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Every subset $B$ of a Lebesgue null set is a Lebesgue measurable set and is itself a null set. In fact the same conclusion follows directly from $m^{\ast }(B)=0$, without assuming measurability in advance. For any $A\subseteq \mathbb{R}$, monotonicity of Lebesgue outer measure gives $m^{\ast }(A\cap B)=0$ and $m^{\ast }(A\setminus B)\le m^{\ast }(A)$. ThereforeThe reverse inequality follows from countable subadditivity of Lebesgue outer measure, applied to the union $A=(A\cap B)\cup (A\setminus B)$. This is Carathéodory's criterion, proving measurability. If $B$ is a subset of an outer-null $N$, its outer measure is zero by the same monotonicity. Thus the restriction of outer measure to measurable sets is a complete measure.

A subset $E\subseteq \mathbb{R}$ is a Lebesgue measurable set when it satisfies Carathéodory's criterion for Lebesgue outer measure: for every subset $A\subseteq \mathbb{R}$,These sets form the Lebesgue sigma-algebra, the completion of the Borel sigma-algebra with respect to Lebesgue measure. Equivalently, $E$ differs from a Borel set by a subset of a Borel null set.

The sigma-algebra of all Lebesgue measurable sets, equivalently the completion of a measure of the Borel sigma-algebra under Lebesgue measure. It contains every subset of every Lebesgue null set. Applying Carathéodory's criterion to Lebesgue outer measure gives another construction of this same sigma-algebra.

A subset $E\subseteq \mathbb{R}$ is a Lebesgue measurable set when it satisfies Carathéodory's criterion: for every $T\subseteq \mathbb{R}$,Suppose $m^{\ast }(A)=0$ and let $B\subseteq A$, allowing $B=A$. The monotonicity of Lebesgue outer measure follows directly from covers: every cover of a larger set covers a smaller set. Thus $m^{\ast }(B)=m^{\ast }(T\cap B)=0$, while $m^{\ast }(T\setminus B)\le m^{\ast }(T)$. ConsequentlyThe reverse inequality is countable subadditivity of Lebesgue outer measure applied to $T=(T\cap B)\cup (T\setminus B)$. Equality follows, even if $m^{\ast }(T)=\infty$. Therefore $A$ and every subset $B\subseteq A$ are Lebesgue measurable sets of measure zero. This is the completeness of Lebesgue measure, obtained without presupposing measurability of the outer-null set.

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.

Translating a Lebesgue measurable set by a real number preserves measurability and its Lebesgue measure. Translation invariance of Lebesgue outer measure and Carathéodory's criterion give the measurability assertion and equality. Applying the result to indicator functions, then simple functions, and finally increasing simple approximations gives ${\int }_{\mathbb{R}}u(x-t)dx={\int }_{\mathbb{R}}u(x)dx$ for every nonnegative Lebesgue measurable function. For a Lebesgue integrable function, apply the identity to its positive part and negative part.