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I am trying to prove the following proposition:

If $A \subseteq \mathbb{R^n}$ is a Jordan region, then $vol(A) = vol(int(A)) = vol(\overline{A})$.

I believe I have a simple proof to show $vol(A) = vol(\overline{A})$. Namely, just that any covering of boxes of $A$ must also cover $\overline{A}$ by compactness of the boxes. Does this line of reasoning make sense?

What I am stuck on is showing that $vol(A) = vol(int(a))$. I have tried a similar box covering argument as above, but a covering of $int(A)$ does not necessarily cover $A$. I also tried a proof by contradiction by supposing there is a grid $G$ such that $V(int(A), G) = vol(A) - \epsilon$ for some $\epsilon > 0$, where $V(int(A), G)$ is the outer sum of the grid G on $int(A)$, but to no avail.

I'm starting to doubt that this statement is true, but I cannot find any counterexamples, so I would really appreciate any guidance on this problem. Thank you!

learning_linalg
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No, this is not true in general. We can have a simple closed curve in the plane (not rectifiable, of course) where the curve itself has positive 2-dimensional Lebesgue measure. Then for the inside $A$ of that curve, we have strict inequality $m(A) < m(\overline{A})$. (where $m$ = 2-dimensional Lebesgue measure.)

Now of course the region $A$ is not Jordan measurable (by Lebesgue's criterion). So one could ask what you mean by $vol(A)$ in that case.

Searching for reference ... Googling "Jordan curve positive measure", I FOUND ONE.

GEdgar
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