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The Lebesgue inner measure approximates a subset of the real line from within by compact sets, while Lebesgue outer measure approximates it from outside by countable interval covers. For a finite closed interval $I$ containing $E$,Indeed, every compact $K\subseteq E$ gives $m^{\ast }(I\setminus E)\le m(I)-m(K)$. Conversely, an open cover of $I\setminus E$ with total length arbitrarily close to its outer measure has compact complement in $I$ lying inside $E$, giving the reverse inequality in the limit.

For $m^{\ast }(E)<\infty$, equality of inner and outer measures holds precisely when $E$ is Lebesgue measurable. For an unbounded set, measurability is equivalently checked on all bounded intersections $E\cap [-n,n]$. Equality $m_{\ast }(E)=m^{\ast }(E)=\infty$ alone is insufficient: adjoining a bounded nonmeasurable subset to a disjoint half-line makes both quantities infinite without making the union measurable.

The Lebesgue outer measure isThe Lebesgue inner measure is $m_{\ast }(E)=\sup \{m(K):K\subseteq E,K\text{ compact}\}$, with $m_{\ast }(\varnothing )=0$. For a bounded set inside a finite interval $I$, this is equivalently $|I|-m^{\ast }(I\setminus E)$.

A set is Lebesgue measurable if it satisfies Carathéodory's criterion:For finite outer measure this is equivalent to equality of inner and outer measure. For a general unbounded set, an equivalent local criterion is $m_{\ast }(E\cap [-n,n])=m^{\ast }(E\cap [-n,n])$ for every positive integer $n$. Merely writing $m_{\ast }(E)=m^{\ast }(E)=\infty$ would not establish measurability.

The Lebesgue outer measure isThe Lebesgue inner measure is $m_{\ast }(E)=\sup \{m(K):K\subseteq E,K\text{ compact}\}$, with $m_{\ast }(\varnothing )=0$. For a bounded set inside a finite interval $I$, this is equivalently $|I|-m^{\ast }(I\setminus E)$.

A set is Lebesgue measurable if it satisfies Carathéodory's criterion:For finite outer measure this is equivalent to equality of inner and outer measure. For a general unbounded set, an equivalent local criterion is $m_{\ast }(E\cap [-n,n])=m^{\ast }(E\cap [-n,n])$ for every positive integer $n$. Merely writing $m_{\ast }(E)=m^{\ast }(E)=\infty$ would not establish measurability.