Improper multiple integrals and absolute convergence on open sets

Let $D\subseteq {\mathbb{R}}^n$ be open. A function $f:D\to \mathbb{R}$ is locally Riemann integrable when its restriction to every compact Jordan set $K\subseteq D$ is Riemann integrable.

For nonnegative $f$, its improper integral is the extended-real supremum of its compact Jordan integrals.

${\int }_{D}f:=\sup \{{\int }_{K}f:K\subseteq D\text{ is compact and Jordan measurable}\}\in [0,+\mathrm{\infty }]\mathrm{.}$

The supremum exists in $\bar{\mathbb{R}}$ by Every subset of $\bar{\mathbb{R}}$ has a least upper bound and a greatest lower bound in $\bar{\mathbb{R}}$, agreeing with the real supremum and infimum on nonempty sets bounded in $\mathbb{R}$ and includes the empty compact set, whose integral is $0$.

For a signed locally Riemann-integrable $f$, set $f^{+}:=(\mathrm{\mid }f\mathrm{\mid }+f)\mathrm{/}2$ and $f^{-}:=(\mathrm{\mid }f\mathrm{\mid }-f)\mathrm{/}2$. These functions are locally integrable by the absolute-value and linearity clauses of Linearity, monotonicity, the absolute-value estimate and coordinate-slice additivity for the Riemann integral in ${\mathbb{R}}^m$, and $\mathrm{\mid }f\mathrm{\mid }=f^{+}+f^{-}$. The function $f$ is absolutely improperly integrable when ${\int }_D\mathrm{\mid }f\mathrm{\mid }<+\mathrm{\infty }$. A locally Riemann-integrable signed function is improperly integrable precisely when the nonnegative improper integral of its absolute value is finite. In that case define

${\int }_{D}f:={\int }_D{f}^{+}-{\int }_D{f}^{-},$

a difference of finite real numbers. Thus this exhaustion-independent signed convention has no $+\mathrm{\infty }-(+\mathrm{\infty })$ branch.

Conditional one-variable improper integrals use a fixed order of approach to their endpoints. The definition here instead requires independence from compact Jordan exhaustions, so signed integrability is absolute in every dimension.