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 $m(U_n)<M+\epsilon$.
Letting $\epsilon>0$ be arbitrary, then there exists $N\in\mathbb{N}$ such that $
m(U_n)<M+\epsilon
$
whenever $n\geq N$. Since $A\subseteq U_n$ for all $n$, then we have
$$
m(A)\leq m(U_n)<M+\epsilon
$$
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!
Best Answer
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 $\sum m(I_n) <m(A)+\epsilon$. 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.