 Solution

The exact criterion isNecessity 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 $N\ge 1$ so that $m(E\setminus [-N,N])<\epsilon /2$. Put $B=[-N,N]\setminus E$. By part (b)(i), there is an open set $V\supseteq B$ with $m(V\setminus B)<\epsilon /2$. DefineIt is closed and bounded, hence a compact set by the Heine-Borel theorem, and $K\subseteq E$. The discarded portion satisfiesThe second term is at most $m(V\setminus B)$. This proves the claimed inner regularity of Lebesgue measure with an arbitrarily small absolute error.