Riemann-integrable half-space extensions of chart coefficients

Compact support of a differential form: Let $M$ be a smooth manifold, possibly with boundary, and $k\ge 0$. For $\omega \in {\mathrm{\Omega }}^k(M)$ define $\mathrm{supp}\omega ={\overline{\{p\in M:{\omega }_p\mathrm{\ne }0\}}}^M,{\mathrm{\Omega }}_c^k(M)=\{\omega \in {\mathrm{\Omega }}^k(M):\mathrm{supp}\omega \text{ is compact}\}\mathrm{.}$ The closure and compactness are in $M$, including its genuine boundary. Zero is the intrinsic zero of each exterior-power fiber, so this definition is independent of trivialization. The zero form has empty support.

Smooth functions on relatively open half-space sets: Near each point of $U$, the function agrees on $U$ with a smooth function on a Euclidean-open neighbourhood. In particular it is continuous on $U$ and smooth on its Euclidean interior. No double or simultaneous selection of extensions is needed.

Compactness can be read using ambient open covers (A subspace is compact exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it); a compact subset of a Hausdorff space is closed (In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones). The induced topology on $K$ is the same whether inherited from $U$, $H^n$, or ${\mathbb{R}}^n$.

Lebesgue's criterion in ${\mathbb{R}}^m$: a bounded function on a closed nondegenerate rectangle is Riemann integrable iff its discontinuity set is null: A bounded real function on a closed nondegenerate rectangle in ${\mathbb{R}}^m$, $m\ge 1$, is Riemann integrable if and only if its discontinuity set is null.

Measure zero and content zero in ${\mathbb{R}}^m$ by countable and finite cube covers: Fix $m\ge 1$. A closed cube is a rectangle ${\prod }_{j<m}[a_j,a_j+\mathrm{\ell }]$ with $\mathrm{\ell }\ge 0$; its volume is ${\mathrm{\ell }}^m$. A set $E\subseteq {\mathbb{R}}^m$ is null when, for every $\epsilon >0$, it is covered by a sequence of closed cubes whose nonnegative volume series converges with sum at most $\epsilon$. It has content zero when such a cover can be finite. The series and finite sums are def-series and def-finite-sum, and their nonnegative bounds use thm-nonnegative-series-bounded-partial-sums and lem-finite-sum-laws. Both properties pass to subsets. Padding a finite cover with degenerate zero-volume cubes proves that content zero implies null. This terminology defines only cover-nullity; it does not define a measure on arbitrary sets.

For every $\epsilon >0$ in a complete ordered field there is a natural $n\ge 1$ with $1\mathrm{/}n<\epsilon$: Let $F$ be a complete ordered field (def-complete-ordered-field) and let $\epsilon \in F$ with $\epsilon >0$. Then there is a natural number $n\ge 1$ such that $\frac{1}{n\cdot 1_F}<\epsilon ,$ where $n\cdot 1_F$ is the canonical natural of $F$ (thm-of-archimedean) and $1\mathrm{/}(n\cdot 1_F)$ is its multiplicative inverse (def-field). As is standard we abbreviate $n\cdot 1_F$ to $n$ and write the conclusion $1\mathrm{/}n<\epsilon$. This is the reciprocal form of the Archimedean property. thm-of-archimedean on its own delivers only the assertion that the canonical naturals are cofinal, $x<n\cdot 1_F$; the form actually used in analysis, that the reciprocals of the naturals get below every positive bound, is the statement above, and it is recorded separately so that no proof has to reconstruct the inversion step in passing.

The Riemann integral of a compactly supported function is independent of its bounding rectangle: Let $n\ge 1$ and let $f:{\mathbb{R}}^n\to \mathbb{R}$ have compact support. If $f$ is Riemann integrable on one closed rectangle whose interior contains its support, then it is integrable on every such rectangle, and all the resulting integrals are equal. This includes the empty-support case.