Change of variables for a $C^1$ map injective and regular only on the interior of a compact Jordan set

Suppose first $D^{\circ }=\mathrm{\varnothing }$. Then $V=\mathrm{\varnothing }$ and, by [F1], $D=\mathrm{\partial }D$, which has content zero by [L2]; so $\mathrm{cont}(D)=0$ by [L11], the parameter integrand is continuous on the compact Jordan $D$ and hence integrable by [L8], and [L7] with [L11] bounds its integral in absolute value by ${\sup }_D\mid h(\psi )\det D\psi \mid \cdot \mathrm{cont}(D)=0$. All three integrals are then $0$ and both assertions hold. Assume $D^{\circ }\mathrm{\ne }\mathrm{\varnothing }$ for the rest of the proof.

The set $\mathrm{\partial }D$ is a closed subset of the compact $D$ by [F1], hence compact by [L13], and it has content zero by [L2] since $D$ is Jordan measurable. So [L3] gives that $\psi [\mathrm{\partial }D]$ is compact and has content zero.

By [F1] the interior $D^{\circ }$ is open and bounded, and $\mathrm{\partial }(D^{\circ })=\overline{D^{\circ }}\setminus D^{\circ }\subseteq D\setminus D^{\circ }=\mathrm{\partial }D$ because $\overline{D^{\circ }}\subseteq \bar{D}=D$. So $\mathrm{\partial }(D^{\circ })$ has content zero by step 1.2 and [F2], and $D^{\circ }$ is Jordan measurable by [L2].

On $D^{\circ }$ the map $\psi$ is injective and $\det D\psi \mathrm{\ne }0$, so [F3] and [L14] make each $D\psi (c)$ invertible, and then [L1] makes $\psi$ carry an open neighbourhood of each $c\in D^{\circ }$ onto an open set. Hence $\psi [D^{\circ }]=V$ is open, and $\psi {\mathrm{\mid }}_{D^{\circ }}:D^{\circ }\to V$ is a bijection whose inverse is $C^1$, in particular continuous, on $V$.

By [L10] the set $\psi [D]$ is compact, hence closed and bounded by [L13]. Since $D=D^{\circ }\cup \mathrm{\partial }D$ by [F1], $\psi [D]=V\cup \psi [\mathrm{\partial }D]$, so $\psi [D]\setminus V\subseteq \psi [\mathrm{\partial }D]$ has content zero by step 1.2 and [F2]. As $V$ is open with $V\subseteq \psi [D]$, we get $\mathrm{\partial }V=\bar{V}\setminus V\subseteq \psi [D]\setminus V$ and $\mathrm{\partial }(\psi [D])=\psi [D]\setminus \mathrm{int}(\psi [D])\subseteq \psi [D]\setminus V$; both therefore have content zero, and [L2] makes $V$ and $\psi [D]$ Jordan measurable.

Apply [L4] to the bounded open Jordan set $D^{\circ }$ of step 2.1, obtaining compact Jordan sets $K_1\subseteq K_2\subseteq \cdots \subseteq D^{\circ }$ with every compact subset of $D^{\circ }$ contained in some $K_j$ and $\mathrm{cont}(D^{\circ }\setminus K_j)\to 0$. By step 2.2 the hypotheses of [L5] hold with $U=D^{\circ }$ and $g=\psi {\mathrm{\mid }}_{D^{\circ }}$, so for each $j$ the set $\psi [K_j]$ is compact and Jordan measurable by [L6] and ${\int }_{\psi [K_j]}h={\int }_{K_j}h(\psi (x))\mid \det D\psi (x)\mid dx,$ both integrals existing because $h$ is continuous on the compact Jordan $\psi [K_j]$, hence integrable there by [L8].

The set $\psi [D]$ is compact and Jordan measurable by step 3.1 and $h$ is continuous on it, so [L8] makes $h$ integrable over $\psi [D]$ and, $\psi [D]$ being compact, $\mid h\mid \le M$ there for some $M\ge 0$. Fix a nondegenerate rectangle $Q\supseteq \psi [D]$. The zero extensions of $h{\mathrm{\mid }}_V$ and of $h$ from $\psi [D]$ differ only on $\psi [D]\setminus V$, which has content zero by step 3.1, so [L12] applied on $Q$ makes the first integrable too, with ${\int }_{V}h={\int }_{\psi [D]}h$ by [F4].

The map $x\mapsto h(\psi (x))\mid \det D\psi (x)\mid$ is continuous on the compact Jordan $D$, hence integrable over $D$ and over each compact Jordan $K_j$ by [L8], and bounded there by some $M^{\mathrm{\prime }}\ge 0$. Fix a nondegenerate rectangle $Q^{\mathrm{\prime }}\supseteq D$. Because $\mathrm{\partial }(D\setminus K_j)\subseteq \mathrm{\partial }D\cup \mathrm{\partial }{K}_j$ — a point outside both boundaries lies either in $\mathrm{int}{K}_j$, whose neighbourhood misses $D\setminus K_j$, or outside $\overline{K_j}$ and inside $\mathrm{int}D$, whose neighbourhood lies in $D\setminus K_j$ — the set $D\setminus K_j$ is Jordan measurable by [L2] and [F1]. The two zero extensions differ only on $D\setminus K_j$ and by at most $M^{\mathrm{\prime }}$, so [L7] and [L11] give $\mid {\int }_{D}h(\psi )\mid \det D\psi \mid -{\int }_{K_j}h(\psi )\mid \det D\psi \mid \mid \le M^{\mathrm{\prime }}\mathrm{cont}(D\setminus K_j)\mathrm{.}$ Now $D\setminus K_j=\mathrm{\partial }D\cup (D^{\circ }\setminus K_j)$ is a union of two disjoint Jordan sets, $\mathrm{\partial }D$ having content zero by step 1.2, so [L9] gives $\mathrm{cont}(D\setminus K_j)=\mathrm{cont}(D^{\circ }\setminus K_j)$, which tends to $0$ by step 3.2. Hence those integrals converge to ${\int }_{D}h(\psi )\mid \det D\psi \mid$.

Apply [L4] to the bounded open Jordan set $V$ of step 3.1, obtaining compact Jordan $C_1\subseteq C_2\subseteq \cdots \subseteq V$ with $\mathrm{cont}(V\setminus C_l)\to 0$ and every compact subset of $V$ inside some $C_l$. Fix $l$. By step 2.2 the inverse of $\psi {\mathrm{\mid }}_{D^{\circ }}$ is continuous, so $(\psi {\mathrm{\mid }}_{D^{\circ }}{)}^{-1}[C_l]$ is a compact subset of $D^{\circ }$ by [L10], and step 3.2 puts it inside some $K_{j(l)}$; applying $\psi$ gives $C_l\subseteq \psi [K_{j(l)}]$ and hence $V\setminus \psi [K_j]\subseteq V\setminus C_l$ for every $j\ge j(l)$, the sets $K_j$ being increasing. Both sets are Jordan measurable by step 3.1, step 3.2 and the boundary inclusion of step 4.2, so [L7] and [L11] give $\mathrm{cont}(V\setminus \psi [K_j])\le \mathrm{cont}(V\setminus C_l)$; letting $l$ grow, $\mathrm{cont}(V\setminus \psi [K_j])\to 0$.

With $M$ and $Q$ as in step 4.1, the zero extensions of $h{\mathrm{\mid }}_V$ and of $h{\mathrm{\mid }}_{\psi [K_j]}$ differ only on $V\setminus \psi [K_j]$ and by at most $M$, so [L7] and [L11] give $\mid {\int }_{V}h-{\int }_{\psi [K_j]}h\mid \le M\mathrm{cont}(V\setminus \psi [K_j])$, which tends to $0$ by step 5.1. Hence ${\int }_{\psi [K_j]}h\to {\int }_{V}h$.

By step 3.2 the two sequences of integrals agree term by term; by step 4.2 the parameter side converges to ${\int }_{D}h(\psi )\mid \det D\psi \mid$ and by step 6.1 the image side converges to ${\int }_{V}h$, so those two numbers are equal, and step 4.1 identifies ${\int }_{V}h$ with ${\int }_{\psi [D]}h$. With step 3.1 this is both assertions of the Statement.