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By countable subadditivity of outer measure, $\mathrm{omes}(A\cup G)\le \mathrm{omes}\left(A\right)+\mathrm{omes}\left(G\right)$. For the reverse inequality, cover $A\cup G$ by open intervals with total length arbitrarily close to $\mathrm{omes}(A\cup G)$. Since $G$ is open, each closed bounded part of $G$ has positive distance from $A$, so the intervals in the cover can be split into a part covering $A$ and a disjoint part covering that closed part of $G$. Taking suprema over closed bounded subsets of $G$, and then the infimum over covers of $A\cup G$, gives the reverse inequality.

Let $G=ℝ\setminus F$. Then $G$ is open, and $A\cap F=\varnothing$ implies $A\subseteq G$. Apply additivity when one set is open to $A\cup F$ inside large bounded intervals and pass to the limit over those intervals. Equivalently, approximate $F$ from inside by closed bounded sets $F\cap [-n,n]$, use the open-set case on their complements, and use order preservation to let $n\to \infty$. This yields $\mathrm{omes}(A\cup F)\ge \mathrm{omes}\left(A\right)+\mathrm{omes}\left(F\right)$, while the opposite inequality is subadditivity.

Let $𝒞$ be the class of Borel sets with this approximation property. Open sets belong to $𝒞$ because they are increasing unions of closed sets $F_n=\{x\in [-n,n]:\mathrm{dist}(x,ℝ\setminus B)\ge 1/n\}$. The class $𝒞$ is closed under complements and countable unions by using countable subadditivity and distributing an error $ϵ/2^k$ over the pieces. Since the Borel $\sigma$-algebra is generated by the open sets, every Borel set belongs to $𝒞$.

Choose, by Borel approximation from below, a closed set $F\subseteq B$ such that $\mathrm{omes}(B\setminus F)<ϵ$. Since $A\cap F=\varnothing$, additivity with a closed set gives $$\mathrm{omes}(A\cup F)=\mathrm{omes}\left(A\right)+\mathrm{omes}\left(F\right).$$ Using subadditivity on $B=F\cup (B\setminus F)$ and on $A\cup B=(A\cup F)\cup (B\setminus F)$, we get the reverse inequality up to $ϵ$. Letting $ϵ\downarrow 0$ proves equality; the other inequality is again subadditivity.

By outer measure is not additive on all subsets, there is a subset $V\subseteq [\mathrm{0,1}]$ for which outer measure fails to be additive on disjoint unions. Since $V\subseteq [\mathrm{0,1}]$, order preservation and the closed interval computation give $\mathrm{omes}\left(V\right)\le 1$. If $V$ were Borel, then outer measure would be additive when one set is Borel, contradicting the construction. Thus $V$ is a non-Borel set of finite outer measure.

The empty set has outer measure $0$. If $A,B\in ℬ$ are disjoint, then additivity when one set is Borel gives finite additivity. For a disjoint sequence $B_1,B_2,\dots \in ℬ$, apply finite additivity to the first $n$ sets and then use order preservation for the lower bound. The upper bound is countable subadditivity. Letting $n\to \infty$ gives countable additivity, so the restriction is a measure.

The definition immediately gives the fourth condition. The approximation theorem for Borel sets gives the second condition from the fourth, and the second implies the third by choosing $F_k\subseteq A$ with $\mathrm{omes}(A\setminus F_k)<1/k$. The third implies the fourth because a Borel set is closed under countable unions. The open-set conditions follow by applying the closed-set conditions to complements and using subadditivity; conversely, if $B\supseteq A$ is Borel and $\mathrm{omes}(B\setminus A)=0$, then a Borel subset of $B$ differing from $A$ by outer measure zero gives the defining condition. Hence all seven conditions are equivalent.

The equivalent characterizations show that complements and countable unions of Lebesgue measurable sets are again Lebesgue measurable, so they form a $\sigma$-algebra. If $A_1,A_2,\dots$ are disjoint Lebesgue measurable sets, choose Borel sets $B_k\supseteq A_k$ with $\mathrm{omes}(B_k\setminus A_k)=0$. Countable additivity on Borel sets and subadditivity over the null differences give countable additivity on $ℒ$. Therefore outer measure restricted to $ℒ$ is a measure.

A point $x\in [\mathrm{0,1}]$ remains in the Cantor set exactly when no stage of the construction removes it. In base $3$, the first stage removes the numbers whose first ternary digit is $1$, the second removes those whose first allowed digit is followed by a $1$, and so on. Thus the points never removed are precisely those admitting a ternary expansion using only $0$ and $2$. At ternary-expansion endpoints, choose the expansion that avoids a trailing string of $2$'s, which keeps the same description valid.

At stage $n$, the removed set is a finite union of open intervals, so its complement is closed; hence $C$, an intersection of closed sets, is closed. The total length removed is $$\sum _{n=1}^{\infty }\frac{2^{n-1}}{3^n}=1,$$ so $C\subseteq [\mathrm{0,1}]$ has Lebesgue measure $0$ by subadditivity and the interval computation. Every nontrivial interval contains a subinterval removed at some finite stage, so no interval with more than one point is contained in $C$.

The base-three description sends $x=0.x_1{x}_2{\cdots }_3\in C$, with each $x_k\in \left\{\mathrm{0,2}\right\}$, to $0.y_1{y}_2{\cdots }_2$, where $y_k=x_k/2$. This produces every binary expansion in $[\mathrm{0,1}]$, so $\mathrm{\Lambda }\left(C\right)=[\mathrm{0,1}]$. If $x\le y$, then the first ternary digit at which they differ is smaller for $x$, so the corresponding binary value is no larger; hence $\mathrm{\Lambda }$ is increasing. On each removed middle-third interval the truncation rule makes $\mathrm{\Lambda }$ constant, and the endpoint values match the limiting Cantor values, so $\mathrm{\Lambda }$ is continuous.

Let $N\subseteq [\mathrm{0,1}]$ be a non-Lebesgue-measurable set, and choose $A\subseteq C$ with $\mathrm{\Lambda }\left(A\right)=N$, possible because $\mathrm{\Lambda }\left(C\right)=[\mathrm{0,1}]$. Since $A\subseteq C$ and $C$ has Lebesgue measure $0$, order preservation gives $\mathrm{omes}\left(A\right)=0$, so $A$ is Lebesgue measurable. But $\mathrm{\Lambda }\left(A\right)=N$ is not Lebesgue measurable by construction.