Set Operations on Lebesgue-Measurable Sets

Why Set Operations Matter

Measurability must be preserved when sets are combined. For example, the set of points that belong to infinitely many sets in a sequence is described using countable unions and intersections. To know that such a set is Lebesgue measurable, it is not enough to check that each set in the sequence is measurable: we need closure results for the operations used to build the new set.

Throughout, the ambient measure space is \((\mathbb{R},\mathcal{L},m)\), where \(\mathcal{L}\) is the Lebesgue sigma-algebra and \(m\) is Lebesgue measure. This measure space is sigma-finite but not finite: \(\mathbb{R}=\bigcup_{k=1}^{\infty}[-k,k]\), and \(m([-k,k])=2k\), while \(m(\mathbb{R})=\infty\). It is complete, although completeness is not needed for the set-operation arguments below. Lebesgue outer measure is denoted by \(m^*\). Recall that outer measure is monotone and countably subadditive, and that a set \(E\subseteq\mathbb{R}\) is Lebesgue measurable precisely when, for every \(A\subseteq\mathbb{R}\),

m^*(A)=m^*(A\cap E)+m^*(A\cap E^c).

The equality is in the extended nonnegative reals, so \(+\infty\) is allowed. Subadditivity always gives the inequality \(m^*(A)\leq m^*(A\cap E)+m^*(A\cap E^c)\); measurability requires the reverse inequality as well. We use this criterion to prove the closure properties rather than assuming that set operations preserve measurability.

Complements, Differences, and Finite Operations

Proposition: In \((\mathbb{R},\mathcal{L},m)\), the complement of a Lebesgue-measurable set is measurable. The intersection and union of two Lebesgue-measurable sets are measurable. Consequently, the difference of two Lebesgue-measurable sets is measurable.

Proof. Suppose \(E\in\mathcal{L}\). The criterion for \(E\) states that for every \(A\subseteq\mathbb{R}\), \(m^*(A)=m^*(A\cap E)+m^*(A\cap E^c)\). The two terms on the right are exactly \(m^*(A\cap E^c)\) and \(m^*(A\cap (E^c)^c)\), in the opposite order. Thus \(E^c\) satisfies the criterion and is measurable.

Now let \(E,F\in\mathcal{L}\), and fix any \(A\subseteq\mathbb{R}\). Apply the criterion first to \(E\) and then to \(F\), using \(A\cap E\) as the test set in the second application:

m^*(A) =m^*(A\cap E)+m^*(A\cap E^c) =m^*(A\cap E\cap F)+m^*(A\cap E\cap F^c)+m^*(A\cap E^c).

The last two sets are disjoint and their union is \(A\cap(E\cap F)^c\). By subadditivity, their outer measures sum to at least the outer measure of that union. Therefore \(m^*(A)\geq m^*(A\cap E\cap F)+m^*(A\cap(E\cap F)^c)\). The reverse inequality follows from subadditivity applied to the partition of \(A\) into its parts inside and outside \(E\cap F\). Hence \(E\cap F\) satisfies the criterion. Since \(E\cup F=(E^c\cap F^c)^c\), closure under complements and intersections gives closure under unions. Finally, \(E\setminus F=E\cap F^c\), so differences are measurable as well. \(\square\)

These operations give an algebra of measurable sets: finite unions, finite intersections, and complements remain measurable. A sigma-algebra also requires closure under countable unions. That stronger property is essential for limit sets.

Countable Unions and Intersections

Theorem: If \(E_1,E_2,\ldots\) are Lebesgue-measurable subsets of \(\mathbb{R}\), then \(\bigcup_{n=1}^{\infty}E_n\) and \(\bigcap_{n=1}^{\infty}E_n\) are Lebesgue measurable in \((\mathbb{R},\mathcal{L},m)\).

Proof. First suppose \(D_1,D_2,\ldots\) are pairwise disjoint measurable sets, and write \(D=\bigcup_{n=1}^{\infty}D_n\). Fix an arbitrary set \(A\subseteq\mathbb{R}\). Repeatedly applying the measurability criterion to \(D_1,\ldots,D_N\) gives, for every positive integer \(N\),

m^*(A)=\sum_{n=1}^{N}m^*(A\cap D_n) +m^*\left(A\setminus\bigcup_{n=1}^{N}D_n\right).

Because \(A\setminus D\subseteq A\setminus\bigcup_{n=1}^{N}D_n\), monotonicity yields \(m^*(A)\geq\sum_{n=1}^{N}m^*(A\cap D_n)+m^*(A\setminus D)\). Letting \(N\) increase gives \(m^*(A)\geq\sum_{n=1}^{\infty}m^*(A\cap D_n)+m^*(A\setminus D)\); the series here is the increasing limit of its finite partial sums and may be infinite. The sets \(A\cap D_n\) are disjoint and their union is \(A\cap D\). Countable subadditivity therefore gives \(m^*(A\cap D)\leq\sum_{n=1}^{\infty}m^*(A\cap D_n)\). Combining the inequalities shows \(m^*(A)\geq m^*(A\cap D)+m^*(A\setminus D)\). The reverse inequality follows from subadditivity, so \(D\) satisfies the criterion and is measurable.

For a general sequence of measurable sets, define \(D_1=E_1\) and \(D_n=E_n\setminus\bigcup_{k=1}^{n-1}E_k\) for \(n\geq2\). The finite-operation result shows that each \(D_n\) is measurable. The sets \(D_n\) are pairwise disjoint: if \(j<n\), then \(D_n\) excludes \(E_j\), whereas \(D_j\subseteq E_j\). Also \(\bigcup_nD_n=\bigcup_nE_n\), since any point in the latter union belongs to the first \(E_n\) in the sequence that contains it, and hence belongs to the corresponding \(D_n\). The disjoint case proves that \(\bigcup_nE_n\) is measurable. Finally, by De Morgan’s identity, \(\bigcap_nE_n=(\bigcup_nE_n^c)^c\); complements and countable unions preserve measurability, so the intersection is measurable. \(\square\)

In particular, the Lebesgue-measurable sets form a sigma-algebra. Restricting \(m^*\) to \(\mathcal{L}\) gives Lebesgue measure \(m\); for pairwise disjoint measurable \(D_n\), countable additivity follows by applying the same finite splitting argument to \(A=\bigcup_nD_n\), and using subadditivity for the reverse inequality. This distinguishes the outer measure used in the criterion from the measure defined on measurable sets.

Worked Example: A Difference of Overlapping Intervals

Let \(A=[0,3]\) and \(B=(1,2]\). Both are Borel sets and hence Lebesgue measurable. Their difference removes the points of \(B\) from \(A\): the point \(1\) remains because \(1\notin B\), while \(2\) is removed because \(2\in B\). Thus \(A\setminus B=[0,1]\cup(2,3]\). The two intervals in this union are disjoint, with lengths \(1-0=1\) and \(3-2=1\), respectively, so \(m(A\setminus B)=1+1=2\).

Check: \(A\cap B=(1,2]\) has measure \(2-1=1\). The disjoint sets \(A\cap B\) and \(A\setminus B\) partition \(A\), and their measures sum to \(1+2=3=m([0,3])\), as required.

Worked Example: Countable Union and Intersection of Nested Intervals

For \(n\geq1\), let \(E_n=(-1/n,1+1/n)\). These are measurable intervals and are nested decreasing: if \(n+1>n\), then \(-1/(n+1)>-1/n\) and \(1+1/(n+1)<1+1/n\), so \(E_{n+1}\subseteq E_n\). Because every \(E_n\subseteq E_1=(-1,2)\) and \(E_1\) is itself in the sequence, \(\bigcup_nE_n=(-1,2)\), whose measure is \(2-(-1)=3\).

To find the intersection, \(x\) belongs to every \(E_n\) exactly when \(x>-1/n\) and \(x<1+1/n\) for every \(n\). These inequalities imply \(x\geq0\) and \(x\leq1\): if \(x<0\), choose \(n\) large enough that \(1/n<-x\), contradicting \(x>-1/n\); if \(x>1\), choose \(n\) large enough that \(1/n<x-1\), contradicting \(x<1+1/n\). Conversely, every \(x\in[0,1]\) satisfies both strict inequalities for every \(n\). Thus \(\bigcap_nE_n=[0,1]\), of measure \(1-0=1\).

Check: the union has the endpoints \(-1\) and \(2\) excluded because it equals \(E_1\). The intersection includes \(0\) and \(1\) because \(0>-1/n\) and \(1<1+1/n\) for every positive \(n\).

Limsup and Liminf of Measurable Sets

Definition: For a sequence \((E_n)\) of subsets of \(\mathbb{R}\), define \(\limsup_n E_n=\bigcap_{N=1}^{\infty}\bigcup_{n\geq N}E_n\) and \(\liminf_n E_n=\bigcup_{N=1}^{\infty}\bigcap_{n\geq N}E_n\). A point belongs to \(\limsup_n E_n\) exactly when it belongs to infinitely many \(E_n\); it belongs to \(\liminf_n E_n\) exactly when it belongs to every \(E_n\) from some index onward.

If every \(E_n\in\mathcal{L}\), then each tail union \(\bigcup_{n\geq N}E_n\) is measurable by countable-union closure, and the intersection of those tail unions is measurable by countable-intersection closure. Thus \(\limsup_nE_n\in\mathcal{L}\). Similarly, each tail intersection and then their union are measurable, so \(\liminf_nE_n\in\mathcal{L}\). These conclusions do not require the sets to be nested or their measures to be finite.

Worked Example: Alternating Intervals

Let \(E_n=(0,1)\) for even \(n\), and \(E_n=(1,2)\) for odd \(n\). Both intervals are Lebesgue measurable. Every tail contains an even index and an odd index, so its union is \((0,1)\cup(1,2)\). Intersecting these identical tail unions gives \(\limsup_nE_n=(0,1)\cup(1,2)\). Neither interval occurs at every index after some \(N\), so every tail intersection is empty and \(\liminf_nE_n=\varnothing\).

Check: each \(x\in(0,1)\) belongs to every even-indexed set and hence infinitely many sets; each \(x\in(1,2)\) belongs to every odd-indexed set and hence infinitely many. The point \(1\) belongs to none of the sets, and no point belongs to all sets in any tail. These membership checks give the same limsup and liminf.

Worked Example: Shrinking Closed Intervals

Let \(E_n=[0,1/n]\) for \(n\geq1\). These sets decrease with \(n\). A point belongs to \(\limsup_nE_n\) if it belongs to infinitely many \(E_n\). The point \(0\) belongs to every \(E_n\). If \(x>0\), choose an integer \(N>1/x\); then for every \(n\geq N\), \(1/n\leq1/N<x\), so \(x\notin E_n\). Negative \(x\) belongs to no \(E_n\). Therefore \(\limsup_nE_n=\{0\}\).

For the liminf, a point must belong to every set in some tail. The point \(0\) does, while every \(x>0\) is excluded from all \(E_n\) with \(n\geq N\), using the same choice of \(N\). Negative points are again excluded. Hence \(\liminf_nE_n=\{0\}\). The set \(\{0\}\) is measurable by the closure results (it is also the intersection of the measurable sets \(E_n\)).

Check: the direct intersection \(\bigcap_nE_n\) is \(\{0\}\), and because the sequence is decreasing, every tail intersection is also \(\{0\}\). Every tail union is \(E_N\), which contains \(0\) and excludes each fixed \(x>0\) once \(N>1/x\). Thus the definitions give the same answers.

A Common Mistake and the Main Lesson

A frequent error is to swap the formulas for limsup and liminf. For the alternating-interval example, the incorrect formula \(\bigcup_N\bigcap_{n\geq N}E_n\) for limsup gives the empty set, because every tail contains both \((0,1)\) and \((1,2)\), whose intersection is empty. But points in either interval occur infinitely often, so the limsup cannot be empty. The correction is to first take the union within each tail, recording points that occur at least once in that tail, and then intersect the tails. For liminf, the order is reversed: tail intersections record points that occur at every index in the tail.

The closure properties proved here explain why these constructions remain within Lebesgue measure theory. Complements and finite operations follow directly from the Carathéodory criterion; countable unions follow by disjointification; countable intersections then follow from complements. The countability matters: these arguments establish closure under countably many operations, not arbitrary uncountable unions.

Check Your Understanding

Use the definitions and closure results above to answer each question.

If \(E,F\in\mathcal{L}\), write \(E\setminus F\) using intersection and complement, and explain why it is measurable.
For \(E_n=[-1/n,1/n]\), determine \(\limsup_nE_n\) and \(\liminf_nE_n\), justifying membership at the endpoints.
If \(E_n=(0,1)\) for odd \(n\) and \(E_n=(2,3)\) for even \(n\), compute both limit sets.
Explain why disjointification is useful in proving that a countable union of measurable sets is measurable.
State which operation comes first in the definitions of limsup and liminf, and interpret each definition in terms of membership in the sequence.