Riemann–Lebesgue comparison for distribution test integrands

Choose $C\ge 0$ with $\mathrm{\mid }f\mathrm{\mid }\le C$, and put $g=f+C\ge 0$. F2 and F3 give ${\int }_Q\mathrm{\mid }f\mathrm{\mid }\le C\mathrm{vol}Q<\mathrm{\infty }$ and ${\int }_{Q}g\le 2C\mathrm{vol}Q<\mathrm{\infty }$, so both are integrable. Borel measurability makes all these integrals defined.

For a finite rectangular grid, list its closed cells $Q_1,\dots ,Q_s$ and set $D_i=Q_i\setminus {\bigcup }_{j<i}{Q}_j$. These Borel sets partition $Q$, contain each cell's interior and lie in its closure, so ${\lambda }_n(D_i)=\mathrm{vol}{Q}_i$ by F2. Put $m_i={\inf }_{Q_i}g$ and $M_i={\sup }_{Q_i}g$. The simple functions $l={\sum }_i{m}_i{\mathbf{1}}_{D_i}$ and $h={\sum }_i{M}_i{\mathbf{1}}_{D_i}$ satisfy $l\le g\le h$ everywhere, including grid faces. F3 identifies their integrals as the grid's lower and upper Darboux sums for $g$, so these sums bracket ${\int }_{Q}g$.

Every tagged sum of $g$ is the corresponding sum of $f$ plus $C\mathrm{vol}Q$. Thus $g$ is Riemann integrable with value $I_R(f)+C\mathrm{vol}Q$. F1 says its supremum of lower Darboux sums and infimum of upper sums have this same value. Taking these bounds in step 2.1 squeezes ${\int }_{Q}g$ to that value. F4 and F2 then give ${\int }_{Q}f={\int }_{Q}g-C\mathrm{vol}Q=I_R(f)$.

Apply the real result separately to the real and imaginary parts for the complex assertion; the bound $\mathrm{\mid }f\mathrm{\mid }\le \mathrm{\mid }\mathrm{Re}f\mathrm{\mid }+\mathrm{\mid }\mathrm{Im}f\mathrm{\mid }$ ensures absolute integrability. For the last clause, let $E\subseteq Q$ be compact Jordan and let $h:E\to \mathbb{R}$ be continuous. If $E$ is empty its zero extension is zero. Otherwise F6 makes $h$ bounded, and compactness makes $E$ closed. Its zero extension $\tilde{h}$ is Borel: for every open $O\subseteq \mathbb{R}$, continuity in the subspace gives $h^{-1}(O)=E\cap G$ for some open $G\subseteq Q$; then ${\tilde{h}}^{-1}(O)=E\cap G$ when $0\mathrm{\notin }O$, while ${\tilde{h}}^{-1}(O)=(Q\setminus E)\cup (E\cap G)$ when $0\in O$. The definition and theorem in F6 say exactly that this bounded zero extension is Riemann integrable on $Q$. The real result therefore applies; treating real and imaginary parts gives the same conclusion for continuous complex integrands, in particular smooth compact test integrands. The zero function has both integrals zero; the constant one has both integrals $\mathrm{vol}Q$. Degenerate boxes are excluded from this statement. No choices of tags over an infinite family were used; Countable Choice is precisely the measure hypothesis in F2. $\mathrm{\square }$