## Volume and Jordan measurable sets

Given a bounded set \(S \subset {\mathbb{R}}^n\) its *characteristic function* or *indicator function* is \[\chi_S(x) := \begin{cases} 1 & \text{ if $x \in S$} \\ 0 & \text{ if $x \notin S$}. \end{cases}\] A bounded set \(S\) is said to be *Jordan measurable* if for some closed rectangle \(R\) such that \(S \subset R\), the function \(\chi_S\) is in \({\mathcal{R}}(R)\). Take two closed rectangles \(R\) and \(R'\) with \(S \subset R\) and \(S \subset R'\), then \(R \cap R'\) is a closed rectangle also containing \(S\). By and , \(\chi_S \in {\mathcal{R}}(R \cap R')\) and hence \(\chi_S \in {\mathcal{R}}(R')\), and furthermore \[\int_R \chi_S = \int_{R'} \chi_S = \int_{R \cap R'} \chi_S.\] We define the *\(n\)-dimensional volume* of the bounded Jordan measurable set \(S\) as \[V(S) := \int_R \chi_S ,\] where \(R\) is any closed rectangle containing \(S\).

A bounded set \(S \subset {\mathbb{R}}^n\) 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 \(R\). If \(x \in \partial S\), then for every \(\delta > 0\), 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 \(x\). If \(x\) is either in the interior of \(S\) or in the complement of the closure \(\overline{S}\), then \(\chi_S\) is either identically 1 or identically 0 in a whole neighbourhood of \(x\) and hence \(\chi_S\) is continuous at \(x\). Therefore, the set of discontinuities of \(\chi_S\) is precisely the boundary \(\partial S\). The proposition then follows.

The proof of the following proposition is left as an exercise.

[prop:jordanmeas] Suppose \(S\) and \(T\) are bounded Jordan measurable sets. Then

- The closure \(\overline{S}\) 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.

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

Given \(\epsilon > 0\), let \(R\) be a closed rectangle that contains \(S\). Let \(P\) be a partition of \(R\) such that \[U(P,\chi_S) \leq \int_R \chi_S + \epsilon = V(S) + \epsilon \qquad \text{and} \qquad L(P,\chi_S) \geq \int_R \chi_S - \epsilon = V(S)-\epsilon.\] Let \(R_1,\ldots,R_k\) be all the subrectangles of \(P\) such that \(\chi_S\) is not identically zero on each \(R_j\). That is, there is some point \(x \in R_j\) such that \(x \in S\). Let \(O_j\) be an open rectangle such that \(R_j \subset O_j\) and \(V(O_j) < V(R_j) + \nicefrac{\epsilon}{k}\). Notice that \(S \subset \bigcup_j O_j\). Then \[U(P,\chi_S) = \sum_{j=1}^k V(R_k) > \left(\sum_{j=1}^k V(O_k)\right) - \epsilon \geq m^*(S) - \epsilon .\] As \(U(P,\chi_S) \leq V(S) + \epsilon\), then \(m^*(S) - \epsilon \leq V(S) + \epsilon\), or in other words \(m^*(S) \leq V(S)\).

Now let \(R'_1,\ldots,R'_\ell\) be all the subrectangles of \(P\) such that \(\chi_S\) is identically one on each \(R'_j\). In other words, these are the subrectangles contained in \(S\). The interiors of the subrectangles \(R'^\circ_j\) are disjoint and \(V(R'^\circ_j) = V(R'_j)\). It is easy to see from definition that \[m^*\Bigl(\bigcup_{j=1}^\ell R'^\circ_j\Bigr) = \sum_{j=1}^\ell V(R'^\circ_j) .\] Hence \[m^*(S) \geq m^*\Bigl(\bigcup_{j=1}^\ell R'_j\Bigr) \geq m^*\Bigl(\bigcup_{j=1}^\ell R'^\circ_j\Bigr) %= %\sum_{j=1}^\ell %m^*(R'^\circ_j) = \sum_{j=1}^\ell V(R'^\circ_j) = \sum_{j=1}^\ell V(R'_j) = L(P,f) \geq V(S) - \epsilon .\] Therefore \(m^*(S) \geq V(S)\) as well.

## Integration over Jordan measurable sets

In one variable there is really only one type of reasonable set to integrate over: an interval. In several variables we have many very simple sets we might want to integrate over and these cannot be described so easily.

Let \(S \subset {\mathbb{R}}^n\) be a bounded Jordan measurable set. A bounded function \(f \colon S \to {\mathbb{R}}\) is said to be Riemann integrable on \(S\) if for a closed rectangle \(R\) such that \(S \subset R\), the function \(\widetilde{f} \colon R \to {\mathbb{R}}\) defined by \[\widetilde{f}(x) = \begin{cases} f(x) & \text{ if $x \in S$}, \\ 0 & \text{ otherwise}, \end{cases}\] is in \({\mathcal{R}}(R)\). In this case we write \[\int_S f := \int_R \widetilde{f}.\]

When \(f\) is defined on a larger set and we wish to integrate over \(S\), then we apply the definition to the restriction \(f|_S\). In particular note that if \(f \colon R \to {\mathbb{R}}\) for a closed rectangle \(R\), and \(S \subset R\) is a Jordan measurable subset then \[\int_S f = \int_R f \chi_S .\]

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## Images of Jordan measurable subsets

Let us prove the following FIXME. We will only need this simple

Suppose \(S \subset {\mathbb{R}}^n\) is a closed bounded Jordan measurable set, and \(S \subset U\) for an open set \(U \subset {\mathbb{R}}^n\). If \(g \colon U \to {\mathbb{R}}^n\) is a one-to-one continuously differentiable mapping such that \(J_g\) is never zero on \(S\). Then \(g(S)\) is Jordan measurable.

Let \(T = g(S)\). We claim that the boundary \(\partial T\) is contained in the set \(g(\partial S)\). Suppose the claim is proved. As \(S\) is Jordan measurable, then \(\partial S\) is measure zero. Then \(g(\partial S)\) is measure zero by . As \(\partial T \subset g(\partial S)\), then \(T\) is Jordan measurable.

It is therefore left to prove the claim. First, \(S\) is closed and bounded and hence compact. By , \(T = g(S)\) is also compact and therefore closed. In particular \(\partial T \subset T\). Suppose \(y \in \partial T\), then there must exist an \(x \in S\) such that \(g(x) = y\). The Jacobian of \(g\) is nonzero at \(x\).

We now use the inverse function theorem . We find a neighbourhood \(V \subset U\) 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 \(g(x) = y \in W\). As \(y \in \partial T\), there exists a sequence \(\{ y_k \}\) in \(W\) with \(\lim y_k = y\) and \(y_k \notin T\). 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 \(\lim x_k = x\). Since \(y_k \notin T = g(S)\), clearly \(x_k \notin S\). Since \(x \in S\), we conclude that \(x \in \partial S\). The claim is proved, \(\partial T \subset g(\partial S)\).

### Exercises

Prove .

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

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