan measurable sets in $\bR^n$ is a field of subsets of $\bR^n$, in the sense of Definition 1.

My approach:

$\textbf{(FS-1)}$ This follows trivially.

$\textbf{(FS-2)}$ Let $A,B\in \cM_n$. If $A\cap B=\emptyset$ then $A\setminus B = A$; so that $A\setminus B\in \cM_n$.

Here's where I'm stuck: if $A\setminus B \ne \emptyset$ then $A\setminus B$ is bounded, since $A$ and $B$ are bounded. But how does one prove that the boundary of $A\setminus B$ has zero content (measure 0)?

$\textbf{(FS-3)}$ Let $A_1, \dots, A_k \in \cM_n$, then $U:=A_1\cup \dots \cup A_k$ is a bounded set, since a finite union of bounded sets is bounded. Choose $\varepsilon_1, \dots, \varepsilon_k>0$, let $\varepsilon=\varepsilon_1+\dots+\varepsilon_k$ and let let $P:=P_{1,i} \cup \dots \cup P_{l,i} $ (for $1\le i \le k$) be an $\varepsilon$-cover of $A_i$. Then vol$(P) \le \sum\limits_{j=1}^k \sum\limits_{i=1}^l \mbox{vol}(P_{j,i})<\varepsilon_1 +\dots+\varepsilon_k=\varepsilon$. Hence, a finite union of Jordan measurable sets is Jordan measurable.

For $A\cap B$, since bd$(A)\cap$ bd$(B)\subset$ bd$(A)\cup$ bd$(B)\subset A\cup B$, and $A\cup B$ is Jordan measurable, $A\cap B$ is also Jordan measurable.

Please let me know if you think my proof is correct or not. Please also let me know how to prove $\textbf{(FS-2)}$ above.

  • $\begingroup$ $A \setminus B$ is bounded by $A$ being bounded alone, as $A \setminus B \subseteq A$. You haven't defined zero-content sets BTW. $\endgroup$ Jul 17 '17 at 7:23

Using ($\partial A= \operatorname{bd}(A)$):

$$\partial (A \cap B) \subseteq \partial A \cap \partial B$$ and $$\partial (A \cup B) \subseteq \partial A \cap \partial B$$

See this answer for proofs... They extend by induction to finite unions and intersections as usual. Then:

$$\partial (A \setminus B) = \partial (A \cap (X\setminus B)) \subseteq \partial A \cup \partial (X\setminus B) = \partial A \cup \partial B$$

and prove the easy but useful lemma that a finite union of zero-content sets has zero-content.

This implies that $\partial (A \setminus B)$ has zero content, when $\partial A$ and $\partial B$ have.

$A \setminus B \subseteq A$ implies $A \setminus B $ is also bounded.

This takes care of the difference condition.

The finite unions and finite intersections also use the finite unions of zero-content sets and the top formulae. Boundedness is indeed clear.

I don't see, however, how $\partial A \cup \partial \subseteq A \cup B $ holds (it's actually false) and how $\partial A \cap \partial B \subset A \cup B$ would help proving Jordan-measurability of anything. You just have to check boundedness and zero-content boundary, which can be done as I did above.


A\B is a subset of A. Given the definition of boundary, what can you say about the boundary of A\B and it's measure. Hint: measures are monotone.

  • $\begingroup$ What does it mean for a measure to be monotone? $\endgroup$
    – sequence
    Jul 17 '17 at 5:35
  • 1
    $\begingroup$ It means that if A is a subset of B, then the measure of A is less than or equal to the measure of B. $\endgroup$ Jul 17 '17 at 7:10
  • $\begingroup$ There is no measure here; there Is the family of zero-content sets that is closed under finite unions (not under countable ones). $\endgroup$ Jul 17 '17 at 8:14

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