Does there exist an open subset $A \subset [0,1]$ such that $m_*(A)\neq m_*(\bar{A})$?

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.

Any help will be appreciated

• Is $m_*$ outer or inner Lebesgue measure? – Alex Ortiz Feb 17 at 4:05
• Sir it is outer measure. Is answer will change if we put inner measure? As i had only studied outer measure. I was thinking till thar inner measure will work same. – MathLover Feb 17 at 4:08
• It is actually inconsequential since both inner and outer measure are subadditive with respect to countable unions; see my answer below. Cheers! – Alex Ortiz Feb 17 at 4:19

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 $$m_*(A) < 1/8 < 1/2 = m_*\big(\overline A\big).$$
• Why not just let $A$ be an open subset of $(0,1)$ containing $\mathbb Q \cap (0,1)$ such that $m(A)<1/2?$ – zhw. Feb 17 at 4:43