Assuming countable choice, Carathéodory measurability of a finite-outer-measure set is equivalent to source-algebra approximation

For the forward direction, suppose $E$ is Carathéodory measurable. Choose by [F1] a cover $(C_k)$ of $E$ with finite cost below ${\mu }^{\ast }(E)+\epsilon \mathrm{/}4$ and put $H={\bigcup }_k{C}_k$. Applying the Carathéodory identity for $E$ to the test set $H$ shows that ${\mu }^{\ast }(H\setminus E)<\epsilon \mathrm{/}4$; the finite total covering cost also has a tail below $\epsilon \mathrm{/}4$, so some finite union $A={\bigcup }_{k<n}{C}_k\in {\mathcal{A}}_0$ satisfies ${\mu }^{\ast }(H\setminus A)<\epsilon \mathrm{/}4$.

For the forward direction, step 1.1 gives $E\mathrm{△}A\subseteq (H\setminus E)\cup (H\setminus A)$ and hence ${\mu }^{\ast }(E\mathrm{△}A)<\epsilon$. For the reverse direction, assume the approximation property, fix a test set $T$, and choose $A\in {\mathcal{A}}_0$ with ${\mu }^{\ast }(E\mathrm{△}A)<\delta$; since ${\mu }^{\ast }$ is an outer measure by [L3] and $A$ is measurable by [L1], subadditivity gives ${\mu }^{\ast }(T\cap E)+{\mu }^{\ast }(T\setminus E)\le {\mu }^{\ast }(T\cap A)+{\mu }^{\ast }(T\setminus A)+2\delta ={\mu }^{\ast }(T)+2\delta$.

For the reverse direction, letting $\delta$ decrease to $0$ in step 2.1 gives ${\mu }^{\ast }(T\cap E)+{\mu }^{\ast }(T\setminus E)\le {\mu }^{\ast }(T)$, and [L2] applied to the outer measure of [L3] gives the opposite inequality; thus the Carathéodory identity holds for every $T$, completing both implications.