What Is Lebesgue Outer Measure?

Lebesgue outer measure gives a notion of length to every subset of the real line, including sets whose measurability has not been established. It does so by covering a set with intervals and asking how small the total length of a cover can be.

This construction is the starting point for Lebesgue measure. Outer measure itself is not countably additive on all subsets; the measurable sets are the ones for which it behaves like a genuine measure.

Definition on the real line

For a set $E\subseteq\mathbb{R}$, its Lebesgue outer measure is

$$m^*(E)=\inf\left\{\sum_{k=1}^{\infty}|I_k|: E\subseteq\bigcup_{k=1}^{\infty} I_k,\ I_k\text{ are intervals}\right\}.$$

Here $|I_k|$ is the length of the interval $I_k$. The infimum is taken over all countable interval covers of $E$. The value may be infinite. One may define the covers using open intervals; allowing other interval types gives the same outer measure, since intervals can be enlarged by arbitrarily small amounts.

The infimum need not be attained by a particular cover. Instead, it is the greatest lower bound of the total lengths of all possible covers. For example, saying $m^*(E)=0$ means that for every $\varepsilon>0$, there is a countable interval cover of $E$ whose total length is less than $\varepsilon$.

In $\mathbb{R}^n$, the analogous construction uses countable covers by boxes and sums their volumes.

Basic properties

Lebesgue outer measure has three key properties:

• Empty set: $m^*(\varnothing)=0$.
• Monotonicity: If $E\subseteq F$, then $m^*(E)\leq m^*(F)$.
• Countable subadditivity: For any sequence of sets $E_1,E_2,\ldots$,

$$m^*\left(\bigcup_{j=1}^{\infty}E_j\right)\leq\sum_{j=1}^{\infty}m^*(E_j).$$

For monotonicity, any cover of $F$ also covers its subset $E$. For countable subadditivity, cover each $E_j$ with intervals whose total length is within a chosen small error of $m^*(E_j)$, then combine all those covers. Taking the errors to zero gives the inequality. If the sum on the right is infinite, the inequality holds automatically.

For an interval $I$, its outer measure is its usual length: $m^*(I)=|I|$. Thus the construction agrees with ordinary length where length is already familiar.

Worked example: the rationals in $[0,1]$

The set $\mathbb{Q}\cap[0,1]$ is countable, so list its elements as $q_1,q_2,\ldots$. Given any $\varepsilon>0$, cover $q_k$ by an open interval $I_k$ of length $\varepsilon/2^k$. These intervals cover every rational in $[0,1]$, and their total length is

$$\sum_{k=1}^{\infty}|I_k|=\sum_{k=1}^{\infty}\frac{\varepsilon}{2^k}=\varepsilon.$$

Since this works for every positive $\varepsilon$, no positive number can be a lower bound for the total lengths of all covers. Therefore

$$m^*(\mathbb{Q}\cap[0,1])=0.$$

A set of outer measure zero is called a null set. This example shows that a set can be dense in an interval yet have outer measure zero: the rationals are dense in $[0,1]$, but their outer measure is zero.

From outer measure to Lebesgue measure

Outer measure is defined for every subset, but it is not a countably additive measure on every subset. The Carathéodory criterion determines which sets behave properly. A set $E\subseteq\mathbb{R}$ is Lebesgue measurable if, for every set $A\subseteq\mathbb{R}$,

$$m^*(A)=m^*(A\cap E)+m^*(A\setminus E).$$

The criterion says that splitting any test set $A$ into the part inside $E$ and the part outside $E$ does not lose or create outer measure. The Lebesgue measurable sets form a sigma-algebra, and restricting $m^*$ to that sigma-algebra gives Lebesgue measure, which is countably additive.

Intervals, open sets, and closed sets are Lebesgue measurable. Every subset of a null set is measurable and has measure zero. These facts make outer measure useful both as a construction and as a way to prove that particular sets have measure zero.

Common mistakes

• Assuming every subset is measurable: Outer measure exists for every subset, but Lebesgue measure is obtained by restricting it to the measurable sets.
• Treating the infimum as a minimum: A cover with total length exactly equal to the outer measure may not exist. Covers arbitrarily close to that value are enough.
• Confusing subadditivity with additivity: Outer measure always satisfies countable subadditivity. Equality for unions of disjoint sets requires suitable measurability conditions; it is not guaranteed for arbitrary subsets.
• Assuming density implies positive length: Dense sets can have outer measure zero, as the rationals in $[0,1]$ demonstrate.