A property holding outside a set of elementary measure zero is exactly a property holding $\lambda$-almost everywhere

Given: A natural number $m\ge 1$, the Axiom of Countable Choice, and a property $P$ of points of ${\mathbb{R}}^m$ with exceptional set $N_0$.

Assuming countable choice, ${\lambda }_1^{\ast }(A)=0$ if and only if $A\subseteq \mathbb{R}$ has measure zero in the covering sense (A subset of $\mathbb{R}$ has Lebesgue outer measure zero if and only if it has measure zero in the sense of countable closed-interval covers, Measure zero (a countable cover by intervals of total length below every $\epsilon$) and content zero (a finite such cover)).

Assuming countable choice, ${\lambda }_m^{\ast }(E)=0$ if and only if $E\subseteq {\mathbb{R}}^m$ is null in the covering sense (A subset of ${\mathbb{R}}^m$ has Lebesgue outer measure zero if and only if it is null in the sense of countable closed-cube covers, Measure zero and content zero in ${\mathbb{R}}^m$ by countable and finite cube covers).

Assuming countable choice, $\mathcal{L}({\mathbb{R}}^m)$ is a sigma-algebra, ${\lambda }_m$ is a complete measure on it and is the restriction of ${\lambda }_m^{\ast }$, and every $S$ with ${\lambda }_m^{\ast }(S)=0$ is Lebesgue measurable of measure $0$ (Assuming countable choice, $\mathcal{L}({\mathbb{R}}^n)$ is a sigma-algebra containing every elementary set and ${\lambda }_n$ is a complete measure extending elementary volume).

Assuming countable choice, ${\lambda }_m^{\ast }$ is an outer measure on ${\mathbb{R}}^m$, hence monotone (Assuming countable choice, Lebesgue outer measure is an outer measure that restricts to elementary volume, Outer measures).

A property $P(x)$ holds $\mu$-almost everywhere if its exceptional set is contained in a measurable $\mu$-null set: there is $N\in \mathcal{A}$ with $\mu (N)=0$ such that $P(x)$ holds for every $x\in X\setminus N$ (Measure-null sets and almost-everywhere statements relative to a measure).