I was wondering if a regular outer measure remains a regular outer measure when the space is a measurable subset of the original space. Rigorously speaking :
Let $X$ be our embedding space and let $μ^∗$ be a regular outer measure and let $H \subset X$ be measurable with $μ(H) < \infty$. For any set $A \subset X$ there is a measurable set $B \subset X$ ($A \subset B$) such that $μ^∗(A)=μ(B)$; which includes this special case : for any $A \cap H \subset X$ there is a measurable set $C \subset X$ such that $μ^∗(A\cap H)=μ(C)$. I was wondering :
1- Is there always exist a measurable set of the form $B \cap H$ such that for $A \cap H$, $μ^∗(A\cap H)=μ(B \cap H)$?
2- There always is a measurable set $C \subset X$ such that $μ^∗(A\cap H)=μ(C)$ by definition. Does necessarily this $C$ equal $B \cap H$ for some $B$?
You know that for any $A \cap H \subset X$ there is a measurable set $C \subset X$ such that $A\cap H \subset C$ and $\mu^∗(A\cap H)=\mu(C)$.
Note that $A\cap H \subset C\cap H \subset C$ and that $ C\cap H $ is measurable. So we have $$ \mu^∗(A\cap H) \leq \mu(C \cap H) \leq \mu(C)$$
Since $\mu^∗(A\cap H)=\mu(C)$, we have $\mu^∗(A\cap H) = \mu(C \cap H) = \mu(C)$.
So for your first question, just take $B=C$.
For your second question, $C$ is not necessarily equal $B \cap H$ for some $B$, but it can also be replaced by $C \cap H$.