## Section10.5Jordan measurable sets

Note: 1 lecture

### Subsection10.5.1Volume 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

\begin{equation*} \chi_S(x) := \begin{cases} 1 & \text{if } x \in S, \\ 0 & \text{if } x \notin S. \end{cases} \end{equation*}

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

\begin{equation*} \int_R \chi_S = \int_{R'} \chi_S = \int_{R \cap R'} \chi_S. \end{equation*}

We define the $$n$$-dimensional volume of the bounded Jordan measurable set $$S$$ as

\begin{equation*} V(S) := \int_R \chi_S , \end{equation*}

where $$R$$ is any closed rectangle containing $$S\text{.}$$

#### 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.

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.

#### Proof.

Given $$\epsilon > 0\text{,}$$ let $$R$$ be a closed rectangle that contains $$S\text{.}$$ Let $$P$$ be a partition of $$R$$ such that

\begin{equation*} U(P,\chi_S) \leq \left( \int_R \chi_S \right) + \epsilon = V(S) + \epsilon \qquad \text{and} \qquad L(P,\chi_S) \geq \left( \int_R \chi_S \right) - \epsilon = V(S)-\epsilon. \end{equation*}

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

\begin{equation*} U(P,\chi_S) = \sum_{j=1}^k V(R_j) > \left(\sum_{j=1}^k V(O_j)\right) - \epsilon \geq m^*(S) - \epsilon . \end{equation*}

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,

\begin{equation*} m^*\Bigl(\bigcup_{j=1}^\ell R'^\circ_j\Bigr) = \sum_{j=1}^\ell V(R'^\circ_j) . \end{equation*}

Hence

\begin{equation*} m^*(S) \geq m^*\Bigl(\bigcup_{j=1}^\ell R'_j\Bigr) \geq m^*\Bigl(\bigcup_{j=1}^\ell R'^\circ_j\Bigr) = \sum_{j=1}^\ell V(R'^\circ_j) = \sum_{j=1}^\ell V(R'_j) = L(P,f) \geq V(S) - \epsilon . \end{equation*}

Therefore $$m^*(S) \geq V(S)$$ as well.

### Subsection10.5.2Integration 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.

#### Definition10.5.4.

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

\begin{equation*} \widetilde{f}(x) := \begin{cases} f(x) & \text{if } x \in S, \\ 0 & \text{otherwise}, \end{cases} \end{equation*}

is in $$\sR(R)\text{.}$$ In this case we write

\begin{equation*} \int_S f := \int_R \widetilde{f}. \end{equation*}

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

\begin{equation*} \int_S f = \int_R f \chi_S . \end{equation*}

#### 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.

### Subsection10.5.3Images 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.

#### 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 one-to-one 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, $$\partial T \subset g(\partial S)\text{.}$$

### Subsection10.5.4Exercises

#### Exercise10.5.2.

Prove that a bounded convex set is Jordan measurable. Hint: Induction on dimension.

#### Exercise10.5.3.

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

\begin{equation*} U := \bigl\{ (x,y) \in \R^2 : a < x < b \text{ and } f(x) < y < g(x) \bigr\} . \end{equation*}
1. Show that $$U$$ is Jordan measurable.

2. 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*}

#### Exercise10.5.4.

Let us construct an example of a non-Jordan 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 $$\sum_{j=1}^\infty (b_j-a_j) < \nicefrac{1}{2}\text{.}$$ Now let $$U := \bigcup_{j=1}^\infty (a_j,b_j)\text{.}$$

1. Show the open intervals $$(a_j,b_j)$$ as above actually exist.

2. Prove $$\partial U = [0,1] \setminus U\text{.}$$

3. Prove $$\partial U$$ is not of measure zero, and therefore $$U$$ is not Jordan measurable.

4. 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.

#### Exercise10.5.5.

Suppose $$K \subset \R^n$$ is a closed measure zero set.

1. If $$K$$ is bounded, prove that $$K$$ is Jordan measurable.

2. If $$S \subset \R^n$$ is bounded and Jordan measurable, prove that $$S \setminus K$$ is Jordan measurable.

3. 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.

#### Exercise10.5.6.

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 $$S$$).

#### Exercise10.5.7.

Prove a version of Corollary 10.4.4, replacing all closed rectangles with closed and bounded Jordan measurable sets.

Named after the French mathematician Marie Ennemond Camille Jordan 2  (1838–1922).
https://en.wikipedia.org/wiki/Camille_Jordan
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