From Open Sets to Borel Sets
Open subsets of \(\mathbb{R}\) are Lebesgue measurable: each is a countable union of intervals, and intervals are measurable. But the Borel sets include much more than open sets. They are obtained by repeatedly taking complements and countable unions, so the key question is whether Lebesgue measurability survives those operations. The Caratheodory criterion answers that question and gives the desired inclusion.
We work on \(\mathbb{R}\) with Lebesgue outer measure \(m^*\), defined on every subset of \(\mathbb{R}\), and the Lebesgue measure space \((\mathbb{R},\mathcal{L},m)\). This measure space is sigma-finite and not finite; completeness is not needed in this argument. We use the basic properties of outer measure: \(m^*(\varnothing)=0\), monotonicity, and countable subadditivity. In particular, if \(C\subseteq D\), then \(m^*(C)\le m^*(D)\), and \(m^*(\bigcup_n C_n)\le\sum_n m^*(C_n)\).
The quantifier “for every \(A\subseteq\mathbb{R}\)” is important: the test set \(A\) need not itself be measurable. The criterion says that cutting any set \(A\) into its portion in \(E\) and its portion outside \(E\) preserves its outer measure exactly.
Why the Measurable Sets Form a Sigma-Algebra
Proof. The empty set is measurable because \(A\cap\varnothing=\varnothing\), \(A\cap\varnothing^c=A\), and \(m^*(\varnothing)=0\). If \(E\) is measurable, then the criterion for \(E^c\) is the same equality with its two terms reversed: $$ m^*(A)=m^*(A\cap E^c)+m^*(A\cap E). $$ Addition of nonnegative extended real numbers is commutative, so \(E^c\) is measurable.
We first prove closure under unions of two measurable sets \(E\) and \(F\). Fix any \(A\subseteq\mathbb{R}\). Apply the criterion for \(E\), then apply the criterion for \(F\) to \(A\cap E^c\). This gives $$ m^*(A)=m^*(A\cap E)+m^*(A\cap E^c\cap F)+m^*(A\cap E^c\cap F^c). $$ Set \(C=A\cap(E\cup F)\) and \(D=A\setminus(E\cup F)\). The first two sets on the right have union \(C\), so subadditivity gives $$ m^*(C)\le m^*(A\cap E)+m^*(A\cap E^c\cap F). $$ The third set is \(D\). Therefore \(m^*(C)+m^*(D)\le m^*(A)\). In the other direction, \(A=C\cup D\), so subadditivity gives \(m^*(A)\le m^*(C)+m^*(D)\). The two inequalities prove the Caratheodory criterion for \(E\cup F\). Thus finite unions are measurable; together with complements, this also gives finite intersections and set differences.
Now let \((E_n)_{n\ge1}\) be a sequence of measurable sets. Define $$ F_1=E_1,\qquad F_n=E_n\setminus\bigcup_{k=1}^{n-1}E_k\quad(n\ge2). $$ The finite-union and difference results show that each \(F_n\) is measurable. The sets \(F_n\) are pairwise disjoint and have the same union \(U=\bigcup_{n\ge1}E_n=\bigcup_{n\ge1}F_n\). Fix any \(A\subseteq\mathbb{R}\), and write \(a_n=m^*(A\cap F_n)\) and \(b=m^*(A\setminus U)\). Repeatedly applying the criterion to the disjoint sets \(F_1,\ldots,F_N\) gives $$ m^*(A)=\sum_{n=1}^N a_n+m^*\left(A\setminus\bigcup_{n=1}^N F_n\right) \ge \sum_{n=1}^N a_n+b. $$ Taking the supremum over \(N\) yields \(\sum_{n\ge1}a_n+b\le m^*(A)\). On the other hand, \(A\cap U=\bigcup_{n\ge1}(A\cap F_n)\), so subadditivity gives \(m^*(A\cap U)\le\sum_{n\ge1}a_n\). Since \(A=(A\cap U)\cup(A\setminus U)\), another application of subadditivity gives $$ m^*(A)\le m^*(A\cap U)+b\le\sum_{n\ge1}a_n+b\le m^*(A). $$ Consequently \(m^*(A)=m^*(A\cap U)+m^*(A\setminus U)\). This holds for every \(A\), so \(U\) is measurable. We have proved closure under countable unions and hence the proposition. \(\square\)
The Borel Inclusion
Proof. By the earlier results on measurability of intervals and open sets as countable unions of intervals, every open subset of \(\mathbb{R}\) belongs to \(\mathcal{L}\). The proposition shows that \(\mathcal{L}\) is a sigma-algebra. By definition, \(\mathcal{B}_{\mathbb{R}}=\sigma(\{\text{open subsets of }\mathbb{R}\})\) is the smallest sigma-algebra containing every open set. Since \(\mathcal{L}\) is one such sigma-algebra, it contains \(\mathcal{B}_{\mathbb{R}}\). \(\square\)
This argument uses a general minimality principle: if a sigma-algebra contains a family of sets, it contains the sigma-algebra generated by that family. Here the family is the open sets, and the containing sigma-algebra is \(\mathcal{L}\). The proof establishes an inclusion; it does not claim that every Lebesgue-measurable set is Borel.
Worked Example: A Union of Open Intervals
Let \(O=(-2,0)\cup(1,3)\). Verify that \(O\) is Borel and Lebesgue measurable.
Both \((-2,0)\) and \((1,3)\) are open intervals, so \(O\) is open. Every open set is one of the generating sets for \(\mathcal{B}_{\mathbb{R}}\), hence \(O\in\mathcal{B}_{\mathbb{R}}\). Each interval is Lebesgue measurable, and the finite-union part of the proposition shows directly that \(O\in\mathcal{L}\). For instance, \(0\notin O\), because both intervals exclude their endpoints; \(1.5\in O\), because \(1<1.5<3\). Check: the interval description and the membership tests agree, and the theorem gives the same measurability conclusion from \(O\in\mathcal{B}_{\mathbb{R}}\).
Worked Example: A Closed Interval by Complementation
Show that \([0,1]\) is Borel and Lebesgue measurable, paying attention to its endpoints.
The complement in \(\mathbb{R}\) is $$ [0,1]^c=(-\infty,0)\cup(1,\infty). $$ Both rays are open, so their union is open and therefore Borel and Lebesgue measurable. A sigma-algebra is closed under complements; thus \([0,1]\), the complement of this open set, is both Borel and Lebesgue measurable. The endpoint check is exact: \(0\) and \(1\) belong to \([0,1]\), while neither belongs to the displayed complement. Check: every real number is either less than \(0\), in the closed interval \([0,1]\), or greater than \(1\); these three disjoint cases account for all of \(\mathbb{R}\).
Worked Example: A Countable Intersection of Open Sets
For each integer \(n\ge1\), let \(G_n=(-1/n,1/n)\). Find \(G=\bigcap_{n=1}^{\infty}G_n\), then identify its measurability.
If \(x=0\), then \(-1/n<0<1/n\) for every \(n\), so \(0\in G\). If \(x\ne0\), choose an integer \(n>1/|x|\). Then \(1/n<|x|\), so \(x\notin(-1/n,1/n)\), and therefore \(x\notin G\). Hence \(G=\{0\}\). Each \(G_n\) is open and belongs to \(\mathcal{L}\). Since complements and countable unions preserve Lebesgue measurability, countable intersections do as well: by De Morgan’s law, $$ \bigcap_{n=1}^{\infty}G_n=\left(\bigcup_{n=1}^{\infty}G_n^c\right)^c. $$ Thus \(G\) is Lebesgue measurable, and it is Borel by the same sigma-algebra argument applied to \(\mathcal{B}_{\mathbb{R}}\). Check: the membership argument excludes every nonzero real number and includes \(0\), exactly as required for \(G=\{0\}\).
A Common Mistake: Borel Does Not Mean Open
A tempting but incorrect shortcut is to assume every Borel set must itself be a countable union of open intervals. That describes open subsets of \(\mathbb{R}\), not all sets obtained by the sigma-algebra operations. For a concrete failure, the equality $$ [0,1]=\bigcup\{I:I\text{ is an open interval contained in }[0,1]\} $$ is false. Every open interval containing \(0\) also contains negative numbers, so no open interval contained in \([0,1]\) contains \(0\); the right-hand side therefore omits \(0\). The correction is to use the operations that define a sigma-algebra: \([0,1]\) is the complement of the open set \((-\infty,0)\cup(1,\infty)\). It is Borel and Lebesgue measurable even though it is not open.
The proof matters because it ensures that the sigma-algebra generated by open sets sits inside the collection of sets on which Lebesgue measure is defined. The inclusion follows from two ingredients, each with a distinct role: intervals make open sets measurable, and the Caratheodory criterion guarantees that measurability persists through the countable set operations used to generate Borel sets.
Check Your Understanding
Use the definitions and closure argument to answer the following questions.
- State the Caratheodory criterion for a set \(E\subseteq\mathbb{R}\) to be Lebesgue measurable.
- Why does measurability of \(E\) imply measurability of \(E^c\)?
- In the countable-union proof, why are the sets \(F_n\) pairwise disjoint, and why is their union the same as \(\bigcup_n E_n\)?
- Explain why closure under complements and countable unions also gives closure under countable intersections.
- Does \(\mathcal{B}_{\mathbb{R}}\subseteq\mathcal{L}\) assert that every Lebesgue-measurable set is Borel? State what the inclusion does assert.