Fubini and Change of Variables

For a cell $I$ of $P$ and a cell $J$ of $R$, put $m_{IJ}:={\inf }_{I\times J}f$ and $M_{IJ}:={\sup }_{I\times J}f$. Regrouping the finite sums over the product grid gives $L(f,P\times R)={\sum }_I\mathrm{vol}(I){\sum }_J{m}_{IJ}\mathrm{vol}(J)$ and the analogous formula with $M_{IJ}$ for the upper sum.

If $x\in I$, then ${\inf }_J{f}_x\ge m_{IJ}$ and ${\sup }_J{f}_x\le M_{IJ}$ for every $J$. Hence ${\mathrm{\ell }}_B(x)\ge {\sum }_J{m}_{IJ}\mathrm{vol}(J)$ and $u_B(x)\le {\sum }_J{M}_{IJ}\mathrm{vol}(J)$. Taking the infimum or supremum over $x\in I$ preserves these inequalities.

Multiply the cellwise inequalities by $\mathrm{vol}(I)$ and sum over $I$. Together with ${\mathrm{\ell }}_B\le u_B$, this gives the displayed chain. Exchanging the coordinate blocks gives the symmetric chain.

Given $\epsilon >0$, [L2] supplies a grid of $A\times B$ with Darboux gap below $\epsilon$. Its coordinate grids form a product grid, and [L1] places the lower and upper Darboux gaps of both ${\mathrm{\ell }}_B$ and $u_B$ inside that same gap.

By [L2], both ${\mathrm{\ell }}_B$ and $u_B$ are integrable. The inequalities in [L1], applied to grids with gaps tending to zero, give ${\int }_A{\mathrm{\ell }}_B\ge {\int }_{A\times B}f$ and ${\int }_A{u}_B\le {\int }_{A\times B}f$; since ${\mathrm{\ell }}_B\le u_B$, all three values are equal. The same argument after exchanging $A$ and $B$ gives the other two equalities.

Suppose $h={\int }_B{f}_x$ outside a content-zero $N$. There $h={\mathrm{\ell }}_B=u_B$. If $M$ bounds $\mathrm{\mid }h\mathrm{\mid },\mathrm{\mid }{\mathrm{\ell }}_B\mathrm{\mid },\mathrm{\mid }{u}_B\mathrm{\mid }$, a finite cube cover of $N$ with arbitrarily small total volume, refined into an outer grid, bounds the upper integral of $\mathrm{\mid }h-{\mathrm{\ell }}_B\mathrm{\mid }$ by $2M$ times that volume. The criterion [L2] therefore makes $h-{\mathrm{\ell }}_B$ integrable with integral $0$, and linearity gives ${\int }_{A}h={\int }_A{\mathrm{\ell }}_B={\int }_{A\times B}f$. The exchanged assertion is identical.

Every coordinate section is continuous, hence integrable by [L2]. Moreover $f$ is uniformly continuous on the compact rectangle by [L3]; therefore integrating in one coordinate produces a continuous function of the remaining coordinates, since the difference of two section integrals is bounded by the interval length times the uniform oscillation of $f$.

Apply [L1] to the first coordinate in a prescribed order. Step 1.1 makes the resulting function continuous, so the same argument applies to the next coordinate. Induction through the finite coordinate list gives the repeated integral.

For $n=1$ the repeated integral is the original integral. At every later stage [L1] preserves its value, so every coordinate order gives ${\int }_{Q}f$.

The product $f$ is continuous and hence integrable by [L2]. For fixed $x$, linearity [L3] gives ${\int }_B{f}_x=a(x){\int }_{B}b$.

Apply [L1] and [L3] once more: ${\int }_{A\times B}f={\int }_A(a(x){\int }_{B}b)=({\int }_{A}a)({\int }_{B}b)$.

Taking $b$ constantly equal to $1$ gives ${\int }_{B}b=\mathrm{vol}(B)$ and yields the coordinate-independent case, including the case ${\int }_{A}a=0$.

A finite subset of ${\mathbb{R}}^p$ has content zero by [L2]: for a given $\epsilon >0$, cover its finitely many points by cubes whose total volume is below $\epsilon$.

Outside $S$ every section is identically zero and has integral zero. Complete the section-integral function by the value $0$ on $S$ and apply [L1]; the resulting outer function is identically zero, so the multiple integral is zero.

If $S$ is empty then $f$ itself is identically zero, and step 2.1 still applies. Exchanging the coordinate blocks proves the symmetric assertion.

Choose nondegenerate rectangles $A\subseteq {\mathbb{R}}^p$ and $B\subseteq {\mathbb{R}}^q$ with $E\subseteq A\times B$, and extend $g$ by zero to $A\times B$. For every $x$, the resulting section is the zero extension to $B$ of $g_x$, and it is identically zero when $E_x$ is empty.

The zero extension is integrable by the definition of the Jordan-set integral. Apply [L1]; outside $N$ its ordinary section integral is exactly $h(x)$, so the exceptional-section clause gives ${\int }_{E}g={\int }_{A}h$.

Enlarging $A$ or $B$ only adds zero to the zero extension. Independence of the Jordan integral from a bounding rectangle [L3] and the content-zero invariance in [L1] therefore prove independence of both factor rectangles and of the assigned values on $N$.

Apply [L1] to the constant-one function on $E$. Its integral over $E$ is $\mathrm{cont}(E)$ by [L2], while the integral over a section is $\mathrm{cont}(E_x)$, again by [L2].

For $E$ and $F$, the two completed sectional-content functions agree outside the union of their exceptional sets, which is content zero. Their integrals are therefore equal, and step 1.1 identifies those integrals with the two total contents.

Empty sections contribute $0$. If either set has content zero, the same formula gives zero on both sides, so no nonemptiness hypothesis is hidden.

The boundary of $K$ is contained in the graphs of $\alpha$ and $\beta$ together with the two vertical endpoint segments. Each is a continuous graph, after exchanging coordinates for the vertical segments, and hence has content zero by [L3]. The set $K$ is closed and bounded, hence compact by [L5], and [L2] makes it Jordan measurable.

The continuous $H$ is integrable on the compact Jordan set by [L4]. Every vertical section is the closed interval $[\alpha (x),\beta (x)]$, and its restriction is continuous, so [L1] gives the displayed formula.

If $\alpha (x)=\beta (x)$, that section is degenerate and contributes $0$; the endpoint sections and all other boundary changes have content zero. Requiring a continuous extension to $K$ supplies boundedness and integrability that continuity only on the open region would not supply.

Extend all functions by zero to one bounding rectangle. The indicators in [L2] are integrable, and the nonnegativity of every $M_i$ makes the following inequality valid both on and off $E$. Monotonicity and finite linearity from [L1] give the cover bound, because each indicator integral is the content of its Jordan set: $h\le {\sum }_{i=1}^N{M}_i{\mathbf{1}}_{E_i}\mathrm{.}$

The restriction of $h$ to each $F_i$ is integrable: away from grid cells meeting $\mathrm{\partial }{F}_i$ its Darboux gap is inherited from $h$, while [L2] makes the total volume of boundary cells arbitrarily small. Hence ${\int }_{F_i}h\ge m_i\mathrm{cont}(F_i)$ by [L1]. Pairwise interior-disjoint Jordan sets intersect only on their content-zero boundaries, so the sum of their zero-extended restrictions equals the restriction to their union outside a content-zero set. The same boundary-cell argument and linearity [L1] therefore add these integrals without overcounting; their union lies in $E$, and $h\ge 0$, giving the lower bound.

The boundary-cell argument in step 1.2 also makes the restriction of a signed integrable $h$ to $F$ integrable. A nonempty $F$ makes ${\sup }_F\mathrm{\mid }h\mathrm{\mid }$ a real number, since $h$ is bounded. On $F$, the inequalities $-\mathrm{\mid }h\mathrm{\mid }\le h\le \mathrm{\mid }h\mathrm{\mid }\le {\sup }_F\mathrm{\mid }h\mathrm{\mid }$ and [L1], together with the indicator identity in [L2], give the last estimate.

Suppose first that $A$ is invertible. For every elementary matrix $E_0$, both $E_0$ and $E_0^{-1}$ are Lipschitz by [L5], so $\mathrm{\partial }(E_{0}F)=E_0(\mathrm{\partial }F)$ for every bounded set $F$. Thus [L4] makes $E_{0}F$ Jordan whenever $F$ is Jordan. Coordinate permutations and nonzero coordinate scalings send rectangular figures to rectangular figures, with content factor $1$ and $\mathrm{\mid }c\mathrm{\mid }$ respectively; applying this to arbitrarily close inner and outer figures from [L8] proves those factors for every bounded Jordan $F$. For a shear adding $c$ times one coordinate to another, take rectangular figures $P\subseteq F\subseteq Q$ from [L8] with $\mathrm{cont}(Q)-\mathrm{cont}(P)$ arbitrarily small. Sections of a rectangular figure parallel to the changed coordinate are finite unions of intervals, hence Jordan at every parameter, and the corresponding sections of $E_{0}P$ and $E_{0}Q$ are their translates by a quantity depending only on the fixed coordinates, so they are finite unions of intervals of the same total length. Both hypotheses of [L3] are therefore met by $P,Q,E_{0}P,E_{0}Q$ but are not claimed for $F$, whose sections need not be Jordan; [L3] gives $\mathrm{cont}(E_{0}P)=\mathrm{cont}(P)$ and $\mathrm{cont}(E_{0}Q)=\mathrm{cont}(Q)$. Since $E_{0}P\subseteq E_{0}F\subseteq E_{0}Q$ and $E_{0}F$ is already known Jordan, its content and that of $F$ are both squeezed between $\mathrm{cont}(P)$ and $\mathrm{cont}(Q)$, so a shear preserves content. These are exactly the absolute determinant factors listed by [L2].

By [L1], write $A$ as a finite product of elementary matrices. Apply step 1.1 successively to $E$: every intermediate image is bounded Jordan, and its content is multiplied by the corresponding absolute determinant factor. The row-operation laws [L2], applied successively from the identity, identify the product of those factors with $\mathrm{\mid }\det A\mathrm{\mid }$. Boundedness follows from [L5].

If $A$ is singular, [L6] reduces it to an echelon matrix with a zero row; [L7] realizes this as an invertible change of codomain coordinates. The transformed range lies in a coordinate hyperplane, whose bounded part fits in slabs of arbitrarily small thickness and therefore has content zero and is Jordan. Applying the invertible case to undo the coordinate change shows that $T(E)$ has content zero and is Jordan; the real criterion in [L1] gives $\det A=0$, so the same formula holds.

The matrix map $T(x)=Ax$ sends $Q$ exactly onto $P(v_1,\dots ,v_n)$.

Apply [L1] to $Q$ and use [L2] to obtain the stated content formula: $\mathrm{cont}(P)=\mathrm{\mid }\det A\mathrm{\mid }\mathrm{cont}(Q)=\mathrm{\mid }\det A\mathrm{\mid }\mathrm{.}$ If the columns are dependent, [L1] simultaneously supplies Jordan measurability and the zero-content conclusion.

Put $R=F-I$ and use [L1] with the Euclidean--sup norm comparison [L4] to obtain the contraction estimate $\mathrm{\parallel }R(x)-R(y){\mathrm{\parallel }}_{\mathrm{\infty }}\le q\mathrm{\parallel }x-y{\mathrm{\parallel }}_{\mathrm{\infty }}$ on the cube. In particular $\mathrm{\parallel }R(x){\mathrm{\parallel }}_{\mathrm{\infty }}\le qr$, so $F(x)=x+R(x)$ lies in $C(a,(1+q)r)$.

Fix $z\in C(a,(1-q)r)$ and define $S_z(x)=z-R(x)$. Step 1.1 gives $S_z(C(a,r))\subseteq C(a,r)$ and makes $S_z$ a $q$-contraction. The cube is a nonempty closed subset of complete Euclidean space, so [L2]--[L3] give $x\in C(a,r)$ with $S_z(x)=x$, equivalently $F(x)=z$. This proves the inner containment.

If $F(x)=F(y)$, then $x-y=R(y)-R(x)$, so step 1.1 gives $\mathrm{\parallel }x-y{\mathrm{\parallel }}_{\mathrm{\infty }}\le q\mathrm{\parallel }x-y{\mathrm{\parallel }}_{\mathrm{\infty }}$. Since $q<1$, $x=y$. The assumptions $r>0$ and $q<1$ are essential to the nondegenerate fixed-point argument.

Normalize at $a$ by the affine map $H(x)=a+A^{-1}(g(x)-g(a))\mathrm{.}$ Choose $0<q<\epsilon$ and a slightly larger cube inside $U$ on which $\mathrm{\parallel }DH-I{\mathrm{\parallel }}_{2\to 2}\le q\mathrm{/}\sqrt{n}$. The mean-value bound in [L4] makes $H-I$ a $q$-contraction in the sup norm, so $H$ is injective and bi-Lipschitz there. The derivative bound also makes every $DH$ invertible; the inverse function theorem in [L4] therefore makes $H$ a homeomorphism on a neighborhood of the smaller positive-radius cube $Q$. Here $H(a)=a$ and $DH(a)=I$; continuity of $Dg$ permits the stated choice inside any prescribed neighborhood.

If $E\subseteq Q$ is Jordan, the homeomorphism in step 1.1 gives $\mathrm{\partial }H(E)=H(\mathrm{\partial }E)$. Compose $H$ on the larger cube with coordinatewise clamping onto that cube to obtain a global Lipschitz map. Since $\mathrm{\partial }E$ is null, [L4] makes $H(\mathrm{\partial }E)$ null and hence makes $H(E)$ Jordan.

Refine inner and outer figures from [L3] into finite unions of sufficiently small, interior-disjoint cubes $P\subseteq E\subseteq R$ with arbitrarily small content gap. After translating at each cube centre, [L2] sandwiches its $H$-image between cubes with factors $(1-q{)}^n$ and $(1+q{)}^n$. Step 2.1 makes those images Jordan, injectivity makes their interiors disjoint, and [L5] adds their contents. Letting the figure gap vanish gives the stronger bounds with $q$; since $q<\epsilon$, these imply the displayed bounds for $H(E)$. Finally $g(E)=g(a)+A(H(E)-a)$, so [L1] multiplies every content by $\mathrm{\mid }\det A\mathrm{\mid }=\mathrm{\mid }\det Dg(a)\mathrm{\mid }$.

Continuity and [L2] make $g(K)$ compact, hence closed and bounded. If $y=g(x)\in \mathrm{\partial }g(K)$, then $x$ cannot lie in the interior of $K$: otherwise [L1], together with global injectivity on $U$, would map a neighborhood of $x$ contained in $K$ onto a neighborhood of $y$ contained in $g(K)$. Thus $\mathrm{\partial }g(K)\subseteq g(\mathrm{\partial }K)\mathrm{.}$

Around each point of the compact set $\mathrm{\partial }K$, choose a closed cube in a slightly larger convex cube inside $U$ on which $Dg$ is bounded. By [L4], $g$ is Lipschitz on the smaller cube. Composing its restriction with coordinatewise clamping onto that cube produces a Lipschitz self-map of ${\mathbb{R}}^n$, so [L3] sends the null set $\mathrm{\partial }K$ inside the cube to a null set. A finite subcover shows that $g(\mathrm{\partial }K)$ is null.

Step 1.1 makes $\mathrm{\partial }g(K)$ a subset of the null set from step 1.2. Since $g(K)$ is bounded, the boundary criterion in [L3] proves it is Jordan measurable.

For each $x\in C$, openness gives a closed grid rectangle $R_x\subseteq U$ with $x\in \mathrm{int}{R}_x$. The interiors cover $C$, so [L1] selects $R_1,\dots ,R_N$. Put $K={\bigcup }_i{R}_i$. Then $C\subseteq \mathrm{int}K\subseteq K\subseteq U$.

The finite union $K$ is closed and bounded, hence compact by [L2]. Its boundary is contained in the union of the boundaries of the $R_i$. Each rectangular face is a continuous coordinate graph over a bounded rectangle and is null by [L3]; a finite union remains null.

The boundary criterion in [L3] now makes $K$ Jordan measurable. Subdividing the finitely many rectangles by their common coordinate endpoints expresses the same set as a finite union of closed cells from one grid.

Enclose $V$ in a rectangle and choose nested dyadic grids whose meshes tend to zero. Let $K_j$ be the union of every closed cell of the $j$th grid that is contained in $V$. Only finitely many cells occur. Every child of a retained cell is retained, so $K_j\subseteq K_{j+1}$; each $K_j$ is compact, Jordan, and contained in $V$.

If compact $C\subseteq V$, the distance from $C$ to the closed complement of $V$ is positive. Once the mesh diameter is smaller than that distance, every grid cell meeting $C$ is contained in $V$, so $C\subseteq K_j$.

Every unretained cell meeting $V$ also meets a mesh-sized neighborhood of $\mathrm{\partial }V$. By [L1], that boundary has content zero; [L2] therefore makes the total volume of all such cells arbitrarily small for fine enough grids. Finite additivity [L3] bounds $\mathrm{cont}(V\setminus K_j)$ by that volume, proving the limit.

The entries of $Dg$ are continuous; [L1] therefore makes $\det Dg$ and $\mathrm{\mid }\det Dg\mathrm{\mid }$ continuous on $K$, and [L6] makes the absolute determinant bounded and Riemann integrable. By [L3], $g(K)$ is also a compact Jordan set.