Null set

Null set

In mathematics, a null set is a set that is negligible in some sense. For different applications, the meaning of "negligible" varies. In measure theory, any set of measure 0 is called a null set (or simply a measure-zero set). More generally, whenever an ideal is taken as understood, then a null set is any element of that ideal.

In some elementary textbooks, null set is taken to mean empty set.

The remainder of this article discusses the measure-theoretic notion.



Let X be a measurable space, let μ be a measure on X, and let N be a measurable set in X. If μ is a positive measure, then N is null (or zero measure) if its measure μ(N) is zero. If μ is not a positive measure, then N is μ-null if N is |μ|-null, where |μ| is the total variation of μ; equivalently, if every measurable subset A of N satisfies μ(A) = 0. For positive measures, this is equivalent to the definition given above; but for signed measures, this is stronger than simply saying that μ(N) = 0.

A nonmeasurable set is considered null if it is a subset of a null measurable set. Some references require a null set to be measurable; however, subsets of null sets are still negligible for measure-theoretic purposes.

When talking about null sets in Euclidean n-space Rn, it is usually understood that the measure being used is Lebesgue measure.


The empty set is always a null set. More generally, any countable union of null sets is null. Any measurable subset of a null set is itself a null set. Together, these facts show that the m-null sets of X form a sigma-ideal on X. Similarly, the measurable m-null sets form a sigma-ideal of the sigma-algebra of measurable sets. Thus, null sets may be interpreted as negligible sets, defining a notion of almost everywhere.

Lebesgue measure

The Lebesgue measure is the standard way of assigning a length, area or volume to subsets of Euclidean space.

A subset N of R has null Lebesgue measure and is considered to be a null set in R if and only if:

Given any positive number ε, there is a sequence {In} of intervals such that N is contained in the union of the In and the total length of the In is less than ε.

This condition can be generalised to Rn, using n-cubes instead of intervals. In fact, the idea can be made to make sense on any topological manifold, even if there is no Lebesgue measure there.

For instance:

  • With respect to Rn, all 1-point sets are null, and therefore all countable sets are null. In particular, the set Q of rational numbers is a null set, despite being dense in R.
  • The Cantor set is an example of a null uncountable set in R.
  • All the subsets of Rn whose dimension is smaller than n have null Lebesgue measure in Rn. For instance straight lines or circles are null sets in R2.
  • Sard's lemma: the set of critical values of a smooth function has measure zero.


Null sets play a key role in the definition of the Lebesgue integral: if functions f and g are equal except on a null set, then f is integrable if and only if g is, and their integrals are equal.

A measure in which all subsets of null sets are measurable is complete. Any non-complete measure can be completed to form a complete measure by asserting that subsets of null sets have measure zero. Lebesgue measure is an example of a complete measure; in some constructions, it's defined as the completion of a non-complete Borel measure.

A subset of the Cantor set which is not Borel measurable

The Borel measure is not complete. One simple construction is to start with the standard Cantor set K, which is closed hence Borel measurable, and which has measure zero, and to find a subset F of K which is not Borel measurable. (Since the Lebesgue measure is complete, this F is of course Lebesgue measurable.)

First, we have to know that every set of positive measure contains a nonmeasurable subset. Let f be the Cantor function, a continuous function which is locally constant on Kc, and monotonically increasing on the unit interval, with f(0) = 0 and f(1) = 1. Obviously, f(Kc) is countable, since it contains one point per component of Kc. Hence f(Kc) has measure zero, so f(K) has measure one. We need a strictly monotonic function, so consider g(x) = f(x) + x. Since g(x) is strictly monotonic and continuous, it is a homeomorphism. Furthermore, g(K) has measure one. Let E \subset g(K) be non-measurable, and let F = g − 1(E). Because g is injective, we have that F \subset K, and so F is a null set. However, if it were Borel measurable, then g(F) would also be Borel measurable (here we use the fact that the preimage of a Borel set by a continuous function is measurable; g(F) = (g − 1) − 1(F) is the preimage of F through the continuous function h = g − 1.) Therefore, F is a null, but non-Borel measurable set.

See also

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