In Chapter $2$ of Royden, #$20$, we are asked to show for a set $E$ of finite outer measure that $E$ is measurable if and only if for each open bounded set $(a,b)$ we have $m^*((a,b))=b-a=m^*((a,b)\cap E) + m^*((a,b)\sim E).$ Note we define $E$ to be measurable if $E$ satisfies Carathéodory's splitting condition.
I have a clear proof, but nowhere did I invoke that $E$ is finite in outer measure. Where is that assumption used in this problem?
For completeness, here is my proof:
Clearly if $E$ is measurable then the claim follows since if we take $A=(a,b)$ in the definition of measurability of a set $E$ then
\begin{equation} \begin{split} m^*(A)& = m^*((a,b))\\ & = b-a\\ & = m^*((a,b)\cap E) + m^*((a,b)\cap E^C)\\ & = m^*((a,b)\cap E) + m^*((a,b)\sim E) \notag \end{split} \end{equation} where the second equality follows from the outer measure of an interval being its length, the third equality follows from the substitution of $A=(a,b)$ into the definition of a measurable set, and finally the fourth equality follows from the set identity $(a,b)\cap E^C=(a,b)\sim E.$ Now we assume for every open bounded interval $(a,b)$ we have \begin{equation} m^*((a,b))=b-a=m^*((a,b)\cap E)+m^*((a,b)\sim E)\tag{1} \end{equation} We want to show $E$ is measurable assuming $(1)$. We thus need to show for every set $A$ of finite outer measure that \begin{equation} \begin{split} m^*(A)& \geq m^*(A\cap E) + m^*(A\cap E^C)\\ & = m^*(A\cap E) + m^*(A\sim E). \notag \end{split} \end{equation} By definition of $m^*(A)$ as the infimum, we need only show that for any countable open cover of $A$ by bounded nonempty open intervals $\{I_k\}_{k=1}^\infty$ we have $$\sum_{k=1}^\infty m^*(I_k)\geq m^*(A\cap E) + m^*(A\sim E).$$ Note \begin{equation} \begin{split} A&\subseteq \bigcup_{k=1}^\infty I_k\implies\\ A\cap E&\subseteq \bigcup_{k=1}^\infty(I_k\cap E)\\ \notag \end{split} \end{equation} and \begin{equation} \begin{split} A\cap E^C&=A\sim E\\ &\subseteq\bigcup_{k=1}^\infty (I_k\cap E^C)\\ & = \bigcup_{k=1}^\infty (I_k\sim E). \notag \end{split} \end{equation} Observe we have for each $I_k=(a_k,b_k)$ assuming $(1)$ that $$m^*(I_k)=b_k-a_k=m^*((a_k,b_k)\cap E)+m^*((a_k,b_k)\sim E),$$ so since $A$ has finite measure the sum is convergent and thus $$\sum_{k=1}^\infty m^*(I_k)=\sum_{k=1}^\infty m^*(I_k\cap E)+\sum_{k=1}^\infty m^*(I_k\sim E).$$ We have by the countable subadditivity and monotonicity of the outer measure that \begin{equation} \begin{split} \sum_{k=1}^\infty m^*(I_k)&=\sum_{k=1}^\infty m^*(I_k\cap E)+\sum_{k=1}^\infty m^*(I_k\sim E)\\ & \geq m^*\left(\bigcup_{k=1}^\infty (I_k\cap E)\right) + m^*\left(\bigcup_{k=1}^\infty (I_k\sim E)\right)\\ & \geq m^*(A\cap E) + m^*(A\sim E). \notag \end{split} \end{equation} Thus $m^*(A)\geq m^*(A\cap E) + m^*(A\cap E^C)$ holds, and hence $E$ is measurable. $\blacksquare$