Proof of Zero Extension Continuity and Box Independence of the Iterated Integral

Throughout, $\Omega$, $D$, $\omega$, $f$, $\tilde f$ are as in the statement, with notions from Continuous n-Form, Support, and Zero Extension on a Euclidean or Half-Space Domain.

Preliminary observation. $\tilde f=0$ on $D\setminus\operatorname{supp}\omega$. Indeed, if $x\in D\setminus\Omega$ this holds by definition of the zero extension; and if $x\in\Omega\setminus\operatorname{supp}\omega$, then by the definition of the support there is an open $W\ni x$ with $\omega_y=0$ for all $y\in W\cap\Omega$; taking $y=x$ gives $f(x)=0$.

Claim 1. Let $x\in D$. If $x\in\Omega$: $\Omega$ is open in $D$ and $\tilde f=f$ on $\Omega$, which is continuous at $x$; agreement on a set open in $D$ around $x$ gives continuity of $\tilde f$ at $x$. If $x\notin\Omega$, then $x\notin\operatorname{supp}\omega$ (as $\operatorname{supp}\omega\subseteq\Omega$), so there is an open $W\ni x$ with $\omega_y=0$ for all $y\in W\cap\Omega$; by the preliminary observation and the definition of $\tilde f$, $\tilde f=0$ on $W\cap D$, so $\tilde f$ is locally constant, hence continuous, at $x$.

Claim 2. $\operatorname{supp}\omega$ is compact, hence bounded by Compact Subset of $\mathbb{R}^n$ is Bounded; choose $R>0$ with $d(x,0)\le R$ for all $x\in\operatorname{supp}\omega$, where $d$ is the Euclidean distance. If $D=\mathbb{R}^n$, the closed box $B=[-R,R]^n$ works. If $D=H^n$, every point of $\operatorname{supp}\omega\subseteq H^n$ has last coordinate in $[0,R]$, so $B=[-R,R]^{n-1}\times[0,R]\subseteq H^n$ works, and any closed box contained in $H^n$ must have $a_n\ge 0$.

Claim 3. First, $\tilde f$ is uniformly continuous on $B$: for every $\varepsilon>0$ there is $\delta>0$ such that $|\tilde f(u)-\tilde f(v)|<\varepsilon$ whenever $u,v\in B$ and $d(u,v)<\delta$. If not, there are $\varepsilon>0$ and points $u_m,v_m\in B$ with $d(u_m,v_m)<1/m$ and $|\tilde f(u_m)-\tilde f(v_m)|\ge\varepsilon$; applying the Bolzano–Weierstrass theorem to each coordinate sequence in turn and passing to successive subsequences yields a subsequence of $(u_m)$ converging coordinatewise to some $w$, whose coordinates lie in the corresponding closed intervals, so $w\in B$; then $(v_m)$ converges to $w$ along the same subsequence, and continuity of $\tilde f$ at $w$ (claim 1) forces $|\tilde f(u_m)-\tilde f(v_m)|\to 0$, a contradiction.

Now proceed by induction on $r$. The function $t_1\mapsto \tilde f(t_1,\dots,t_n)$ is continuous on $[a_1,b_1]$ in the sense of Continuity on a Closed Interval, hence Riemann integrable, so $G_1$ is defined. Moreover, from the definition of the Riemann integral via upper and lower sums, if two integrands on $[a_1,b_1]$ differ by less than $\varepsilon$ at every point, their integrals differ by at most $(b_1-a_1)\varepsilon$. Hence, by uniform continuity of $\tilde f$ on $B$, $G_1$ is (uniformly) continuous on $[a_2,b_2]\times\cdots\times[a_n,b_n]$. Repeating this argument at each stage, every integrand is continuous, every $G_r$ is defined and uniformly continuous in its remaining variables, and $G_n\in\mathbb{R}$ is well defined.

Claim 4. Let $B$ and $B'$ be closed boxes contained in $D$ with $\operatorname{supp}\omega\subseteq B\cap B'$. Let $B''$ be the coordinatewise hull, whose $j$th interval is $[\min(a_j,a_j'),\max(b_j,b_j')]$; then $B\cup B'\subseteq B''$, $\operatorname{supp}\omega\subseteq B''$, and $B''\subseteq D$ (in the half-space case both $a_n\ge 0$ and $a_n'\ge 0$, so the minimum is $\ge 0$). It therefore suffices to show: if $B\subseteq B''$ are boxes in $D$ both containing $\operatorname{supp}\omega$, the iterated integrals agree. Enlarge one coordinate interval at a time. At stage $r$, by additivity on adjacent intervals,

$$\int_{a_r''}^{b_r''} = \int_{a_r''}^{a_r} + \int_{a_r}^{b_r} + \int_{b_r}^{b_r''},$$

applied to the integrand $G_{r-1}(t_r,\dots,t_n)$. If $t_r\notin[a_r,b_r]$, then every point $(t_1,\dots,t_n)$ of $D$ with that $r$th coordinate lies outside $B\supseteq\operatorname{supp}\omega$, so $\tilde f$ vanishes at all such points by the preliminary observation; hence all the inner iterated integrals vanish (the Riemann integral of the zero function is $0$, immediately from the definition), so $G_{r-1}(t_r,\dots,t_n)=0$ on the two outer intervals. Thus enlarging the $r$th interval does not change the value, and after $n$ stages the iterated integral over $B''$ equals that over $B$. The same holds for $B'$, so the values for $B$ and $B'$ coincide. $\blacksquare$