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I want to prove that if $K$ is a compact and convex set of $\mathbb{R}^n$ and $Z=\{z_1, \cdots, z_n\}$ is a finite set of points of $\mathbb{R}^n$ then $$\operatorname{vol}(Z+K)=\operatorname{vol}(\operatorname{int}(Z+K))=\operatorname{vol}(Z+\operatorname{int}K)$$ I know the first equality is true in general so I think I just need to prove the second one, but I don't really know how to do it.

At first I thought that $\operatorname{int}(Z+K)$ was equal to $Z+\operatorname{int}K$, but as it is shown here, this is not true in general but it suggested me that whenever this is not true, it seems that the difference between the sets is of measure zero so if this was true, and then these two volumes would be indeed equal, but again I don't know if this is true or how to prove it.

Any ideas?

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1 Answer 1

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$Z+\text{int}(K)\subset Z+K$, so any difference between the first and third terms would come from points in the latter set but not the former.

But observe that $Z+\text{int}(K)$ is the translates of points in the interior of $K$, while $Z+K$ is the translates of points anywhere in $K$, so $(Z+K)\setminus Z+\text{int}(K)\subset Z+(K\setminus\text{int}(K))$. But the RHS of this consists of finitely many copies of a set of measure $0$ (namely, the boundary of $K$), so it is also of measure $0$.

More precisely:

$$\text{Vol}(Z+(K\setminus \text{int}(K)))\le\sum_{i=1}^n\text{Vol}(z_i+(K\setminus \text{int}(K)))=\sum_{i=1}^n\text{Vol}(K\setminus \text{int}(K))=\sum_{i=1}^n0=0$$

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