Does there exist an open subset $A \subset [0,1]$ such that $m_*(A)\neq m_*(\bar{A})$?
I was thinking we could approximate any set from inside by a closed set . This need not true from outside.
So I was expecting there should be some counterexample to the above statement.
Please help me to construct such an example.
Any help will be appreciated
Enumerate the rational numbers in $[1/4,3/4]$ as $\{r_n:n\in\mathbf{Z}_{>0}\}$, and let $\epsilon < 1/8$. For each $n$, choose an interval $I_n$ centered at $r_n$ of width $\epsilon/2^n$, and put $A = \bigcup_{n=1}^\infty I_n$. Note that $A$ is an open subset of $\mathbf R$ contained in $[0,1]$ because it is a union of open intervals. Since $A$ contains the rationals in $[1/4,3/4]$, density of $\mathbf Q$ in $\mathbf R$ implies that the closure $\overline A$ contains the closed interval $[1/4,3/4]$.
By countable subadditivity (of Lebesgue outer or inner measure, this is an inconsequential point for the matter at hand), $$ m_*(A) \le \sum_{n=1}^\infty m_*(I_n) = \sum_{n=1}^\infty \frac{\epsilon}{2^n} = \epsilon < 1/8. $$ On the other hand, by monotonicity, $$ m_*\big(\overline A\big) \ge m_*\big([1/4,3/4]\big) = 1/2, $$ so $$ m_*(A) < 1/8 < 1/2 = m_*\big(\overline A\big). $$