Why Open Sets Break into Interval Components
On the real line, an open set may have many separated pieces, but each piece has no gaps: between any two of its points, every intermediate point belongs to the same piece. The key fact is that these pieces are open intervals, and that there are at most countably many of them. Countability is essential here: it lets us use the closure of Lebesgue measurable sets under countable unions to conclude that every open set is measurable.
Throughout, sets lie in \(\mathbb{R}\), and measure statements refer to \((\mathbb{R},\mathcal{L},m)\), where \(\mathcal{L}\) is the Lebesgue sigma-algebra and \(m\) is Lebesgue measure. This measure space is complete and sigma-finite, but not finite. We write \(m^*\) for Lebesgue outer measure. An open interval may be bounded or unbounded, and the empty open set will be treated separately.
This definition captures the idea of staying inside \(U\) while moving along the line from one point to the other. The closed segment in the definition ensures that a missing point separates the points into different components, even if the gap consists of just one point.
The Countable Disjoint Decomposition
Proof. Assume first that \(U\) is nonempty. The relation \(\sim\) is reflexive because \(x\in U\) implies \([x,x]=\{x\}\subseteq U\). It is symmetric because the segment between \(x\) and \(y\) does not depend on their order. To check transitivity, suppose \(x\sim y\) and \(y\sim z\). The union of the segments between \(x,y\) and between \(y,z\) contains the segment between \(x,z\). Both segments lie in \(U\), so \([ \min(x,z),\max(x,z)]\subseteq U\), giving \(x\sim z\). Thus \(\sim\) is an equivalence relation on \(U\), and its classes partition \(U\) into disjoint nonempty sets.
Let \(C\) be one class. If \(x,z\in C\) and \(y\) lies between \(x\) and \(z\), then \([x,z]\subseteq U\), and the segment between \(x\) and \(y\) lies in \([x,z]\). Hence \(x\sim y\), so \(y\in C\). Thus \(C\) is convex. It is also open: for any \(x\in C\), openness of \(U\) gives an \(\varepsilon>0\) such that \((x-\varepsilon,x+\varepsilon)\subseteq U\). For any \(y\) in that neighborhood, the segment between \(x\) and \(y\) lies in the neighborhood, so \(x\sim y\) and \(y\in C\).
We verify that every nonempty open convex set \(C\subseteq\mathbb{R}\) is an open interval, allowing infinite endpoints. Set \(a=\inf C\) and \(b=\sup C\), with \(a=-\infty\) or \(b=+\infty\) permitted. If \(a<t<b\), the definitions of infimum and supremum provide \(u,v\in C\) with \(u<t<v\). Convexity gives \(t\in C\). Conversely, every \(t\in C\) satisfies \(a\le t\le b\); if either endpoint is finite, openness rules out its inclusion, since a neighborhood of that endpoint would contain points outside the corresponding infimum or supremum. Therefore \(C=(a,b)\), with the usual interpretation for infinite endpoints.
It remains to prove that there are at most countably many classes. Every nonempty open interval contains a rational number. Fix an enumeration of the rational numbers. Assign to each class the first rational in that enumeration that belongs to the class. Distinct classes are disjoint, so they receive distinct rationals. This gives an injection from the set of classes into a countable set. The classes are therefore at most countable. They cover \(U\) and are pairwise disjoint by the equivalence-class construction. Finally, the classes are uniquely determined by \(\sim\), so no different collection of maximal interval components can give another such decomposition. If \(U=\varnothing\), the union with no intervals is the required decomposition. \(\square\)
The intervals in the theorem are maximal inside \(U\): if an interval contained one of them and also a point of \(U\) outside it, the interval would contain every point between those points. That would place the outside point in the same equivalence class, a contradiction. Maximality explains why two intervals that merely touch at a missing endpoint cannot be combined into one component.
Why the Decomposition Gives Measurability
We use two facts from the earlier material on Caratheodory measurability. First, a set \(A\subseteq\mathbb{R}\) is Lebesgue measurable when it satisfies the Caratheodory criterion $$ m^*(E)=m^*(E\cap A)+m^*(E\setminus A)\qquad\text{for every }E\subseteq\mathbb{R}. $$ Second, the sets satisfying this criterion form a sigma-algebra, and intervals are Lebesgue measurable. These are standard consequences of Caratheodory’s criterion established in the earlier material.
Proof. If \(U=\varnothing\), it is measurable. Otherwise the theorem writes \(U\) as a finite or countable union of open intervals. Each interval is Lebesgue measurable, and the Caratheodory-measurable sets form a sigma-algebra, hence are closed under countable unions. Thus \(U\in\mathcal{L}\). \(\square\)
Because the component intervals are disjoint, countable additivity of \(m\) also gives an exact measure formula. If the components are \(I_j=(a_j,b_j)\), then $$ m(U)=\sum_{j\in J}m(I_j)=\sum_{j\in J}(b_j-a_j), $$ where an interval with an infinite endpoint has measure \(+\infty\). For an empty union the sum and measure are both zero. The sum of nonnegative terms may itself be \(+\infty\).
Worked Example: A Finite Collection of Components
Let \(U=(-2,-1)\cup(0,2)\cup(3,5)\). The intervals are pairwise disjoint: the first ends at \(-1<0\), and the second ends at \(2<3\). They are the components, since each is an interval and the gaps between them are not in \(U\). The rational numbers \(-\tfrac32\), \(1\), and \(4\) give distinct witnesses to the three components.
Their lengths are $$ (-1)-(-2)=1,\qquad 2-0=2,\qquad 5-3=2. $$ Since \(U\) is a finite union of disjoint measurable intervals, additivity gives \(m(U)=1+2+2=5\). Check: the three component lengths sum to \(1+2+2=5\), and the gaps \([-1,0]\) and \([2,3]\) do not contribute to the measure of \(U\), because they are not in \(U\).
Worked Example: Infinitely Many Components with Infinite Measure
Define \(U=\{x\in\mathbb{R}:\inf_{k\in\mathbb{Z}}|x-k|<\tfrac14\}\), the set of points at distance less than \(\tfrac14\) from an integer. The strict inequality means that membership holds precisely when there exists an integer \(k\) such that \(|x-k|<\tfrac14\). Therefore $$ U=\bigcup_{k\in\mathbb{Z}}(k-\tfrac14,k+\tfrac14). $$
Each interval is open and has length \((k+\tfrac14)-(k-\tfrac14)=\tfrac12\). They are disjoint: if \(k<\ell\) are integers, then \(\ell-k\ge1\), so the right endpoint \(k+\tfrac14\) is strictly less than the left endpoint \(\ell-\tfrac14\), since \((\ell-\tfrac14)-(k+\tfrac14)=\ell-k-\tfrac12\ge\tfrac12\). Thus these intervals are the components of \(U\), and countable additivity gives \(m(U)=\sum_{k\in\mathbb{Z}}\tfrac12=+\infty\).
Check: among the indices \(-N,-N+1,\ldots,N\), there are \(2N+1\) intervals, so their total measure is \((2N+1)\tfrac12=N+\tfrac12\). These finite partial unions are contained in \(U\), and \(N+\tfrac12\) grows without bound as \(N\) increases. This confirms that \(m(U)=+\infty\).
Worked Example: Infinitely Many Components with Finite Measure
For each integer \(n\ge1\), set \(I_n=(2^{-n},3\cdot2^{-(n+1)})\), and let \(U=\bigcup_{n=1}^{\infty}I_n\). These intervals are disjoint. In fact, the right endpoint of \(I_{n+1}\) is \(3\cdot2^{-(n+2)}\), while the left endpoint of \(I_n\) is \(2^{-n}\); their difference is $$ 2^{-n}-3\cdot2^{-(n+2)}=2^{-(n+2)}>0. $$ Thus \(I_{n+1}\) lies strictly to the left of \(I_n\), with a gap between them. The same ordering separates any two distinct intervals.
The length of \(I_n\) is $$ 3\cdot2^{-(n+1)}-2^{-n}=(\tfrac32-1)2^{-n}=2^{-(n+1)}. $$ The intervals are measurable and disjoint, so countable additivity gives $$ m(U)=\sum_{n=1}^{\infty}2^{-(n+1)}=\frac12. $$ Indeed, the first \(N\) lengths sum to \(\frac12(1-2^{-N})\), by the finite geometric-series formula, and this tends to \(\frac12\). Check: for \(N=1\), the formula gives \(\frac12(1-\frac12)=\frac14\), the length of \(I_1=(\frac12,\frac34)\). For \(N=2\), it gives \(\frac12(1-\frac14)=\frac38\), equal to \(\frac14+\frac18\). These partial sums increase to the stated measure.
A Common Pitfall: Unions Need Not Be Disjoint
Do not add interval lengths just because a set has been written as a union of intervals. Countable additivity applies to pairwise disjoint measurable sets, not to an arbitrary list of overlapping intervals. For example, the incorrect calculation $$ m\big((0,2)\cup(1,3)\big)=2+2=4 $$ counts the overlap \((1,2)\) twice. The union is exactly \((0,3)\), so its measure is \(3-0=3\). The correction is either to measure the union directly as \((0,3)\), or to rewrite it as the disjoint union \((0,2)\cup[2,3)\), whose lengths are \(2\) and \(1\).
There is a related set-theoretic caution: disjoint intervals that touch at an excluded endpoint do not form one interval. For example, \((0,1)\cup(1,2)\) is not \((0,2)\), because \(1\) belongs to \((0,2)\) but not to the union. The two open intervals remain separate components even though their lengths happen to sum to the length of \((0,2)\). Exact decomposition and exact measure are related, but they are not the same claim.
The component decomposition is useful because it separates the geometry of an open set from the measure calculation. First identify its maximal disjoint intervals; then use measurability of intervals and countable additivity. This method covers bounded or unbounded intervals and works whether there are finitely many components, infinitely many components of finite total length, or infinite total measure.
Check Your Understanding
Use the component definition and the countable decomposition theorem to justify each answer.
- Decompose \(U=(-\infty,0)\cup(0,\infty)\) into its interval components. Why is it not one interval?
- For \(U=\bigcup_{n\ge1}(n,n+\tfrac13)\), verify that the displayed intervals are disjoint and find \(m(U)\).
- Explain why assigning a rational to each component proves that the number of components is at most countable.
- Give an example of two overlapping intervals for which adding their lengths overcounts the measure of their union, and correct the calculation.
- What feature of Caratheodory-measurable sets allows the interval decomposition to prove that every open set is measurable?