Regular distribution from a locally integrable function

For open $\mathrm{\Omega }\subseteq {\mathbb{R}}^n$, write $f\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(\mathrm{\Omega })$ when $f:\mathrm{\Omega }\to \mathbb{C}$ is Lebesgue measurable and ${\int }_K\mathrm{\mid }f\mathrm{\mid }<\mathrm{\infty }$ for every compact $K\subseteq \mathrm{\Omega }$. This extends A locally integrable function on ${\mathbb{R}}^n$: equivalently, require integrability on every ball whose closure is a compact subset of $\mathrm{\Omega }$. Such balls finitely cover each compact $K$, and conversely their closures are compact; on all of ${\mathbb{R}}^n$, any ball lies in a larger compact closed ball.

For a test $\phi$ as in Test function space d of an open set, define the regular functional

$⟨u_f,\phi ⟩={\int }_{\mathrm{\Omega }}f(x)\phi (x)dx\mathrm{.}$

This is well-defined since $\mathrm{\mid }\int f\phi \mathrm{\mid }\le {\sup }_K\mathrm{\mid }\phi \mathrm{\mid }{\int }_K\mathrm{\mid }f\mathrm{\mid }$ for $K=\mathrm{supp}\phi$. It is complex-linear in $f$ and in $\phi$ and depends only on the almost-everywhere class of $f$. For an empty support the integral is zero. The subsequent embedding theorem establishes continuity, so that this is a distribution, and injectivity modulo almost-everywhere equality. That theorem states the Countable Choice cost of its injectivity proof; no choice is needed to define the displayed pairing.