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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.

The completion of a measure extends a measure space $(X,\mathcal{F},\mu )$ so that every subset of a measurable null set is measurable. Its completed sigma-algebra isand the extension is $\overline{\mu }(A\cup N)=\mu (A)$. This is well-defined: two such representations differ only inside a measurable null set, so their measurable parts have the same measure. The extension agrees with $\mu$ on $\mathcal{F}$ and is a complete measure.^[ref]

For example, completing the Borel measure given by length on the real line gives Lebesgue measure on the Lebesgue measurable sets, including every subset of every Borel null set. Completion of a measure changes which sets are measurable; it does not identify distinct sets. If $A$ and $B$ differ only on a null set, they can still be different sets in the completed sigma-algebra. Equality modulo null sets is an additional quotient construction.

A measurable set $N$ is a null set for a measure $\mu$ when $\mu (N)=0$. A property holds almost everywhere when its exceptions lie in a null set. With a complete measure, every subset of a null set is measurable and is itself a null set.

A set $H$ is Lebesgue measurable when the Caratheodory measurability criterion holds:Subadditivity always gives the inequality with left side at most right side. For $H=\{x_i>t\}$, split each covering box $Q_j$ across the plane $x_i=t$. Its two pieces have total volume $|Q_j|$. The piece with a closed cut face can be enlarged by arbitrarily little into an open box; hence its outer measure is at most its corresponding volume, just as for the other piece. Subadditivity applied separately on the two sides consequently givesFor finite outer measure take a covering sum arbitrarily close to ${\mathcal{L}}^{\ast }(E)$; infinite outer measure follows already from subadditivity. Thus every coordinate half space is measurable.

Carathéodory's theorem says that the measurable sets form a sigma-algebra and the outer measure restricted to it is a complete measure. Coordinate half spaces generate all open boxes and then all open sets, using countably many rational-endpoint boxes. They therefore generate the Borel sigma-algebra. Hence every Borel set is Lebesgue measurable.