11.5: Jordan Measurable Sets

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.

FIXME

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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