$\Theta \rho ϵ\eta \Pi \alpha \tau \pi$

First suppose $F=[a,b]$ with $a<b$, and let $𝒞$ be an open cover of $[a,b]$. Define $$D:=\{d\in [a,b]:[a,d]\text{ has a finite subcover from }𝒞\}.$$ Since $a$ is contained in some member of $𝒞$, we have $a\in D$, so $D\ne \varnothing$. Also $D\subseteq [a,b]$, so $D$ is bounded above. Let $s=\sup D$, using the supremum.

Choose $G\in 𝒞$ with $s\in G$. Since $G$ is open, there exists $\delta \in {ℝ}^{+}$ such that $(s-\delta ,s+\delta )\subseteq G$. By the definition of $s=\sup D$, some $d\in D$ satisfies $s-\delta <d\le s$. Since $d\in D$, there are $G_1,\dots ,G_n\in 𝒞$ such that $$[a,d]\subseteq G_1\cup \cdots \cup G_n.$$ Then for every $d^{\prime }\in [s,s+\delta )\cap [a,b]$, $$[a,d^{\prime }]\subseteq G_1\cup \cdots \cup G_n\cup G.$$ Thus every such $d^{\prime }$ is in $D$. If $s<b$, this puts points of $D$ larger than $s$, contradicting that $s$ is an upper bound. Hence $s=b$, and the displayed containment with $d^{\prime }=b$ gives a finite subcover of $[a,b]$.

Now let $F\subseteq ℝ$ be closed and bounded, and let $𝒞$ be an open cover of $F$. Choose $a,b\in ℝ$ with $F\subseteq [a,b]$. Since $F$ is closed, $ℝ\setminus F$ is open. Therefore $𝒞\cup \{ℝ\setminus F\}$ is an open cover of $[a,b]$. By the interval case, finitely many of these sets cover $[a,b]$. Removing $ℝ\setminus F$, if it appears, leaves finitely many members of $𝒞$ that cover $F$. Thus $𝒞$ has a finite subcover.

For every $ϵ\in {ℝ}^{+}$, the open interval $(a-ϵ,b+ϵ)$ covers $[a,b]$. Hence by the definition of outer measure, $$\mathrm{omes}\left([a,b]\right)\le \ell \left((a-ϵ,b+ϵ)\right)=b-a+2ϵ.$$ Since $ϵ\in {ℝ}^{+}$ was arbitrary, $\mathrm{omes}\left([a,b]\right)\le b-a$.

For the reverse inequality, let $I_1,I_2,\dots$ be any sequence of open intervals with $$[a,b]\subseteq \bigcup _{k=1}^{\infty }{I}_k.$$ By the Heine-Borel theorem, finitely many of these intervals cover $[a,b]$. Thus there is $n\in {\mathbb{N}}_1$ such that $$[a,b]\subseteq I_1\cup \cdots \cup I_n$$ after relabeling if needed.

We claim that ${\sum }_{k=1}^n\ell \left(I_k\right)\ge b-a$. This is proved by induction on $n$. For $n=1$, one open interval covering $[a,b]$ must have length at least $b-a$. For the induction step, suppose $I_1,\dots ,I_{n+1}$ cover $[a,b]$. Relabel so $b\in I_{n+1}=(c,d)$. If $c\le a$, then $\ell \left(I_{n+1}\right)\ge b-a$. Otherwise $a<c<b$, and $I_1,\dots ,I_n$ cover $[a,c]$. By the induction hypothesis, $$\sum _{k=1}^n\ell \left(I_k\right)\ge c-a.$$ Therefore $$\sum _{k=1}^{n+1}\ell \left(I_k\right)\ge (c-a)+(d-c)=d-a\ge b-a.$$ Hence every open-interval cover of $[a,b]$ has total length at least $b-a$, so taking the infimum gives $b-a\le \mathrm{omes}\left([a,b]\right)$. Combining the two inequalities proves the result.

Define a relation on $[\mathrm{0,1}]$ by $x\sim y$ if and only if $x-y\in \mathbb{Q}$. This is an equivalence relation. Choose a set $V\subseteq [\mathrm{0,1}]$ containing exactly one representative from each equivalence class. Let $$Q_0:=\mathbb{Q}\cap [-\mathrm{1,1}].$$ For each $q\in Q_0$, consider the translate $q+V$.

If $q,r\in Q_0$ and $q\ne r$, then $(q+V)\cap (r+V)=\varnothing$. Indeed, if $q+v=r+w$, then $v-w=r-q\in \mathbb{Q}$, so $v\sim w$. Since $V$ contains only one representative from each equivalence class, $v=w$, and then $q=r$, a contradiction.

Also, $$[\mathrm{0,1}]\subseteq \bigcup _{q\in Q_0}(q+V)\subseteq [-\mathrm{1,2}].$$ The left inclusion holds because every $x\in [\mathrm{0,1}]$ is equivalent to some $v\in V$, so $x-v\in \mathbb{Q}\cap [-\mathrm{1,1}]$. The right inclusion holds because $q\in [-\mathrm{1,1}]$ and $V\subseteq [\mathrm{0,1}]$.

Suppose, for contradiction, that outer measure were additive for every pair of disjoint subsets of $ℝ$. Then it would be additive for every finite disjoint union. By translation invariance, each $q+V$ has outer measure $\mathrm{omes}\left(V\right)$. Since every finite union of these disjoint translates is contained in $[-\mathrm{1,2}]$, order preservation and the outer measure of a closed interval give $$n\mathrm{omes}\left(V\right)\le 3$$ for every $n\in {\mathbb{N}}_1$. Hence $\mathrm{omes}\left(V\right)=0$.

Because $Q_0$ is countable, countable subadditivity gives $$\mathrm{omes}\left(\bigcup _{q\in Q_0}(q+V)\right)\le \sum _{q\in Q_0}\mathrm{omes}(q+V)=0.$$ But this union contains $[\mathrm{0,1}]$, so by order preservation and the closed interval computation, its outer measure is at least $1$, a contradiction. Therefore outer measure cannot be additive on all disjoint pairs of subsets of $ℝ$.

Assume, for contradiction, that such a function $\mu$ exists. Use the same representative set $V\subseteq [\mathrm{0,1}]$ and the same countable set $Q_0=\mathbb{Q}\cap [-\mathrm{1,1}]$ from the proof of outer measure is not additive. The sets $q+V$, for $q\in Q_0$, are pairwise disjoint, their union contains $[\mathrm{0,1}]$, and their union is contained in $[-\mathrm{1,2}]$.

By translation invariance of $\mu$, each set $q+V$ has the same $\mu$-value as $V$. Since every finite union of these sets is contained in $[-\mathrm{1,2}]\subseteq (-\mathrm{2,3})$, countable additivity, monotonicity, and the extension-of-length property imply $$n\mu \left(V\right)=\mu \left(\bigcup _{j=1}^n(q_j+V)\right)\le \mu \left((-\mathrm{2,3})\right)=5$$ for every finite list $q_1,\dots ,q_n$ of distinct elements of $Q_0$. Hence $\mu \left(V\right)=0$.

Now use countable additivity on the disjoint sequence of all translates $q+V$, $q\in Q_0$. We get $$\mu \left(\bigcup _{q\in Q_0}(q+V)\right)=\sum _{q\in Q_0}\mu (q+V)=0.$$ But $\left(\mathrm{0,1}\right)$ is contained in that union, so monotonicity, which follows from countable additivity, gives $$1=\mu \left(\left(\mathrm{0,1}\right)\right)\le \mu \left(\bigcup _{q\in Q_0}(q+V)\right)=0,$$ a contradiction. Therefore no countably additive translation-invariant extension of interval length can be defined on all subsets of $ℝ$.