Let $X$ be a set, and $\rho$ an outer measure on $X$. Let $\mathcal{M}(\rho)$ be the set of $\rho$-measurable sets. Then the restriction $\rho|_{\mathcal{M}(\rho)}$ is a measure on $\mathcal{M}(\rho)$. By Hahn-Kolmogorov extension theorem, $\rho|_{\mathcal{M}(\rho)}$ can be extended to an outer measure $\rho|_{\mathcal{M}(\rho)}^\ast$ on $X$. (As the infimum of the sum of the $\rho$-measures of countably many $\rho$-measurable sets covering a set.) My question is whether the two outer measures $\rho$ and $\rho|_{\mathcal{M}(\rho)}^\ast$ agree.
I can see that $\rho\leq\rho|_{\mathcal{M}(\rho)}^\ast$, but I doubt that they are equal in general.
I can show that they are equal if $\rho$ is itself the extension of a (pre-)measure, for example if $\rho$ is the Lebesgue outer measure.
In general, $\rho$ and $\rho|_{\mathcal{M}(\rho)}^\ast$ are not equal. Here is a counter-example.
Let $X=\Bbb N$ and let $k \in \Bbb N$, $k \geq 2$. Let us define $\rho$ by, for all $A \subseteq X$,
$$ \rho(A) = \left \{ \begin{matrix} \# A & \text{ if }\# A \leq k \\ k & \text{ if } \# A > k \end{matrix} \right. $$ where $\# A$ is the number of elements in $A$ if $A$ is finite and $\# A$ is $+\infty$ if $A$ is infinite.
It is easy to see that $\rho$ is in fact an outer measure.
Now given any $B \subseteq X$, such that $B \neq \emptyset$ and $B \neq X$, then $B$ and $B^c$ are non-empty sets and, at least of them is infinite. Suppose $B$ is infinite. Let $E = B \cup \{e\}$ where $e \in B^c$. So $$ \rho(E)=k < k + 1= \rho(E \cap B) + \rho(E \cap B^c) $$ So $B$ is not measurable. Now, suppose $B^c$ is infinite. In a similar way define $E = B^c \cup \{e\}$ where $e \in B$. Then $$ \rho(E)=k < 1 + k= \rho(E \cap B) + \rho(E \cap B^c) $$ So $B$ is not measurable. So, in both case, $B$ is not measurable. So, we have that $\mathcal{M}(\rho)=\{\emptyset, X \}$ and then the measure $\rho|_{\mathcal{M}(\rho)}$ is given by $\rho|_{\mathcal{M}(\rho)}(\emptyset)=0$ and $\rho|_{\mathcal{M}(\rho)}(X)=k$. It follows immediately that $$ \rho|_{\mathcal{M}(\rho)}^\ast(A) = \left \{ \begin{matrix} 0 & \text{ if } A = \emptyset \\ k & \text{ if } A \neq \emptyset \end{matrix} \right. $$
So $\rho \neq \rho|_{\mathcal{M}(\rho)}^\ast$. In fact, given any $e \in X$, we have $$ \rho(\{e\})=1 < 2 \leqslant k = \rho|_{\mathcal{M}(\rho)}^\ast(\{e\})$$