# 10.5: Jordan Measurable Sets

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## 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

1. The closure $$\overline{S}$$ is Jordan measurable.
2. The interior $$S^\circ$$ is Jordan measurable.
3. $$S \cup T$$ is Jordan measurable.
4. $$S \cap T$$ is Jordan measurable.
5. $$S \setminus T$$ 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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