Section 10.5 Jordan measurable sets
Note: 1 lecture
Subsection 10.5.1 Volume and Jordan measurable sets
Given a set \(S \subset \R^n\text{,}\) its characteristic function or indicator function \(\chi_S \colon \R^n \to \R\) is defined by
A bounded set \(S\) is Jordan measurable^{ 1 } if for some closed rectangle \(R\) such that \(S \subset R\text{,}\) the function \(\chi_S\) is Riemann integrable, that is, \(\chi_S \in \sR(R)\text{.}\) Take two closed rectangles \(R\) and \(R'\) with \(S \subset R\) and \(S \subset R'\text{,}\) then \(R \cap R'\) is a closed rectangle also containing \(S\text{.}\) By Proposition 10.1.13 and Exercise 10.1.7, \(\chi_S \in \sR(R \cap R')\) and so \(\chi_S \in \sR(R')\text{.}\) Thus
We define the \(n\)dimensional volume of the bounded Jordan measurable set \(S\) as
where \(R\) is any closed rectangle containing \(S\text{.}\)
Proposition 10.5.1.
A bounded set \(S \subset \R^n\) is Jordan measurable if and only if the boundary \(\partial S\) is a measure zero set.
Proof.
Suppose \(R\) is a closed rectangle such that \(S\) is contained in the interior of \(R\text{.}\) If \(x \in \partial S\text{,}\) then for every \(\delta > 0\text{,}\) the sets \(S \cap B(x,\delta)\) (where \(\chi_S\) is 1) and the sets \((R \setminus S) \cap B(x,\delta)\) (where \(\chi_S\) is 0) are both nonempty. So \(\chi_S\) is not continuous at \(x\text{.}\) If \(x\) is either in the interior of \(S\) or in the complement of the closure \(\widebar{S}\text{,}\) then \(\chi_S\) is either identically 1 or identically 0 in a whole neighborhood of \(x\) and hence \(\chi_S\) is continuous at \(x\text{.}\) Therefore, the set of discontinuities of \(\chi_S\) is precisely the boundary \(\partial S\text{.}\) The proposition follows.
Proposition 10.5.2.
Suppose \(S\) and \(T\) are bounded Jordan measurable sets. Then
The closure \(\widebar{S}\) is Jordan measurable.
The interior \(S^\circ\) is Jordan measurable.
\(S \cup T\) is Jordan measurable.
\(S \cap T\) is Jordan measurable.
\(S \setminus T\) is Jordan measurable.
The proof of the proposition is left as an exercise. Next, we find that the volume that we defined above coincides with the outer measure we defined above.
Proposition 10.5.3.
If \(S \subset \R^n\) is Jordan measurable, then \(V(S) = m^*(S)\text{.}\)
Proof.
Given \(\epsilon > 0\text{,}\) let \(R\) be a closed rectangle that contains \(S\text{.}\) Let \(P\) be a partition of \(R\) such that
Let \(R_1,R_2,\ldots,R_k\) be all the subrectangles of \(P\) such that \(\chi_S\) is not identically zero on each \(R_j\text{.}\) That is, there is some point \(x \in R_j\) such that \(x \in S\) (i.e., \(\chi_S(x)=1\)). Let \(O_j\) be an open rectangle such that \(R_j \subset O_j\) and \(V(O_j) < V(R_j) + \nicefrac{\epsilon}{k}\text{.}\) Notice that \(S \subset \bigcup_j O_j\text{.}\) Then
As \(U(P,\chi_S) \leq V(S) + \epsilon\text{,}\) then \(m^*(S)  \epsilon \leq V(S) + \epsilon\text{,}\) or in other words \(m^*(S) \leq V(S)\text{.}\)
Let \(R'_1,R'_2,\ldots,R'_\ell\) be all the subrectangles of \(P\) such that \(\chi_S\) is identically one on each \(R'_j\text{.}\) In other words, these are the subrectangles contained in \(S\text{.}\) The interiors of the subrectangles \(R'^\circ_j\) are disjoint and \(V(R'^\circ_j) = V(R'_j)\text{.}\) Via Exercise 10.3.16,
Hence
Therefore \(m^*(S) \geq V(S)\) as well.
Subsection 10.5.2 Integration over Jordan measurable sets
In \(\R\) there is only one reasonable type of set to integrate over: an interval. In \(\R^n\) there are many kinds of sets. The ones that work with the Riemann integral are the Jordan measurable sets.
Definition 10.5.4.
Let \(S \subset \R^n\) be a bounded Jordan measurable set. A bounded function \(f \colon S \to \R\) is said to be Riemann integrable on \(S\), or \(f \in \sR(S)\text{,}\) if for a closed rectangle \(R\) such that \(S \subset R\text{,}\) the function \(\widetilde{f} \colon R \to \R\) defined by
is in \(\sR(R)\text{.}\) In this case we write
When \(f\) is defined on a larger set and we wish to integrate over \(S\text{,}\) then we apply the definition to the restriction \(f_S\text{.}\) As the restriction can be defined by the product \(f\xi_S\text{,}\) and the product of Riemann integrable sets is Riemann integrable, \(f_S\) is automatically Riemann integrable. In particular, if \(f \colon R \to \R\) for a closed rectangle \(R\text{,}\) and \(S \subset R\) is a Jordan measurable subset, then
Proposition 10.5.5.
If \(S \subset \R^n\) is a bounded Jordan measurable set and \(f \colon S \to \R\) is a bounded continuous function, then \(f\) is integrable on \(S\text{.}\)
Proof.
Define the function \(\widetilde{f}\) as above for some closed rectangle \(R\) with \(S \subset R\text{.}\) If \(x \in R \setminus \widebar{S}\text{,}\) then \(\widetilde{f}\) is identically zero in a neighborhood of \(x\text{.}\) Similarly if \(x\) is in the interior of \(S\text{,}\) then \(\widetilde{f} = f\) on a neighborhood of \(x\) and \(f\) is continuous at \(x\text{.}\) Therefore, \(\widetilde{f}\) is only ever possibly discontinuous at \(\partial S\text{,}\) which is a set of measure zero, and we are finished.
Subsection 10.5.3 Images of Jordan measurable subsets
Finally, images of Jordan measurable sets are Jordan measurable under nice enough mappings. For simplicity, we assume that the Jacobian determinant never vanishes.
Proposition 10.5.6.
Suppose \(U \subset \R^n\) is open and \(S \subset U\) is a compact Jordan measurable set. Suppose \(g \colon U \to \R^n\) is a onetoone continuously differentiable mapping such that the Jacobian determinant \(J_g\) is never zero on \(S\text{.}\) Then \(g(S)\) is bounded and Jordan measurable.
Proof.
Let \(T := g(S)\text{.}\) By Lemma 7.5.5, the set \(T\) is also compact and so closed and bounded. We claim \(\partial T \subset g(\partial S)\text{.}\) Suppose the claim is proved. As \(S\) is Jordan measurable, then \(\partial S\) is measure zero. Then \(g(\partial S)\) is measure zero by Proposition 10.3.10. As \(\partial T \subset g(\partial S)\text{,}\) then \(T\) is Jordan measurable.
It is therefore left to prove the claim. As \(T\) is closed, \(\partial T \subset T\text{.}\) Suppose \(y \in \partial T\text{,}\) then there must exist an \(x \in S\) such that \(g(x) = y\text{,}\) and by hypothesis \(J_g(x) \not= 0\text{.}\) We use the inverse function theorem (Theorem 8.5.1). We find a neighborhood \(V \subset U\) of \(x\) and an open set \(W\) such that the restriction \(f_V\) is a onetoone and onto function from \(V\) to \(W\) with a continuously differentiable inverse. In particular, \(g(x) = y \in W\text{.}\) As \(y \in \partial T\text{,}\) there exists a sequence \(\{ y_k \}\) in \(W\) with \(\lim y_k = y\) and \(y_k \notin T\text{.}\) As \(g_V\) is invertible and in particular has a continuous inverse, there exists a sequence \(\{ x_k \}\) in \(V\) such that \(g(x_k) = y_k\) and \(\lim x_k = x\text{.}\) Since \(y_k \notin T = g(S)\text{,}\) clearly \(x_k \notin S\text{.}\) Since \(x \in S\text{,}\) we conclude that \(x \in \partial S\text{.}\) The claim is proved, \(\partial T \subset g(\partial S)\text{.}\)
Subsection 10.5.4 Exercises
Exercise 10.5.1.
Prove Proposition 10.5.2.
Exercise 10.5.2.
Prove that a bounded convex set is Jordan measurable. Hint: Induction on dimension.
Exercise 10.5.3.
Let \(f \colon [a,b] \to \R\) and \(g \colon [a,b] \to \R\) be continuous functions and such that for all \(x \in (a,b)\text{,}\) \(f(x) < g(x)\text{.}\) Let
Show that \(U\) is Jordan measurable.

If \(\varphi \colon U \to \R\) is Riemann integrable on \(U\text{,}\) then
\begin{equation*} \int_U \varphi = \int_a^b \int_{g(x)}^{f(x)} \varphi(x,y) \, dy \, dx . \end{equation*}
Exercise 10.5.4.
Let us construct an example of a nonJordan measurable open set. Start in one dimension. Let \(\{ r_j \}\) be an enumeration of all rational numbers in \((0,1)\text{.}\) Let \((a_j,b_j)\) be open intervals such that \((a_j,b_j) \subset (0,1)\) for all \(j\text{,}\) \(r_j \in (a_j,b_j)\text{,}\) and \(\sum_{j=1}^\infty (b_ja_j) < \nicefrac{1}{2}\text{.}\) Now let \(U := \bigcup_{j=1}^\infty (a_j,b_j)\text{.}\)
Show the open intervals \((a_j,b_j)\) as above actually exist.
Prove \(\partial U = [0,1] \setminus U\text{.}\)
Prove \(\partial U\) is not of measure zero, and therefore \(U\) is not Jordan measurable.
Show that \(W := \bigl( U \times (0,2) \bigr) \cup \bigl( (0,1) \times (1,2) \bigr)\) is a connected bounded open set in \(\R^2\) that is not Jordan measurable.
Exercise 10.5.5.
Suppose \(K \subset \R^n\) is a closed measure zero set.
If \(K\) is bounded, prove that \(K\) is Jordan measurable.
If \(S \subset \R^n\) is bounded and Jordan measurable, prove that \(S \setminus K\) is Jordan measurable.
Construct a bounded Jordan measurable \(S \subset \R^n\) and a bounded \(T \subset \R^n\) of measure zero, such that neither \(T\) nor \(S \setminus T\) is Jordan measurable.
Exercise 10.5.6.
Suppose \(U \subset \R^n\) is open and \(K \subset U\) is compact. Find a compact Jordan measurable set \(S\) such that \(S \subset U\) and \(K \subset S^\circ\) (\(K\) is in the interior of \(S\)).
Exercise 10.5.7.
Prove a version of Corollary 10.4.4, replacing all closed rectangles with closed and bounded Jordan measurable sets.
https://en.wikipedia.org/wiki/Camille_Jordan