# Regular outer measure resricted to a measurable subset

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 $$C \cap H$$.