Analytic quasiconformality gives both quadrilateral modulus bounds

A continuous planar homeomorphism with finite partial derivatives almost everywhere is totally differentiable almost everywhere. Here are the needed details. On an interior rectangle, Egorov and Lusin give a compact set $E$ outside arbitrarily small area on which the two partials are continuous and their directional difference quotients converge uniformly. The measurable error is the supremum over rational $0<\mathrm{\mid }t\mathrm{\mid }<1\mathrm{/}n$; continuity in nonzero $t$ makes this the full supremum. By [F2], almost every $p\in E$ is a density-one point of both coordinate slices. Translate $p$ to $0$, and put $A(x+iy)=f(0)+x{f}_x(0)+y{f}_y(0)$. Uniform quotients and continuous partials give $\mathrm{\mid }f(w)-A(w)\mathrm{\mid }\le 3\epsilon \mathrm{\mid }w\mathrm{\mid }$ whenever the horizontal or vertical projection of $w$ belongs to the corresponding slice of $E$, for sufficiently small $w$. For any fixed $z$ with $r=\mathrm{\mid }z\mathrm{\mid }$, slice density supplies rectangle sides at coordinates within $\epsilon r$ on either side of each coordinate of $z$. Every boundary point $w$ of that rectangle has one of the good projections. Openness of $f$ makes the maximum of $\mathrm{\mid }f(w)-A(z)\mathrm{\mid }$ occur on the boundary: $A(z)$ is a fixed constant, and an open image cannot have an interior maximum of distance from it. Thus $\mathrm{\mid }f(z)-A(z)\mathrm{\mid }\le 3\epsilon (1+\sqrt{2}\epsilon )r+\sqrt{2}\epsilon (\mathrm{\mid }{f}_x(0)\mathrm{\mid }+\mathrm{\mid }{f}_y(0)\mathrm{\mid })r$. Let $\epsilon \downarrow 0$. Taking compact sets with excluded area tending to zero proves the assertion; [F1] makes it applicable to $f$. No maximum principle for a variable affine difference is used.

We record the precise exceptional-curve argument. On each compact interior neighborhood mollify a continuous $W^{1,2}$ map $u$ to smooth $u_n$ with uniform convergence and derivative convergence in $L^2$. Choose a subsequence with ${\sum }_n\mathrm{\parallel }D{u}_n-Du{\mathrm{\parallel }}_2<\mathrm{\infty }$ and put $G={\sum }_n\mathrm{\mid }D{u}_n-Du\mathrm{\mid }+\mathrm{\mid }Du\mathrm{\mid }$. This is in $L^2$. Rectifiable curves with ${\int }_{\gamma }G=\mathrm{\infty }$ have modulus zero, because $G\mathrm{/}m$ is admissible on them with area tending to zero. On every remaining curve, the derivatives of $u_n\circ \gamma$ converge in $L^1$ to $Du(\gamma ){\gamma }^{\mathrm{\prime }}$ along arc-length parameterization; uniform convergence and the fundamental theorem of calculus identify the limit as the derivative of the absolutely continuous path $u\circ \gamma$. Borel null representatives are handled by an infinite density on their null set. A countable compact exhaustion and positive summable multiples of the local barriers give one $L^2$ barrier for all bad compact restrictions. If $u$ has finite total energy on a bounded domain and extends continuously to its boundary, include $\mathrm{\mid }Du\mathrm{\mid }$ in the barrier. Good boundary-joining paths then have finite derivative integral on the whole interval; local absolute continuity and endpoint continuity imply absolute continuity including endpoints. For an AC path, its variation on intervals is the integral of its speed, first by partitions and the fundamental theorem and then by differentiation. Uniqueness of measures and simple approximation therefore give the weighted arc-length identity for every nonnegative Borel density. This proves the chain-rule length inequality outside a modulus-zero family, not merely on coordinate lines.

Every marked Jordan quadrilateral rectifies to a rectangle. Map its interior to the upper half-plane with three marked boundary points normalized to $0,1,\mathrm{\infty }$; the fourth is $a<0$, by [F4]. On the upper half-plane take a primitive $P$ of $1\mathrm{/}\sqrt{z(z-1)(z-a)}$, choosing the analytic square root by a logarithm. Its derivative is nonzero. The inverse-square-root singularities at $a,0,1$ are integrable; at infinity the derivative is $O(\mathrm{\mid }z{\mathrm{\mid }}^{-3\mathrm{/}2})$, so the primitive has a common finite limit there. Each of the four real boundary intervals maps monotonically to a straight side; the directions alternate by $\pi \mathrm{/}2$. Equality of the two limits at infinity closes the polygon, and its four positive side lengths give a rectangle with equal opposite sides. On upper half-disks indented about the three branch points, image boundaries converge to that rectangle boundary. The argument principle [F4] gives exactly one preimage for every interior point and none for an exterior point. Thus $P$ is a conformal bijection, extending homeomorphically to the four sides.

Define the positive locally finite Borel measure $\nu (E)=\mathrm{\mid }f(E)\mathrm{\mid }$. At a differentiability point with invertible derivative $A$, the expansion $f(p+v)=f(p)+Av+o(\mathrm{\mid }v\mathrm{\mid })$ traps the image of a radius-$r$ disk between the ellipses $A(D_r)$ enlarged and contracted by $o(r)$. The inner containment follows by Jordan separation, since the image boundary is an $o(r)$ perturbation of the ellipse and has the same winding on its contracted interior. Hence its area ratio tends to $\mathrm{\mid }\det A\mathrm{\mid }$. At a singular derivative the image is contained in an $o(r)$ neighborhood of a bounded segment or point, so that ratio tends to zero. Since $f$ preserves orientation, a nonsingular derivative has positive determinant: interpolation to $Av$ on a small boundary circle preserves the local orientation sign. Measure differentiation [F3] therefore identifies $\nu$'s absolutely continuous density with $J_f$. Its singular part is nonnegative, giving ${\int }_E{J}_f\le \mathrm{\mid }f(E)\mathrm{\mid }$ for every relatively compact Borel $E$. Simple approximation and exhaustion give $\int (g\circ f)J_f\le \int g$ for every nonnegative Borel target function. This is only the lower area inequality.

The rectangle map and its inverse have finite conformal energy: nonnegative change of variables on compact interior exhaustions gives $\int \mathrm{\mid }{\varphi }^{\mathrm{\prime }}{\mathrm{\mid }}^2=\mathrm{\mid }Q\mathrm{\mid }$ and the corresponding finite rectangle area for its inverse. Their boundary extensions are continuous. Step 1.2 therefore transports boundary-joining families outside modulus-zero exceptional families; the weighted length identity and the conformal area identity give invariance in both directions. Consequently [F5] transfers the rectangle joining-family values, equality with its straight foliation, and transverse reciprocity to every Jordan quadrilateral. This explicitly extends the compact-interior conformal-invariance supplier to the boundary families used here.

Removing a zero-modulus family does not change modulus: add to any good-family admissible density a bad-family admissible density with arbitrarily small $L^2$ norm and use Minkowski [F6]. A density of infinite length on all bad curves and arbitrarily small norm is obtained by summing a sequence of bad-family admissible densities with summable norms. Apply step 1.2 to $f$ and a target admissible density $\tau$. Its pullback $\rho =(\tau \circ f)\mathrm{\parallel }Df{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$ satisfies $A(\rho )\le KA(\tau )$ by step 2.1 and [F1]. Weighted length on every good source curve is at least the target image length. Adding a vanishing-cost bad-family density gives $\mu (\mathrm{\Gamma })\le K\mu (f\mathrm{\Gamma })$, equivalently $\lambda (f\mathrm{\Gamma })\le K\lambda (\mathrm{\Gamma })$, for the compact-trace families used here. This establishes one direction only.

Rectify the source $Q$ and take its straight joining foliation $\mathcal{F}$. In conformal coordinates, $f\circ \varphi$ has finite energy on the whole rectangle: step 2.1 and $\mathrm{\parallel }Df{\mathrm{\parallel }}^2\le K{J}_f$ bound it by a constant times $\mathrm{\mid }f(Q)\mathrm{\mid }$ after conformal change of variables. Sobolev coordinate change follows by mollification on compact subsets, the classical chain rule and $L^2$ convergence there. Fubini gives a Borel null set $N$ of bad leaf parameters; good leaves are AC on the entire closed interval with the weighted chain rule. The bad leaves have modulus zero: if the leaf width is $w$, the density $w^{-1}{\mathbf{1}}_{(0,w)\times O}$ for an open $O\supset N$ has length one on each bad leaf and area $\mathrm{\mid }O\mathrm{\mid }\mathrm{/}w\to 0$. Removing them leaves the source foliation's value unchanged. The full target joining family contains the image good foliation. Step 3.1, or its identical good-leaf pullback, thus gives $\lambda ({\mathrm{\Gamma }}_{f(Q)})\le \lambda (f{\mathcal{F}}_{\mathrm{g}\mathrm{o}\mathrm{o}\mathrm{d}})\le K\lambda ({\mathcal{F}}_{\mathrm{g}\mathrm{o}\mathrm{o}\mathrm{d}})=K\lambda ({\mathrm{\Gamma }}_Q)$. No claim that images of bad leaves have zero modulus is needed.

Repeat step 4.1 for the transverse source foliation. By step 2.2 the two joining-family extremal lengths multiply to one in both quadrilaterals. Inverting the transverse upper bound therefore gives $\lambda ({\mathrm{\Gamma }}_{f(Q)})\ge K^{-1}\lambda ({\mathrm{\Gamma }}_Q)$. Their values are finite and positive because the rectifying rectangles are nondegenerate. Taking reciprocals gives both displayed modulus bounds. The second bound came from transverse reciprocity, not an assumed inverse-null property.