Integrals over General Regions

We prove the theorem in two parts using the Riemann integrability criterion (upper/lower sum characterization), which is the standard tool inherited from Multiple Integrals over Rectangles and does not itself depend on anything proved here — so there is no circularity.

Part 1: Jordan measurable $\Rightarrow \chi_D$ integrable. Assume $\partial D$ has content zero. Fix $\varepsilon > 0$. By definition of content zero, choose closed rectangles $S_1,\dots,S_k$ with

We may enlarge each $S_i$ slightly (multiplying side lengths by $1+\delta$ for small $\delta$) so that the $S_i$ still cover $\partial D$, the sum of volumes is still $<\varepsilon$, and — crucially — we can choose a partition $P$ of $R$ whose grid lines include the endpoints of the (enlarged) $S_i$'s sides, so that each $S_i$ is a union of closed sub-rectangles of $P$. Let $P$ be such a partition, and classify each sub-rectangle $R_j$ of $P$ into three types:

• Interior type: $R_j \subseteq D^{\circ}$ (interior of $D$). Here $\chi_D \equiv 1$ on $R_j$, so $\sup_{R_j}\chi_D = \inf_{R_j}\chi_D = 1$.
• Exterior type: $R_j \cap \overline{D} = \emptyset$. Here $\chi_D \equiv 0$ on $R_j$, so $\sup = \inf = 0$.
• Boundary type: $R_j \cap \partial D \neq \emptyset$. Here $\sup_{R_j}\chi_D - \inf_{R_j}\chi_D \le 1$ (this oscillation is at most $1$ since $\chi_D$ only takes values $0,1$).

Every sub-rectangle not entirely in $D^\circ$ or entirely outside $\overline D$ meets $\partial D$ (since $D^\circ \cup \partial D \cup (\mathbb{R}^n \setminus \overline D)$ partitions $\mathbb{R}^n$, and a connected rectangle meeting both $D$ and its complement must meet $\partial D$). Hence every boundary-type $R_j$ is contained in some $S_i$ (by our refinement), so the total volume of boundary-type sub-rectangles is at most $\sum_i \operatorname{vol}(S_i) < \varepsilon$. Therefore

Since $\varepsilon>0$ was arbitrary, by the Riemann integrability criterion $\chi_D$ is integrable on $R$.

Part 2: $\chi_D$ integrable $\Rightarrow$ Jordan measurable. Assume $\chi_D$ is Riemann integrable on $R$. Fix $\varepsilon > 0$; by the integrability criterion there is a partition $P$ of $R$ with

As above, classify sub-rectangles of $P$ as interior, exterior, or boundary type (relative to $D$). For interior and exterior sub-rectangles, $\sup - \inf = 0$; for the rest, $\sup - \inf$ could be $0$ or $1$, but in any case

because every sub-rectangle meeting $\partial D$ meets both $D$ and its complement (it contains boundary points, which are limits of points in $D$ and limits of points outside $D$), forcing $\sup_{R_j}\chi_D = 1 > 0 = \inf_{R_j}\chi_D$ on any such sub-rectangle — wait, we must be careful: a sub-rectangle could meet $\partial D$ yet have $\inf = \sup$ only if $\chi_D$ is constant on it, which cannot happen on a sub-rectangle meeting $\partial D$, since $\partial D$ is precisely the set of points every neighborhood of which meets both $D$ and $D^c$; so every closed sub-rectangle meeting $\partial D$ has nonempty interior meeting both $D$ and $D^c$ (as long as the sub-rectangle has nonempty interior, which we may assume by choosing $P$ with no degenerate cells), giving oscillation exactly $1$ there. Hence

These sub-rectangles cover $\partial D$ (since $\partial D \subseteq R$ and $P$ partitions $R$ into these sub-rectangles), and their total volume is $< \varepsilon$. As $\varepsilon>0$ was arbitrary, $\partial D$ has content zero, i.e. $D$ is Jordan measurable.

Part 3: integrability of $f$ over $D$. Suppose $D$ is Jordan measurable and $f: D \to \mathbb{R}$ is bounded and continuous except on a set $Z \subseteq D$ of content zero. We show $f_D$ is Riemann integrable on $R$ by Lebesgue's criterion: a bounded function on a rectangle is Riemann integrable iff its set of discontinuities has measure zero (content zero implies measure zero, and for our purposes a finite union of content-zero sets is again content zero, hence measure zero). The discontinuities of $f_D$ on $R$ are contained in $Z \cup \partial D$: at any point of $D^\circ \setminus Z$, $f_D = f$ agrees with the continuous $f$ on a neighborhood inside $D^\circ$, so $f_D$ is continuous there; at any point outside $\overline{D}$, $f_D \equiv 0$ near that point, so it's continuous there. So discontinuities can only occur on $Z \cup \partial D$. Since $Z$ has content zero (hypothesis) and $\partial D$ has content zero ($D$ Jordan measurable), their union has content zero (a finite union of content-zero sets has content zero: cover each by rectangles of total volume $<\varepsilon/2$ and combine). A fortiori this union has measure zero, so by Lebesgue's criterion $f_D$ is Riemann integrable on $R$, i.e. $f$ is integrable over $D$. In particular if $f$ is continuous on all of $D$ and $D$ is compact, take $Z = \emptyset$: continuity on the compact set $D$ (with the subspace topology) does not by itself immediately give continuity of $f_D$ at boundary points, but boundary points of $D$ lie in $\partial D$, which is already accounted for and has content zero regardless of $f$'s values there. So the conclusion holds. $\blacksquare$

Step 1: $D$ is Jordan measurable. The graph $\Gamma_1 = \{(x, g_1(x)) : x \in [a,b]\}$ has content zero: since $g_1$ is continuous on the compact interval $[a,b]$, it is uniformly continuous, so given $\varepsilon>0$ choose $\delta>0$ with $|g_1(x)-g_1(x')|<\varepsilon/(2(b-a))$ whenever $|x-x'|<\delta$. Partition $[a,b]$ into $N$ subintervals of length $<\delta$; over each subinterval $I_k$, the graph of $g_1$ lies in a rectangle $I_k \times [\,m_k, m_k + \varepsilon/(2(b-a))\,]$ where $m_k = \min_{I_k} g_1$ (which exists by compactness/continuity). The total area of these $N$ rectangles is

The same argument gives a cover of $\Gamma_2$ (graph of $g_2$) of total area $<\varepsilon/2$. Also, the left and right edges $\{a\}\times[g_1(a),g_2(a)]$ and $\{b\}\times[g_1(b),g_2(b)]$ are (possibly degenerate) segments, each coverable by a single rectangle of arbitrarily small area (width $\eta \to 0$). Since $\partial D \subseteq \Gamma_1 \cup \Gamma_2 \cup (\text{two vertical segments})$, a finite union of content-zero sets, $\partial D$ has content zero. Step 2: reduction to the iterated integral. Let $R = [a,b] \times [c,d]$ enclose $D$, with $c \le \min g_1$, $d \ge \max g_2$ (exist by compactness). By Fubini's Theorem applied to the bounded, Riemann-integrable function $f_D$ on $R$ (integrability guaranteed by the theorem above, since $f$ is continuous on the compact Jordan measurable $D$),

provided the inner integral exists for (at least almost) every $x$; here we verify it exists for every $x \in [a,b]$. Fix $x$. As a function of $y$, $f_D(x,\cdot)$ equals $f(x,y)$ for $y \in [g_1(x), g_2(x)]$ and $0$ elsewhere on $[c,d]$; it has at most two discontinuities (at $y = g_1(x)$ and $y=g_2(x)$, where the extension may jump), hence is Riemann integrable on $[c,d]$ (bounded, finitely many discontinuities $\Rightarrow$ measure zero discontinuity set $\Rightarrow$ integrable, by Lebesgue's criterion for one variable). Moreover

since $f_D$ vanishes outside $[g_1(x), g_2(x)]$ and the integral is additive over subintervals of $[c,d]$ (splitting $[c,d]$ into $[c,g_1(x)]$, $[g_1(x),g_2(x)]$, $[g_2(x),d]$ and noting the outer two contribute $0$). Substituting into the Fubini identity gives exactly