Areas of Elementary Plane Figures

Translation by $a$ sends every axis-parallel rectangle $[u,v]$ bijectively to $[u+a,v+a]$, preserves all side lengths and volumes by [L2], preserves containment and interior-disjointness, and sends finite covers or inner families for $E$ to families of the same total volume for $E+a$; this also covers the empty family, the empty set, and rectangles with zero side length.

Step 1.1 gives outer content of $E+a$ at most that of $E$ and inner content of $E+a$ at least that of $E$; applying the same argument to translation by $-a$ gives the reverse inequalities, so both respective contents are equal.

Equality of the two contents for $E$ is therefore equivalent to equality of the two contents for $E+a$, and when these equalities hold their common values agree.

Apply [L2] to the constant function $H=1$ on the compact Jordan set supplied by [L1]; by [L3], the left side is $\mathrm{cont}(K)$, while the right side is ${\int }_a^b\left({\int }_{\alpha (x)}^{\beta (x)}1dy\right)dx$.

The inner integral is $\beta (x)-\alpha (x)$, including a zero contribution when the two graphs coincide, so step 1.1 is exactly [L4] and proves the formula.

In the case $r=0$, the disc is the singleton $\{(0,0)\}=[(0,0),(0,0)]$, a degenerate rectangle, so [L2] gives $\mathrm{cont}(D_0)=0=\pi 0^2$.

In the case $r>0$, the disc is the region on $[-r,r]$ between the continuous graphs $y=-\sqrt{r^2-x^2}$ and $y=\sqrt{r^2-x^2}$; [L3] identifies its Jordan content with its Riemann graph area, which [L1] evaluates as $\pi r^2$.

The cases $r=0$ and $r>0$ exhaust $r\ge 0$, and each gives $\mathrm{cont}(D_r)=\pi r^2$.

The set $E$ lies in the unit square, hence in a sufficiently large ball about the origin, so it is bounded by [L1].

The unit square is both a one-rectangle inner family and a one-rectangle outer cover of itself, each of total volume $1$ by [L5]; the inner-versus-outer inequality in [L4] therefore forces both contents to equal $1$, so the square does not have content zero.

By coordinatewise use of [L2], every ball centred at a point of $[0,1{]}^2$ meets $E$ and also meets its complement, including at the four sides; no point outside the closed square is adherent to $E$. Thus [L3] gives $\mathrm{\partial }E=[0,1{]}^2$.

By steps 2.1 and 1.2, the boundary of $E$ does not have content zero, so [L6] shows that $E$ is not Jordan measurable.

The bounded set $E$ has no Jordan area, contradicting the universal Statement.

Put $e:=v\mathrm{/}\parallel v{\parallel }_2$. Then $(e)$ is an orthonormal basis of $\mathbb{R}v$, so [L1] and [L2] give $P_{\mathbb{R}v}w=(⟨w,v⟩\mathrm{/}\parallel v{\parallel }_2^2)v$ and $d(w,\mathbb{R}v)=\parallel w-P_{\mathbb{R}v}w{\parallel }_2$.

Writing $v=(v_1,v_2)$ and $w=(w_1,w_2)$, inner-product expansion of step 1.1 gives $\parallel v{\parallel }_2^{2}d(w,\mathbb{R}v{)}^2=\parallel v{\parallel }_2^2\parallel w{\parallel }_2^2-⟨v,w{⟩}^2=(v_1{w}_2-v_2{w}_1{)}^2=\det [vw{]}^2$.

Both $\parallel v{\parallel }_{2}d(w,\mathbb{R}v)$ and $\mathrm{\mid }\det [vw]\mathrm{\mid }$ are nonnegative, so equality of their squares in step 2.1 and [L4] give the claimed identity.

Specialize [L1] to $n=2$ and use [L3] to translate the origin-based parallelepiped by $p$; it gives Jordan measurability and $\mathrm{cont}(P(p;v,w))=\mathrm{\mid }\det [vw]\mathrm{\mid }$, including dependent or zero spanning vectors.

When $v\mathrm{\ne }0$, substitute [L2] into the determinant formula of step 1.1 to obtain content equal to base length times perpendicular height.

The standard triangle $S=T((0,0),(1,0),(0,1))$ is the region $0\le x\le 1$, $0\le y\le 1-x$; [L1], [L5], [L6], and [L7] give $\mathrm{cont}(S)={\int }_0^1(1-x)dx=1\mathrm{/}2$.

Let $M$ have columns $B-A$ and $C-A$. Then $T(A,B,C)=A+M(S)$, so [L2], [L3], and step 1.1 give $\mathrm{cont}(T(A,B,C))=\mathrm{\mid }\det M\mathrm{\mid }\mathrm{/}2=\frac{1}{2}\mathrm{\mid }\det [B-AC-A]\mathrm{\mid }$, with the singular cases included.

If $A\mathrm{\ne }B$, apply [L4] with $v=B-A$ and $w=C-A$ in step 2.1 to obtain the half-base-times-height formula.

For the forward implication from collinearity to zero content, dependence in [L3] makes one of the two displacement vectors a scalar multiple of the other, including when either is zero; the two columns then have determinant zero, so [L1] and [L2] give content zero.

For the converse implication, suppose the content is zero. By [L1] and [L2], $\det [B-AC-A]=0$. If $B-A=0$ the list is dependent by [L3]; otherwise one coordinate of $B-A=(u_1,u_2)$ is nonzero, and the equation $u_1{v}_2-u_2{v}_1=0$ for $C-A=(v_1,v_2)$ shows by division in that nonzero coordinate that $C-A$ is a scalar multiple of $B-A$. Thus [L3] gives collinearity.

Choose a direction $e$ such that the scalar projections $⟨v_i,e⟩$ of the finitely many boundary vertices are pairwise distinct. Such a direction exists because only finitely many directions perpendicular to a difference $v_i-v_j$ are forbidden. Use the projection onto $e$ as the horizontal coordinate and the perpendicular coordinate as the vertical one. No boundary edge is vertical in these coordinates.

Between two consecutive vertex projections, every boundary edge that crosses the open vertical slab is the graph of an affine function, and disjoint boundary edges have a fixed vertical order throughout the slab. On a vertical line in the slab, the compact set $P$ can change between membership and nonmembership only at one of these boundary crossings. Near such a crossing the boundary is a single straight segment and, because the crossing point is not an interior point of $P$, the two local sides cannot both lie in $P$; since $P=\overline{\mathrm{int}P}$, exactly the side belonging to $P$ is filled. Starting below the bounded set and moving upward therefore expresses the part of $P$ in the slab as a finite union of closed bands between consecutive affine boundary graphs.

Take the closures of all those bands in the finitely many closed slabs. Insert every boundary--wall intersection as a subdivision vertex and use their common finite refinement on each vertical wall. Before this common refinement, the closure of a nonempty band is a convex triangle or quadrilateral: it is bounded by two vertical segments and two affine graphs that retain their vertical order. The common refinement may add collinear vertices to a vertical side, so each resulting two-dimensional cell is a convex polygon whose boundary carries all wall-subdivision vertices. These cells cover $P$, and two distinct cells meet only in a union of full consecutive wall segments, a common vertex, or not at all.

Choose one interior point in each two-dimensional cell. Join it to every boundary vertex of that cell in cyclic order. Convexity keeps every joining segment in the cell, and each consecutive boundary pair with the interior point forms a nondegenerate triangle. These triangles cover the cell. On a shared vertical wall, both adjacent cells use the same refined boundary segments, so triangles from opposite sides meet in a full common refined segment, a common endpoint, or not at all; within one cell the fan triangles meet in a full radial edge, the chosen interior point, or not at all. The resulting finite family is therefore face-to-face, covers $P$, and is a triangulation by [L1].

Choose a triangulation by [L1]. Each of its finitely many closed triangular faces is Jordan measurable by [L2].

Distinct faces meet only in a common edge, a common vertex, or not at all. An edge is a graph of a continuous affine function after interchanging coordinates if necessary, so [L4] gives it content zero. A common vertex or the empty set is a subset of such a graph, so [L5] gives it content zero as well. Thus every face intersection has content zero.

Apply [L3] repeatedly to the finite face family using step 1.2. The union $P$ is Jordan measurable and its content is the sum of all face contents.

The same argument applies to any triangulation, and every resulting sum equals the intrinsic number $\mathrm{cont}(P)$, so the sum is independent of the triangulation.

Choose a triangulation by [L0] and orient every triangular face counterclockwise. By [L1] and [L2], the polygon content is the sum over faces of one half of the signed boundary-edge expression $\det [p,q]+\det [q,r]+\det [r,p]$.

By [L3], the finite face sum may be regrouped by oriented edges. Every internal edge is traversed once in each direction by its two incident faces, so its two determinant terms cancel.

Only the counterclockwise boundary edges remain. A triangulation may subdivide the polygon edge from $v_i$ to $v_{i+1}$ at ordered points $p_j=(1-t_j)v_i+t_j{v}_{i+1}$, but direct bilinearity gives $\det [p_j,p_{j+1}]=(t_{j+1}-t_j)\det [v_i,v_{i+1}]$, so [L3] makes the subdivision sum equal $\det [v_i,v_{i+1}]$. This includes the final edge from $v_{n-1}$ to $v_0$; expanding the determinants gives the stated cyclic shoelace sum.