Compact convex windows admit rectangle approximations and have negligible translated boundary

Choose $x_0\in \mathrm{int}W$ and $r>0$ with the open ball $B(x_0,r)\subseteq W$. For $0<t<1$ put $W_t=x_0+(1-t)(W-x_0)$. If $w\in W$ and $\mathrm{\parallel }z\mathrm{\parallel }<tr$, then $x_0+z\mathrm{/}t\in B(x_0,r)\subseteq W$ and $(1-t)w+t(x_0+z\mathrm{/}t)=(1-t)w+t{x}_0+z\in W$ by convexity; thus every point of $W_t$ has an open ball of radius $tr$ contained in $W$, so $W_t\subseteq \mathrm{int}W$.

If $0<t^{\mathrm{\prime }}<t<1$, put $w^{\mathrm{\prime }}=\frac{1-t}{1-t^{\mathrm{\prime }}}w+\frac{t-t^{\mathrm{\prime }}}{1-t^{\mathrm{\prime }}}{x}_0$. The two coefficients are nonnegative and sum to one, so $w^{\mathrm{\prime }}\in W$ by convexity, and $x_0+(1-t)(w-x_0)=x_0+(1-t^{\mathrm{\prime }})(w^{\mathrm{\prime }}-x_0)\mathrm{.}$ Thus $W_t\subseteq W_{t^{\mathrm{\prime }}}$. Moreover, every $y\in \mathrm{int}W$ lies in some $W_t$: for $y\mathrm{\ne }{x}_0$, choose $s>0$ small enough that $w=x_0+(1+s)(y-x_0)\in W$, and then $y=x_0+(1-t)(w-x_0)$ for $t=s\mathrm{/}(1+s)$; $x_0$ lies in every $W_t$. Hence for any sequence $t_n\downarrow 0$, $W_{t_n}\uparrow \mathrm{int}W$.

By [F2] and [F3], $\mathrm{\mid }{W}_{t_n}\mathrm{\mid }=(1-t_n{)}^d\mathrm{\mid }W\mathrm{\mid }$ and the sets increase to $\mathrm{int}W$, so [F3] gives $\mathrm{\mid }\mathrm{int}W\mathrm{\mid }=\mathrm{\mid }W\mathrm{\mid }$. Thus the compact boundary $\mathrm{\partial }W=W\setminus \mathrm{int}W$ is a null set. Since $W$ contains a nonempty open ball and is bounded, [F1] gives $0<\mathrm{\mid }W\mathrm{\mid }<\mathrm{\infty }$.

Given $ϵ>0$, choose $t>0$ so that $\mathrm{\mid }W\mathrm{\mid }-\mathrm{\mid }{W}_t\mathrm{\mid }=(1-(1-t{)}^d)\mathrm{\mid }W\mathrm{\mid }<ϵ$. Choose a grid of half-open cubes of side $\delta <tr\mathrm{/}\sqrt{d}$. Each cube that intersects $W_t$ is contained in $W$, because its points lie within distance at most $\sqrt{d}\delta <tr$ of a point whose radius-$tr$ ball lies in $W$. Only finitely many grid cubes intersect bounded $W_t$; their union $R$ is disjoint, finite, contains $W_t$, and is contained in $W$. Therefore $\mathrm{\mid }W\setminus R\mathrm{\mid }\le \mathrm{\mid }W\setminus W_t\mathrm{\mid }<ϵ$.

Fix $ϵ>0$ and choose $R\subseteq W$ as in step 4.1, a finite union of disjoint rectangles. Translation invariance and the symmetric-difference inclusion give $\mathrm{\mid }(W+h)\mathrm{△}W\mathrm{\mid }\le 2\mathrm{\mid }W\setminus R\mathrm{\mid }+\mathrm{\mid }(R+h)\mathrm{△}R\mathrm{\mid }\mathrm{.}$ For each half-open rectangle, its overlap with a translate by $h$ has volume tending to the rectangle's volume as $h\to 0$, by the box-volume formula; the same limit holds for their finite union. Thus ${\mathrm{lim\hspace{0.17em}sup}}_{h\to 0}\mathrm{\mid }(W+h)\mathrm{△}W\mathrm{\mid }\le 2ϵ$, and arbitrariness of $ϵ$ gives convergence to zero.

For $a>0$, [F2] gives $\mathrm{\mid }aW\mathrm{\mid }=a^d\mathrm{\mid }W\mathrm{\mid }$ and $\mathrm{\mid }(aW+x)\mathrm{△}aW\mathrm{\mid }=a^d\mathrm{\mid }(W+x\mathrm{/}a)\mathrm{△}W\mathrm{\mid }$. Dividing and applying step 5.1 with $h=x\mathrm{/}a\to 0$ proves the asserted boundary ratio.

Steps 3.1, 4.1, 5.1, and 6.1 prove positive finite volume, the finite inner rectangle approximation, dilation scaling, and the translated-boundary limit. Countable Choice is used only through [F1]–[F2]; no AC or DC is used.