Exact Outer Measure of Open Intervals

From Covering Bounds to Exact Length

Lebesgue outer measure is defined by minimizing the total length of countable interval covers. For an open bounded interval, one cover is already available: the interval itself. This gives an upper bound immediately. To know that this bound is exact, we also need the lower bound established for every interval cover: covering an interval of length \(b-a\) cannot have total length less than \(b-a\). The two bounds together determine the outer measure without first needing to show that the interval is measurable.

The lower bound we use is the length lower bound for interval covers: if \(a<b\) and \((a,b)\subseteq\bigcup_{n=1}^{\infty}I_n\), where each \(I_n\) is a bounded open interval, then \(\sum_{n=1}^{\infty}|I_n|\ge b-a\). This result applies whether the cover has finitely or infinitely many intervals and whether or not those intervals overlap. In particular, it applies to every cover considered in the definition of \(m^*((a,b))\).

The Exact Outer Measure

Proof. First, the single interval \((a,b)\) covers itself and has length \(b-a\). Since outer measure is an infimum over cover costs, this gives \(m^*((a,b))\le b-a\). Conversely, take any countable cover \((a,b)\subseteq\bigcup_{n=1}^{\infty}I_n\) by bounded open intervals. By the length lower bound for interval covers, \(\sum_{n=1}^{\infty}|I_n|\ge b-a\). Thus every cost in the infimum defining \(m^*((a,b))\) is at least \(b-a\), so \(m^*((a,b))\ge b-a\). Combining the inequalities gives \(m^*((a,b))=b-a\). \(\square\)

This argument uses both sides of the infimum. A particular cover supplies an upper bound, while a lower bound for every admissible cover supplies a lower bound for the infimum. The direct cover does more than approximate the infimum: its cost equals the infimum, so here the smallest cover cost is attained.

Proof. The length lower bound gives \(\sum_{n=1}^{\infty}|I_n|\ge b-a\) for every such cover. By the theorem, \(b-a=m^*((a,b))\). The cover consisting of \(I_1=(a,b)\), with empty intervals added if needed, has total length \(|(a,b)|=b-a\). Hence this cover reaches the lower bound. \(\square\)

Endpoint Membership and a Common Pitfall

Open endpoints matter when checking whether an interval cover actually covers its target. For instance, the two intervals \((0,2)\) and \((2,4)\) do not cover \((0,4)\): neither interval contains \(2\), even though every other point of \((0,4)\) lies in one of them. Their lengths add to \(2+2=4\), but this does not make them a cover. A covering argument must verify set inclusion, not just compare lengths.

A corrected cover is \((0,\tfrac94)\) together with \((\tfrac74,4)\). If \(0<x\le\tfrac74\), then \(x<\tfrac94\), so \(x\) is in the first interval. If \(\tfrac74<x<4\), then \(x\) is in the second interval. These cases include every \(x\in(0,4)\); in particular, the intervals overlap and cover the previously missed point \(2\). Their lengths are \(\tfrac94\) and \(4-\tfrac74=\tfrac94\), giving total cost \(\tfrac92\), which is an upper bound, not the exact outer measure. The direct cover and the lower bound together still give \(m^*((0,4))=4\).

A related error is to treat the total length of a cover as though it were the length of the union. In the two-interval example above, the cost used in the outer-measure definition is \(|I_1|+|I_2|=\tfrac{13}{2}\); overlap does not authorize subtracting a length from that sum. The definition takes an infimum of sums of interval lengths. To prove an exact value, use an actual cover for the upper bound and a bound that applies to every cover for the lower bound.

The result identifies the precise outer-measure cost of covering an open bounded interval. This is useful whenever an argument estimates a set by interval covers: a cover of \((a,b)\) cannot have total length below \(b-a\), and the interval itself realizes that value. The conclusion here concerns outer measure \(m^*\); it does not require treating outer measure as a measure or using measurability.