# Proof Check: Outer Regularity of Lebesgue Measure

I am trying to prove that for any bounded set $$A$$ in the borel $$\sigma$$ algebra, that for the Lebesgue measure $$m$$ $$m(A)=\inf\{m(U)|U\text{ is open and }A\subseteq U\}$$ here is my attempt.

Let $$U$$ be an open set containing $$A$$. Then $$m(A)\leq m(U)$$ by monotonicity. Then $$m(A)$$ is a lower bound on the set $$\{m(U)|U\text{ is open and }A\subseteq U\}$$. Denote the infimum by $$M$$. By definition of infimum, there exists a sequence of open sets $$(U_n)_{n\in\mathbb{N}}$$ such that $$A\subset U_n$$ for each $$n$$, and for all $$\epsilon>0$$ there exists an $$N\in\mathbb{N}$$ such that for all $$n\geq N$$, we have $$m(U_n).

Letting $$\epsilon>0$$ be arbitrary, then there exists $$N\in\mathbb{N}$$ such that $$m(U_n) whenever $$n\geq N$$. Since $$A\subseteq U_n$$ for all $$n$$, then we have $$m(A)\leq m(U_n) which shows that $$M-m(A)<\epsilon$$, and since $$\epsilon$$ was arbitrary, then we must have equality.

Just looking to see if there are any mistakes in my proof. Thanks!

• How do you get $M-m(A) <\epsilon$? You see to have $m(A)-M <\epsilon$ which is of no use. – Kavi Rama Murthy Apr 18 at 8:49
• Should I have $m(U_n)-\epsilon<M$ rather than what I have said? That way I can say $M-m(A)< \epsilon$? @KaviRamaMurthy – JB071098 Apr 18 at 8:54
• No. For proving regularity you need the fact that $m(A)=M$. You have only stated that $m(A)$ is a lower bound. This is not enough to prove regularity. – Kavi Rama Murthy Apr 18 at 8:59
• @KaviRamaMurthy Could you show me how? Even a hint would be appreciated. – JB071098 Apr 18 at 9:02

For any Lebesgue measurable set $$A$$ we have $$m(A)=\inf \{\sum m(I_n): A \subset \cup_n I_n, I_n\, \text {is an open interval}\,\}$$. This comes from the construction of Lebesgue measure. Once you have this regularity is easy. Let $$\epsilon >0$$ and choose open intervals $$I_n$$ such that $$A \subset \cup_n I_n$$ and $$\sum m(I_n) . Let $$U=\cup_n I_n$$. Then $$U$$ is open, $$A \subset U$$ and $$m(A) \leq m(U) \leq m(A)+\epsilon$$ from which regularity follows easily.