Half-Open Intervals and Endpoint Independence

Why One Endpoint Does Not Change the Length

An interval such as \([a,b)\) differs from the open interval \((a,b)\) by one point, while \((a,b]\) differs from it by the other endpoint. Since a single point has Lebesgue outer measure zero, we expect these intervals to have the same outer measure. To establish equality, however, we need both bounds: a cover gives an upper bound, and inclusion of the open interval gives a lower bound.

Throughout, all sets are subsets of \(\mathbb{R}\), and \(m^*\) denotes Lebesgue outer measure on \(\mathbb{R}\), not a measure restricted to a sigma-algebra. It is defined by taking the infimum of the total lengths over countable covers by open intervals. We consider finite real numbers \(a<b\); degenerate and unbounded intervals are not part of the claim here. We use the previously established facts that \(m^*\) is monotone and countably subadditive, that \(m^*(\{x\})=0\) for each \(x\in\mathbb{R}\), and that \(m^*((a,b))=b-a\) for \(a<b\). These are facts about outer measure, so no measurability assumption on the sets is needed.

Proof. First consider \([a,b)\). Since \((a,b)\subseteq[a,b)\), monotonicity and the exact outer measure of open intervals give $$ m^*((a,b))=b-a\le m^*([a,b)). $$ For the upper bound, fix any \(\varepsilon>0\). The open interval \((a-\varepsilon,b)\) contains every point of \([a,b)\), and its length is \(b-(a-\varepsilon)=b-a+\varepsilon\). It is therefore an admissible cover of \([a,b)\), so $$ m^*([a,b))\le b-a+\varepsilon $$ for every \(\varepsilon>0\). If \(m^*([a,b))>b-a\), choosing \(\varepsilon\) strictly between \(0\) and \(m^*([a,b))-(b-a)\) would contradict this inequality. Hence \(m^*([a,b))\le b-a\), and the lower and upper bounds give \(m^*([a,b))=b-a\).

Now consider \((a,b]\). We have \((a,b)\subseteq(a,b]\), so monotonicity again yields \(b-a\le m^*((a,b])\). For any \(\varepsilon>0\), the open interval \((a,b+\varepsilon)\) contains \((a,b]\) and has length \((b+\varepsilon)-a=b-a+\varepsilon\). Thus \(m^*((a,b])\le b-a+\varepsilon\) for every \(\varepsilon>0\). The same contradiction argument gives \(m^*((a,b])\le b-a\), hence \(m^*((a,b])=b-a\). The corresponding open and closed intervals already have outer measure \(b-a\) by the exact outer-measure calculations for those intervals. All four choices of endpoint inclusion therefore give the same value. \(\square\)

Where the Endpoint Enters the Argument

There are two complementary ways to see the role of an endpoint. The cover proof above makes the included endpoint fit inside an open interval whose length exceeds \(b-a\) by only \(\varepsilon\). Because \(\varepsilon\) can be any positive number, the endpoint creates no positive excess in the infimum. Alternatively, the set difference between \([a,b)\) and \((a,b)\) is exactly \(\{a\}\), and the set difference between \((a,b]\) and \((a,b)\) is exactly \(\{b\}\). Each difference is a singleton of outer measure zero.

Countable subadditivity makes this second observation precise as an upper bound. For instance, \([a,b)=(a,b)\cup\{a\}\), so $$ m^*([a,b))\le m^*((a,b))+m^*(\{a\})=(b-a)+0=b-a. $$ This alone gives only the upper bound. To conclude equality, we must also use \((a,b)\subseteq[a,b)\) and monotonicity, which give \(m^*([a,b))\ge b-a\). The two methods express the same endpoint independence, but the upper and lower bounds must not be confused.

A Common Error: Treating an Upper Bound as Equality

Consider the claim \(m^*([0,2))=2\). A tempting calculation is $$ m^*([0,2))\le m^*((0,2))+m^*(\{0\})=2+0=2, $$ followed by the conclusion that \(m^*([0,2))=2\). The displayed inequality is valid, but it proves only \(m^*([0,2))\le2\). An upper bound does not rule out a smaller value.

The correction is to add the lower bound: since \((0,2)\subseteq[0,2)\), monotonicity gives \(m^*([0,2))\ge m^*((0,2))=2\). Combining the inequalities gives equality. Equivalently, the proof by covers supplies an upper bound \(2+\varepsilon\) for every \(\varepsilon>0\), while the contained open interval supplies the lower bound \(2\).

Another error is to say that the half-open interval is itself covered by the open interval \((a,b)\). It is not: \((a,b)\) omits \(a\) and \(b\), whereas \([a,b)\) includes \(a\), and \((a,b]\) includes \(b\). The endpoint must actually be accounted for, either by extending the open covering interval by \(\varepsilon\) or by adding the singleton and then proving a separate lower bound.

The result is specifically an equality of outer measures. It does not require first establishing that the half-open intervals are measurable. The proof uses properties of outer measure on all subsets of \(\mathbb{R}\), together with the already established exact values for open intervals and singletons. Measurability of intervals is a separate question.