Degenerate and Unbounded Intervals

Degenerate and Unbounded Intervals

For a bounded interval with distinct endpoints, its outer measure is its length, regardless of whether either endpoint is included. Two boundary cases require different arguments. If the endpoints coincide, endpoint inclusion determines whether the interval is a singleton or empty. If an interval is unbounded, there is no finite endpoint difference to use; instead, it contains bounded intervals of arbitrarily large length. These arguments compute outer measure directly, without needing to treat an interval as a measurable set.

Throughout, \(m^*\) denotes Lebesgue outer measure on \(\mathbb{R}\). It is defined by taking the infimum of the sums of lengths of countable interval covers: \(m^*(E)=\inf\{\sum_{k=1}^{\infty}|I_k|:E\subseteq\bigcup_{k=1}^{\infty}I_k\}\), where the \(I_k\) are intervals allowed by the interval-cover convention and \(|I_k|\) is their length. Outer measure takes values in \([0,\infty]\); here we compute \(m^*\), not a measure on a sigma-algebra.

We use two earlier results from the material on Lebesgue outer measure. First, monotonicity says that \(A\subseteq B\) implies \(m^*(A)\le m^*(B)\). Second, if \(u<v\) are finite real numbers, every interval with endpoints \(u,v\), under any choice of endpoint inclusion, has outer measure \(v-u\). In particular, open intervals have their endpoint difference as outer measure. We will use these facts rather than re-establish the bounded-interval covering estimates.

Intervals with Coincident Endpoints

For a finite real number \(a\), the interval \([a,a]\) contains exactly one point, namely \(a\). The intervals \((a,a)\), \([a,a)\), and \((a,a]\) contain no points. Thus, a degenerate interval is either a singleton or the empty set, depending on its endpoint convention. Both have outer measure zero, but the covering argument for the singleton is worth making explicit.

Proof. Outer measure is nonnegative, so \(m^*(\{a\})\ge0\). For each \(\varepsilon>0\), the open interval \((a-\varepsilon/2,a+\varepsilon/2)\) contains \(a\) and has length \((a+\varepsilon/2)-(a-\varepsilon/2)=\varepsilon\). This is a one-interval cover, so \(m^*(\{a\})\le\varepsilon\). If \(m^*(\{a\})\) were a positive number \(c\), taking \(\varepsilon=c/2\) would give \(c\le c/2\), a contradiction. Therefore \(m^*(\{a\})=0\). The empty set has outer measure zero because the empty family covers it with total length zero. Each degenerate interval listed is either \(\{a\}\) or empty, which proves the consequence. \(\square\)

Why Every Nonempty Unbounded Interval Has Infinite Outer Measure

An interval is unbounded above if it contains real numbers as large as desired, and unbounded below if it contains real numbers as small as desired. For intervals, an infinite endpoint makes the direction of unboundedness explicit: for example, \((a,\infty)\) and \([a,\infty)\) are unbounded above. A key point is that endpoint inclusion does not affect the argument. When the finite endpoint is excluded, we can still use bounded open intervals strictly inside the ray.

Proof. Fix any real number \(L>0\). If \(J\) is unbounded above, choose \(c\in J\). Since \(J\) is an interval and is unbounded above, it contains a bounded open interval of length \(L\). More explicitly, if \(J\) has a finite lower endpoint \(a\), then \((a,a+L)\subseteq J\). If \(J\) has no finite lower endpoint, unboundedness above and the interval property also give a segment of length \(L\) inside \(J\): choose \(c\in J\), then choose \(d\in J\) with \(d>c+L\); every point between \(c\) and \(d\) lies in \(J\), so \((c,c+L)\subseteq J\). In either case, monotonicity and the outer measure of a bounded open interval give \(m^*(J)\ge L\).

If \(J\) is unbounded below, the corresponding argument uses a bounded open interval to its left. If \(J\) has a finite upper endpoint \(b\), then \((b-L,b)\subseteq J\). If it has no finite upper endpoint, choose \(c\in J\) and then \(d\in J\) with \(d<c-L\); the interval property gives \((c-L,c)\subseteq J\). Again monotonicity gives \(m^*(J)\ge L\).

Thus, in either direction, \(m^*(J)\ge L\) for every \(L>0\). No finite value can satisfy these inequalities: if \(m^*(J)=C<\infty\), taking \(L=C+1\) would give \(C\ge C+1\). Therefore \(m^*(J)=\infty\). \(\square\)

For a ray whose finite endpoint is \(a\), one can use an even more direct choice of subinterval. For every \(L>0\), the interval \((a,a+L)\) lies in either \((a,\infty)\) or \([a,\infty)\). For a leftward ray, \((b-L,b)\) lies in either \((-\infty,b)\) or \((-\infty,b]\). These inclusions make clear why including or excluding the finite endpoint does not change the answer.

Endpoint Choices and a Common Pitfall

For bounded intervals with distinct finite endpoints, endpoint inclusion does not change outer measure. The same is true for a nonempty unbounded interval: the bounded open intervals used in the proof lie strictly inside it, so adding or removing its finite endpoint cannot alter the infinite outer measure. At coincident endpoints, however, endpoint inclusion determines whether the set is empty or a singleton. Both possibilities have outer measure zero, but they are not the same set.

Common mistake: subtracting endpoints when one endpoint is infinite. Writing “the length of \([2,\infty)\) is \(\infty-2=\infty\)” may suggest the answer, but it does not prove the outer measure calculation from interval covers. The correction is to use monotonicity: for every \(L>0\), \((2,2+L)\subseteq[2,\infty)\) and \(m^*((2,2+L))=L\). Hence \(m^*([2,\infty))\ge L\) for every \(L>0\), which proves that its outer measure is infinite.

A second error is to identify \([a,a]\) with the empty interval. It is the singleton \(\{a\}\), while \((a,a)\) is empty. Their outer measures agree, but only because a singleton can be covered by intervals of arbitrarily small length—not because it has no points.

The two methods in this tutorial are useful beyond the particular examples. An arbitrarily short cover gives an upper bound that can force outer measure to be zero when paired with nonnegativity. Conversely, arbitrarily long bounded subsets give lower bounds that force outer measure to be infinite. In both arguments, the crucial step is to state what the cover or subset is and verify its length before taking a limit or drawing a conclusion.