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I'm working on counterexample here. Can we construct two bounded non empty open sets $A,B$ with $A \subset B$ that are $\lambda(A)=\lambda(B)$ but $\overline{A}\ne\overline{B}$? Here $\lambda$ is the Lebesgue measure, thank you.

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No. If $A\subseteq B$, but $\operatorname{cl}A\ne\operatorname{cl}B$, there must be a point $x\in(\operatorname{cl}B)\setminus\operatorname{cl}A$. Let $U=B\setminus\operatorname{cl}A$; then $x\in\operatorname{cl}U$, so $U$ is a non-empty open subset of $B$ disjoint from $A$, and $\lambda(B)>\lambda(A)$.

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If $b$ is in the closure of $B$ but not the closure of $A$, then there exists an open ball $V$ around $b$ that does not intersect $A$. But that ball must intersect $B$ and the intersection is an open set. Thus, you have an open set that is in $B$ but not $A$. Now, open sets typically have positive measure and so I think this implies that $\lambda(A) \neq \lambda(B)$. So the problem cannot be done.

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