= Solution
The exact criterion is
$$
\boxed{m(E)<\infty.}
$$
Necessity follows already from one approximation with error less than one: a <compact set> $K\subseteq\mathbb R$ is bounded, hence has finite <Lebesgue measure>, so $m(E)=m(K)+m(E\setminus K)<\infty$.
Conversely, suppose $m(E)<\infty$. By <continuity from above of a measure>, the decreasing sets $E\setminus[-N,N]$ have measures tending to zero. Choose $N\geq1$ so that $m(E\setminus[-N,N])<\varepsilon/2$. Put $B=[-N,N]\setminus E$. By part (b)(i), there is an <open set> $V\supseteq B$ with $m(V\setminus B)<\varepsilon/2$. Define
$$
K=[-N,N]\setminus V.
$$
It is closed and bounded, hence a <compact set> by the <Heine-Borel theorem>, and $K\subseteq E$. The discarded portion satisfies
$$
m(E\setminus K)\leq m(E\setminus[-N,N])+m(E\cap[-N,N]\cap V)<\varepsilon.
$$
The second term is at most $m(V\setminus B)$. This proves the claimed <inner regularity of Lebesgue measure> with an arbitrarily small absolute error.
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