 Solution

First suppose $m(E)<\infty$. The interval-cover definition of Lebesgue outer measure, together with $m^{\ast }(E)=m(E)$, provides bounded open intervals $I_j$ covering $E$ such that ${\sum }_j|I_j|<m(E)+\epsilon$. If the definition uses intervals that are not open, enlarge each by an amount whose total is smaller than the available error. For the open set $U={\bigcup }_j{I}_j$, countable subadditivity and additivity of Lebesgue measure giveThis subtraction is valid because both measures are finite.

For arbitrary $E$, including $m(E)=\infty$, set $E_n=E\cap [n,n+1)$ for $n\in \mathbb{Z}$, and choose positive weights $w_n=2^{-|n|}/3$, whose sum is one. Each $E_n$ has finite Lebesgue measure, so the preceding argument gives an open set $U_n\supseteq E_n$ with $m(U_n\setminus E_n)<\epsilon w_n$. Then $U={\bigcup }_{n\in \mathbb{Z}}{U}_n$ is open, contains $E$, andThus the required form of regularity of Lebesgue measure does not require $E$ itself to have finite measure.