What Is the Difference Between Lebesgue-Null and Jordan-Null Sets?

A null set is a set that can be covered by boxes whose total volume is as small as desired. The key difference between the Lebesgue and Jordan versions is whether the cover may use countably many boxes or must use finitely many.

That difference matters: every Jordan-null set is Lebesgue null, but the reverse implication can fail even for bounded sets. The rationals in the unit interval are the standard example.

The two definitions

For a set $E \subseteq \mathbb{R}^n$, being Lebesgue null means that for every $\varepsilon>0$, there is a countable collection of boxes covering $E$ whose total volume is less than $\varepsilon$:

$$E \subseteq \bigcup_{k=1}^{\infty} B_k, \qquad \sum_{k=1}^{\infty} \operatorname{vol}(B_k)<\varepsilon.$$

Boxes here may be intervals in one dimension, rectangles in two dimensions, and their higher-dimensional analogues. Equivalent formulations use open boxes.

Classical Jordan nullity (also called having Jordan content zero) uses the same idea but requires a finite cover. In the usual definition, $E$ must also be bounded: for every $\varepsilon>0$, finitely many boxes cover $E$ and the sum of their volumes is less than $\varepsilon$.

The definitions immediately give one implication. A finite cover is a countable cover, so

$$\text{Jordan null} \implies \text{Lebesgue null}.$$

But a countable cover need not have a finite subcover of similarly small total volume. That is why the converse is not automatic.

Why the rationals are the key example

Let $E=\mathbb{Q}\cap[0,1]$. The rationals are countable, so enumerate them as $q_1,q_2,\ldots$. Given $\varepsilon>0$, cover $q_k$ by an interval of length less than $\varepsilon/2^{k+1}$. The total length of these intervals is less than $\varepsilon$. Thus $E$ is Lebesgue null.

However, $E$ is not Jordan null. Any finite collection of closed intervals that covers every rational in $[0,1]$ must also cover the closure of those rationals, which is all of $[0,1]$. The sum of the intervals’ lengths must therefore be at least $1$. It cannot be made arbitrarily small.

This illustrates the role of the closure. Finite covers cannot avoid covering limit points of the set, because a finite union of closed boxes is closed.

A useful test for bounded sets

For a bounded set $E \subseteq \mathbb{R}^n$, the following criterion is often the quickest way to compare the notions:

$$E \text{ is Jordan null} \quad\Longleftrightarrow\quad \overline{E} \text{ is Lebesgue null}.$$

For the forward direction, suppose finite box covers of $E$ have arbitrarily small total volume. Take the closures of the boxes in such a cover. Their union is closed and contains $E$, so it also contains $\overline{E}$. These closures have the same volumes as the original boxes. Enlarge them slightly to open boxes, increasing the total volume by as little as desired. The resulting open boxes cover $\overline{E}$ with arbitrarily small total volume.

For the reverse direction, suppose $\overline{E}$ is Lebesgue null. Cover it by countably many open boxes with total volume less than $\varepsilon$. Since $E$ is bounded, its closure is compact. Compactness gives a finite subcover, and the total volume of that finite collection is no greater than the total for the original countable cover. Hence $E$ has finite covers of arbitrarily small total volume and is Jordan null.

The criterion explains the rational example: $\overline{\mathbb{Q}\cap[0,1]}=[0,1]$, which is not Lebesgue null.

Worked comparison: a compact null set

Consider the middle-thirds Cantor set $C\subseteq[0,1]$. At stage $m$ of its construction, it is contained in $2^m$ closed intervals, each of length $3^{-m}$. Their total length is

$$2^m 3^{-m}=\left(\frac{2}{3}\right)^m.$$

As $m$ increases, this quantity tends to zero. The cover at every stage is finite, so $C$ is Jordan null as well as Lebesgue null. This example contrasts with the rationals: the Cantor set is closed, and its closure has measure zero.

Common mistakes

• Assuming bounded Lebesgue-null sets are automatically Jordan null. The rationals in $[0,1]$ are bounded and Lebesgue null, but their closure is the whole interval.
• Ignoring the number of boxes. A countable cover with small total volume establishes Lebesgue nullity, not by itself Jordan nullity.
• Calling every null set Jordan null without checking boundedness. Classical Jordan content is defined for bounded sets. Some texts use local or extended variants, so check the convention if the set is unbounded.
• Confusing Jordan nullity with Jordan measurability. A bounded set is Jordan measurable when its boundary is Jordan null. A Jordan-null set is itself a set of Jordan content zero; these are related but different statements.

In short, Lebesgue nullity concerns the size of the set under countable covers. Jordan nullity adds a finite-cover requirement, which makes the closure decisive for bounded sets.