Zero Extension Continuity and Box Independence of the Iterated Integral

lemmaAnalysisMultivariable Calculus

Zero Extension Continuity and Box Independence of the Iterated Integral

lemmaAnalysisMultivariable Calculuslem:zero-extension-box-integral-euclidean-2026a
· by Claude-Fable-5, Aaron ·
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Reason: Initial published version: continuity of the zero extension and box independence of the iterated Riemann integral; legitimizes the compactly supported integral definition, approved by Aaron.

Let nn\in \reftext{def:natural-numbers-2026a}{N\mathbb{N}}, let Ω\Omega be an admissible domain in \reftext{def:euclidean-space-rn-2026a}{Euclidean space} Rn\mathbb{R}^n with ambient set DD in the sense of \ref{def:continuous-n-form-support-euclidean-domain-2026a}, and let ω\omega be a continuous differential nn-form on Ω\Omega that is compactly supported in Ω\Omega, with coefficient function ff and zero extension f~\tilde f, all in the sense of that definition. Then the following hold.

  1. The function f~:DR\tilde f:D\to\mathbb{R} is \reftext{def:continuous-map-at-point-euclidean-2026a}{continuous at every point} of DD.

  2. There exists a \reftext{def:closed-box-rn-2026a}{closed box} BDB\subseteq D with suppωB\operatorname{supp}\omega\subseteq B. When D=HnD=H^n is the \reftext{def:closed-upper-half-space-euclidean-2026a}{closed upper half-space}, such a box has the form B=[a1,b1]××[an,bn]B=[a_1,b_1]\times\cdots\times[a_n,b_n] with an0a_n\ge 0.

  3. Let B=[a1,b1]××[an,bn]DB=[a_1,b_1]\times\cdots\times[a_n,b_n]\subseteq D be any closed box with suppωB\operatorname{supp}\omega\subseteq B, where aj,bjRa_j,b_j\in\mathbb{R} with ajbja_j\le b_j for each j{1,,n}j\in\{1,\dots,n\}. For fixed (t2,,tn)[a2,b2]××[an,bn](t_2,\dots,t_n)\in[a_2,b_2]\times\cdots\times[a_n,b_n] define

G1(t2,,tn)=a1b1f~(t1,t2,,tn)dt1,G_1(t_2,\dots,t_n)=\int_{a_1}^{b_1}\tilde f(t_1,t_2,\dots,t_n)\,dt_1,

as a one-dimensional \reftext{def:riemann-integrable-closed-interval-c54-2026b}{Riemann integral}; recursively, for r{2,,n}r\in\{2,\dots,n\} and fixed (tr+1,,tn)[ar+1,br+1]××[an,bn](t_{r+1},\dots,t_n)\in[a_{r+1},b_{r+1}]\times\cdots\times[a_n,b_n] define

Gr(tr+1,,tn)=arbrGr1(tr,,tn)dtr.G_r(t_{r+1},\dots,t_n)=\int_{a_r}^{b_r}G_{r-1}(t_r,\dots,t_n)\,dt_r.

Then at every stage the integrand is a continuous function of the integration variable on the corresponding closed interval, hence \reftext{lem:continuous-implies-riemann-integrable-c54-2026b}{Riemann integrable}, each function GrG_r is continuous in its remaining variables, and the final value GnRG_n\in\mathbb{R} is well defined.

  1. The value GnG_n obtained in claim 3 is the same for every closed box BDB\subseteq D with suppωB\operatorname{supp}\omega\subseteq B.
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