Definitions

# 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.

## Definition

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.

## Properties

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 e, 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 e.
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:

## Uses

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

## See also

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