Measurability of Intervals

From Interval Covers to Measurability

The outer measure of an interval records the infimum of the total lengths of its admissible interval covers. Measurability asks for something stronger: every test set must split additively across the set being tested. For intervals, the key estimate is that splitting one covering interval into its parts inside and outside a given interval never increases its total length. Applying that estimate to a cover proves the Caratheodory criterion.

Throughout, \(m^*\) is Lebesgue outer measure on \(\mathbb{R}\), and \((\mathbb{R},\mathcal{L},m)\) is the Lebesgue measure space: \(\mathcal{L}\) consists of the Caratheodory-measurable sets, and \(m\) is the restriction of \(m^*\) to \(\mathcal{L}\). In the definition of \(m^*(E)\), a cover is a countable family of open intervals whose union contains \(E\), and its cost is the sum of their lengths. Thus

We use the interval-cover results established earlier in the course: for any interval \(K\), regardless of whether its finite endpoints are included, \(m^*(K)\) equals its length; a singleton and the empty interval have outer measure zero, and an unbounded interval has infinite outer measure. These results include the lower bounds obtained from arbitrary interval covers. Countable subadditivity of \(m^*\) is also used below.

Subadditivity always gives \(m^*(A)\le m^*(A\cap E)+m^*(A\setminus E)\). The substantial part of the criterion is proving the reverse inequality for every \(A\). We first establish the estimate that supplies it.

The Cover-Splitting Estimate

Proof. The intersection \(J\cap I\) is an interval, possibly empty or a singleton. The set \(J\setminus I\) is the union of at most two intervals, possibly with some endpoints included. To see the latter fact, if \(J\cap I\) is nonempty, every point of \(J\) to the left of \(J\cap I\) lies in at most one interval component of \(J\setminus I\), and every point to its right lies in at most one; if the intersection is empty, \(J\setminus I=J\). By the interval outer-measure formulas, the outer measure of each of these intervals is its length, with length zero for a singleton or the empty set. The lengths of the disjoint pieces cannot sum to more than \(v-u\): they occupy portions of \((u,v)\), and any endpoints they share or omit have length zero. Applying finite subadditivity to the at most two components of \(J\setminus I\) therefore gives \(m^*(J\setminus I)\) at most their total length. Adding the length of \(J\cap I\) proves the stated bound. \(\square\)

This is an inequality, not an assertion that the two outer measures always add to the length of \(J\). Subadditivity and the interval formulas are enough for the inequality; no additivity of outer measure on these pieces is being assumed.

Proof. Fix an arbitrary set \(A\subseteq\mathbb{R}\). The set \(A\) is the union of \(A\cap I\) and \(A\setminus I\), so countable subadditivity gives \(m^*(A)\le m^*(A\cap I)+m^*(A\setminus I)\). If \(m^*(A)=+\infty\), this inequality forces the sum on the right to be \(+\infty\), and hence gives equality. Suppose instead that \(m^*(A)<+\infty\). For every \(\varepsilon>0\), the definition of outer measure gives a countable open-interval cover \(A\subseteq\bigcup_k J_k\) such that \(\sum_k|J_k|<m^*(A)+\varepsilon\). Empty or unbounded covering intervals can be omitted: an empty interval covers no points, and an unbounded interval has infinite length and cannot occur in a cover of finite total cost.

Since \(A\cap I\subseteq\bigcup_k(J_k\cap I)\) and \(A\setminus I\subseteq\bigcup_k(J_k\setminus I)\), countable subadditivity and the lemma give

The two series contain nonnegative terms, so regrouping them by the same index does not change their sum. Because the inequality holds for every \(\varepsilon>0\), it follows that \(m^*(A\cap I)+m^*(A\setminus I)\le m^*(A)\). Indeed, if that sum were strictly larger than the finite number \(m^*(A)\), choosing \(\varepsilon\) smaller than their difference would contradict the displayed inequality. Combining the two inequalities proves the Caratheodory equality for this arbitrary \(A\). Thus \(I\in\mathcal{L}\), and \(m(I)=m^*(I)\) is its interval length by the earlier interval-cover formulas, including the value \(+\infty\) for unbounded intervals. \(\square\)

Computing Measures from Covers

Unbounded Intervals and a Covering Pitfall

The theorem includes unbounded intervals. For example, \([0,\infty)\) is Lebesgue measurable and has measure \(+\infty\), since its interval length is infinite. This does not require applying the finite-cover estimate in the theorem to an unbounded covering interval: the proof first treats sets \(A\) of finite outer measure, for which a cover of finite total cost can be chosen, and handles \(m^*(A)=+\infty\) directly by subadditivity.

A common mistake is to use a closed interval as though it were an admissible open cover. For instance, the calculation “\([-2,3]\) covers itself with length \(3-(-2)=5\), so \(m^*([-2,3])\le5\)” uses a cover that is not allowed in the definition given here: \([-2,3]\) is not open. The corrected upper bound uses \((-2-\delta,3+\delta)\), whose length is \(5+2\delta\), and then lets \(\delta\) decrease to zero. The lower bound of \(5\) comes from the interval-cover estimate, not from treating a closed interval as an admissible cover.

Endpoint inclusion changes which points belong to an interval, but it does not change its length or its Lebesgue measure. The criterion proof explains more than this numerical fact: it verifies that each interval type splits every test set additively in outer measure. That verification is what permits its outer measure to be regarded as Lebesgue measure.