Half-Open Interval Length as a Premeasure

From Interval Length to a Premeasure

Half-open intervals are particularly convenient building blocks: they can be joined at endpoints without overlapping, and finite unions of them form a ring of sets. To use their usual lengths as a starting point for measure construction, one must check more than the formula for a single interval. In particular, the value assigned to a finite union must be independent of its representation, and the resulting set function must be countably additive whenever the union still belongs to the ring.

Let $\mathcal{R}$ consist of the empty set and all finite unions of bounded half-open intervals $[a,b)\subset\mathbb{R}$, where $a<b$. The empty interval $[a,a)$, if it occurs in a representation, is simply the empty set. A ring of sets is a nonempty family closed under finite unions and set differences. The family $\mathcal{R}$ is a ring: intersections and differences of two half-open intervals are finite unions of half-open intervals or empty, and the same holds for finite unions of such intervals. We work on $\mathbb{R}$ with the Borel sigma-algebra $\mathcal{B}_{\mathbb{R}}$ when referring to measurability; every set in $\mathcal{R}$ is Borel.

Definition: For $E\in\mathcal{R}$, write $E$ as a finite disjoint union of nonempty half-open intervals $[a_j,b_j)$, and define $\lambda(E)=\sum_j(b_j-a_j)$. Set $\lambda(\varnothing)=0$. We will prove that this value is independent of the representation. A premeasure on a ring is a nonnegative set function, zero on the empty set, that is countably additive whenever a countable pairwise disjoint family in the ring has its union in the ring.

Canonical Decompositions and Finite Additivity

Any finite union of half-open intervals can be represented as a disjoint union of intervals by sorting all the endpoints and looking at the consecutive half-open cells between them. This also gives a canonical representation: merge any two intervals that meet at an endpoint. The resulting nonempty intervals are separated by actual gaps, so distinct components have a positive distance between their adjacent endpoints. We call these the separated components of the set.

Here is the endpoint-partition argument that makes the length formula unambiguous. Take any two finite interval representations of the same set and list all their endpoints in increasing order as $t_0<t_1<\cdots<t_r$. On each cell $[t_{i-1},t_i)$, membership in each represented set is constant: no interval endpoint lies strictly inside the cell. Thus the two representations include exactly the same cells. The length of the set is the sum of the widths $t_i-t_{i-1}$ of its included cells, regardless of which representation was used. Consequently $\lambda$ is well-defined.

Proposition: The function $\lambda$ on $\mathcal{R}$ is finitely additive and monotone. In particular, if $A,B\in\mathcal{R}$ are disjoint, then $\lambda(A\cup B)=\lambda(A)+\lambda(B)$; and if $A\subseteq B$, then $\lambda(A)\le\lambda(B)$.

Proof. Given finitely many sets in $\mathcal{R}$, sort all endpoints of their interval representations. On each resulting cell $[t_{i-1},t_i)$, membership in each set is constant. If $A$ and $B$ are disjoint, no cell belongs to both, so the cells belonging to $A\cup B$ are exactly the cells belonging to one of $A$ or $B$, counted once. Summing their widths proves finite additivity. If $A\subseteq B$, every cell belonging to $A$ also belongs to $B$; the widths are nonnegative, so summing gives $\lambda(A)\le\lambda(B)$. The empty set contributes no cells and has length zero. $\square$

Worked Example: A Finite Decomposition with Adjacent Pieces

Consider $E=[-2,1)\cup[1,4)$. The two intervals are disjoint: the shared endpoint $1$ belongs to the second interval but not the first. Their lengths are $1-(-2)=3$ and $4-1=3$, so finite additivity gives $\lambda(E)=3+3=6$. They are adjacent, so the canonical separated-component representation merges them: $E=[-2,4)$, whose length is $4-(-2)=6$. The two representations agree.

Check: the endpoint partition with endpoints $-2,1,4$ has cells $[-2,1)$ and $[1,4)$, both included; their widths sum to $3+3=6$.

A Covering Estimate for Countable Additivity

To pass from finite additivity to countable additivity, we need a way to compare a finite union of intervals with a countable interval cover. The estimate below uses only finite endpoint partitions and the compactness of closed bounded intervals. When the countable family is finite, the same proof applies with a finite sum.

Lemma: Suppose $K$ is a finite union of bounded closed intervals and $K\subseteq\bigcup_{i} [u_i,v_i)$, where the covering intervals are bounded and their total length $\sum_i(v_i-u_i)$ is finite. Then the sum of the lengths of the separated interval components of $K$ is at most $\sum_i(v_i-u_i)$.

Proof. If $K=\varnothing$, its length is zero and the claim follows from nonnegativity. Otherwise enumerate the covering intervals by positive integers (a finite family can be extended by empty intervals). Fix $\varepsilon>0$ and expand the $i$-th interval to the open interval $(u_i-\eta_i,v_i+\eta_i)$, choosing $\eta_i=\varepsilon 2^{-i-2}$. These open intervals still cover $K$, since each contains the original interval. The set $K$ is compact, so the Heine–Borel theorem gives a finite subcover. The total increase in widths, even when summed over the entire countable family, is $\sum_i2\eta_i=\varepsilon/2\le\varepsilon$.

For a finite family of intervals, the endpoint-partition argument shows that the length of their union is at most the sum of their lengths: on every cell, the indicator of the union is at most the sum of the indicators of the intervals, and multiplying by the cell width and summing proves the inequality. Also, because the finite open subcover contains $K$, its union has length at least the sum of the lengths of the separated components of $K$; this follows by partitioning at all endpoints and observing that every cell of $K$ is included in the union. Therefore the length of $K$ is at most the sum of the widths in the finite subcover. Those widths are bounded above by the total original covering length plus $\varepsilon$. Since $\varepsilon>0$ was arbitrary, the length of $K$ is at most $\sum_i(v_i-u_i)$. $\square$

Countable Additivity on the Ring

Theorem: The set function $\lambda$ is a premeasure on $\mathcal{R}$. That is, if $E_n\in\mathcal{R}$ are pairwise disjoint and $E=\bigcup_{n=1}^{\infty}E_n$ also belongs to $\mathcal{R}$, then $\lambda(E)=\sum_{n=1}^{\infty}\lambda(E_n)$.

Proof. If $E=\varnothing$, every $E_n$ is empty, so both sides are zero. Now suppose $E\ne\varnothing$. Finite additivity and monotonicity show that for each $N$, $\sum_{n=1}^{N}\lambda(E_n)=\lambda(\bigcup_{n=1}^{N}E_n)\le\lambda(E)$, since the $E_n$ are disjoint and their finite union is contained in $E$. Taking the increasing limit of the partial sums gives $\sum_{n=1}^{\infty}\lambda(E_n)\le\lambda(E)$.

For the reverse inequality, write $E$ in its canonical separated-component form as a finite union of intervals $[a_j,b_j)$, $1\le j\le r$. For each $n$, use its separated-component form as well. Across all $n$, these components form a countable family of bounded half-open intervals, pairwise disjoint because the $E_n$ are pairwise disjoint and the components within each $E_n$ are disjoint. The sum of their lengths is $T=\sum_n\lambda(E_n)$.

Choose numbers $\delta_j>0$ so small that $2\delta_j<b_j-a_j$, and form the compact set $K=\bigcup_{j=1}^{r}[a_j+\delta_j,b_j-\delta_j]$. It is contained in $E$, hence in the union of all the component intervals of the $E_n$. The covering lemma gives $\sum_{j=1}^{r}(b_j-a_j-2\delta_j)\le T$. For each $j$, let $\delta_j$ decrease to zero; because there are only finitely many $j$, the left side tends to $\sum_{j=1}^{r}(b_j-a_j)=\lambda(E)$. Thus $\lambda(E)\le T$, which combined with the first inequality proves the required equality. $\square$

Worked Example: A Countable Partition of a Half-Open Interval

Define $J_0=[0,\tfrac12)$ and, for $n\ge1$, define $J_n=[1-2^{-n},1-2^{-(n+1)})$. Each $J_n$ belongs to $\mathcal{R}$. For $n\ge1$, the right endpoint of $J_n$ is the left endpoint of $J_{n+1}$, and the intervals are half-open, so they are disjoint. Also $J_0$ ends at the left endpoint $1-2^{-1}=\tfrac12$ of $J_1$, so it is disjoint from $J_1$.

The intervals cover $[0,1)$: $J_0$ covers $[0,\tfrac12)$, and the intervals $J_n$, $n\ge1$, cover $[\tfrac12,1)$ with successive endpoints $1-2^{-n}$ increasing to $1$. Their lengths are $\lambda(J_0)=\tfrac12$ and, for $n\ge1$, $\lambda(J_n)=(1-2^{-(n+1)})-(1-2^{-n})=2^{-(n+1)}$. Therefore $$ \sum_{n=0}^{\infty}\lambda(J_n)=\frac12+\sum_{n=1}^{\infty}2^{-(n+1)} =\frac12+\frac{1/4}{1-1/2}=1 =\lambda([0,1)). $$ Check: the first three interval lengths are $\tfrac12,\tfrac14,\tfrac18$, and the finite partial sum through $J_N$ is $1-2^{-(N+1)}$, which increases to $1$.

Worked Example: Splitting a Bounded Interval into Equal Pieces

For a positive integer $q$, the intervals $I_k=[k/q,(k+1)/q)$, $k=0,\ldots,q-1$, are pairwise disjoint and their union is $[0,1)$. Each has length $(k+1)/q-k/q=1/q$. Thus the finite additivity result gives $\sum_{k=0}^{q-1}\lambda(I_k)=q(1/q)=1=\lambda([0,1))$. For $q=4$, the four lengths are each $1/4$, and their sum is $1/4+1/4+1/4+1/4=1$. The endpoint $1$ is in none of these intervals and is also not in $[0,1)$, so there is no endpoint discrepancy.

Check: the union is $[0,\tfrac14)\cup[\tfrac14,\tfrac12)\cup[\tfrac12,\tfrac34)\cup[\tfrac34,1)=[0,1)$, and both calculations give length $1$.

What the Premeasure Statement Does—and Does Not—Say

The countable additivity condition for a premeasure on a ring is conditional: the disjoint union must itself belong to the ring. The ring $\mathcal{R}$ contains only bounded finite unions, so it is not a sigma-algebra on $\mathbb{R}$. For example, the union of $[-n,n)$, $n\ge1$, is all of $\mathbb{R}$, which is not in $\mathcal{R}$. The theorem therefore does not assert that $\lambda$ is already a measure on a sigma-algebra. Constructing an extension to a sigma-algebra is a separate step.

A frequent error is to add the lengths of intervals that overlap, or to apply countable additivity to a family that is not disjoint. For example, $[0,\tfrac12)$ and $[0,\tfrac12)$ are the same set, not two disjoint pieces of $[0,\tfrac12)$. Adding their lengths gives $1$, but the length of their union is only $\tfrac12$. Another endpoint check matters in decompositions: $[0,1)$ and $[1,2)$ are disjoint and adjacent, while $[0,1]$ and $[1,2)$ overlap at $1$. The half-open convention resolves endpoint membership precisely; it does not permit counting an endpoint twice.

Check Your Understanding

In each question, pay attention to whether the sets are disjoint and whether their union belongs to $\mathcal{R}$.

Find the separated-component representation and length of $[0,2)\cup[3,5)\cup[5,7)$.
For $q=5$, write down the lengths of $[k/5,(k+1)/5)$, $k=0,\ldots,4$, and check their sum against $\lambda([0,1))$.
Why can the premeasure theorem not be applied to the family $[-n,n)$, $n\ge1$, as a pairwise disjoint family whose union is in $\mathcal{R}$?
Give an example of two half-open intervals that are adjacent and disjoint, and one pair that overlap at an endpoint.
Explain why a representation-independent length formula is needed before countable additivity can be stated for $\lambda$.