= Solution
First suppose $m(E)<\infty$. The interval-cover definition of <Lebesgue outer measure>, together with $m^*(E)=m(E)$, provides bounded open intervals $I_j$ covering $E$ such that $\sum_j|I_j|<m(E)+\varepsilon$. 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> give
$$
m(U\setminus E)=m(U)-m(E)\leq\sum_j|I_j|-m(E)<\varepsilon.
$$
This 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 $m(U_n\setminus E_n)<\varepsilon w_n$. Then $U=\bigcup_{n\in\mathbb Z}U_n$ is open, contains $E$, and
$$
\boxed{m(U\setminus E)\leq\sum_{n\in\mathbb Z}m(U_n\setminus E_n)<\varepsilon.}
$$
Thus the required form of <regularity of Lebesgue measure> does not require $E$ itself to have finite measure.
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