#### 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.

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) \coloneqq
\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) \coloneqq \int_R \chi_S ,
\end{equation*}

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

A bounded set \(S \subset \R^n\) is Jordan measurable if and only if the boundary \(\partial S\) is a measure zero set.

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.

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.

If \(S \subset \R^n\) is Jordan measurable, then \(V(S) = m^*(S)\text{.}\)

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.

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.

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) \coloneqq
\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 \coloneqq \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 functions 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*}

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 \(S\text{.}\)

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.

We say some property *for almost every \(x\)* if it holds for all \(x\) except on a set of measure zero. We can also just say that it happens *almost everywhere*. For example, we say \(f \colon S \to \R\) and \(g \colon S \to \R\) are equal almost everywhere if there exists a measure zero set \(E \subset S\) such that \(f(x) = g(x)\) for all \(x \in S \setminus E\text{.}\)

Many of the standard properties of the integral just carry over easily since we are really integrating over a rectangle. Furthermore, we can make some of the statements to be *almost everywhere*. Proofs of the following three propositions left as exercises.

Suppose \(S \subset \R^n\) is a bounded Jordan measurable set and \(f \colon S \to \R\) and \(g \colon S \to \R\) are Riemann integrable on \(S\text{,}\) and \(\alpha \in \R\text{.}\) Then

- If \(f = 0\) almost everywhere, then \(\int_S f = 0\text{.}\)
- If \(f = g\) almost everywhere, then \(\int_S f = \int_S g\text{.}\)
- \(f+g\) is Riemann integrable on \(S\) and \(\int_S (f+g) = \int_S f + \int_S g\text{.}\)
- \(\alpha f\) is Riemann integrable on \(S\) and \(\int_S \alpha f = \alpha \int_S f\text{.}\)
- If \(f(x) \leq g(x)\) for almost every \(x\text{,}\) then \(\int_S f \leq \int_S g\text{.}\)

We also have additivity.

Suppose \(A \subset \R^n\) and \(B \subset \R^n\) are disjoint bounded Jordan measurable sets and \(f \colon A \cup B \to \R\) is such that the restrictions \(f|_A\) and \(f|_B\) are Riemann integrable on \(A\) and \(B\) respectively. Then \(f\) is Riemann integrable on \(A \cup B\) and

\begin{equation*}
\int_{A \cup B} f = \int_A f + \int_B f .
\end{equation*}

Finally, to integrate over non-rectangular regions using Fubini’s theorem, the typical way is to cut the region into simpler pieces that can be described by two graphs. We state the theorem in the plane, but similar statements can be made in more variables. The proof is again left as an exercise.

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 \coloneqq \bigl\{ (x,y) \in \R^2 : a < x < b \text{ and } f(x) < y < g(x) \bigr\} .
\end{equation*}

See Figure 10.13. Then \(U\) is Jordan measurable, and if \(\varphi \colon U \to \R\) is Riemann integrable on \(U\text{,}\) then

\begin{equation*}
\int_U \varphi =
\int_a^b \int_{f(x)}^{g(x)} \varphi(x,y) \, dy \, dx .
\end{equation*}

Finally, images of Jordan measurable sets are Jordan measurable under nice enough mappings. For simplicity, we assume that the Jacobian determinant never vanishes.

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 one-to-one continuously differentiable mapping such that the Jacobian determinant \(J_g\) is never zero on \(S\text{.}\) Then \(g(S)\) is bounded and Jordan measurable.

Let \(T \coloneqq g(S)\text{.}\) 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 \(\{ y_k \}_{k=1}^\infty\) in \(W\) with \(\lim_{k\to\infty} y_k = y\) and \(y_k \notin T\text{.}\) As \(g|_V\) is invertible and in particular has a continuous inverse, there exists a sequence \(\{ x_k \}_{k=1}^\infty\) in \(V\) such that \(g(x_k) = y_k\) and \(\lim_{k\to\infty} x_k = x\text{.}\) 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, \(\partial T \subset
g(\partial S)\text{.}\)

Prove Proposition 10.5.2.

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

Prove Proposition 10.5.8. That is,

- Show that \(U\) is Jordan measurable.
- Prove that \(\int_U \varphi = \int_a^b \int_{f(x)}^{g(x)} \varphi(x,y) \, dy \, dx\text{.}\)

Let us construct an example of a non-Jordan measurable open set. Start in one dimension. Let \(\{ r_j \}_{j=1}^\infty\) 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 \(U \coloneqq
\bigcup_{j=1}^\infty (a_j,b_j)\text{.}\)

- 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 \(W \coloneqq \bigl( U \times (0,2) \bigr) \cup \bigl( (0,1) \times (1,2) \bigr)\) is a connected bounded open set in \(\R^2\) that is not Jordan measurable.

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.

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 \(S\)).

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

Prove Proposition 10.5.6.

Prove Proposition 10.5.7.

Named after the French mathematician Marie Ennemond Camille Jordan (1838–1922).