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Measure Theory · Tutorial 837 of 1000

Caratheodory Measurability

Learn how the Carathéodory splitting condition identifies measurable sets and turns outer measure into a countably additive measure.

Advanced 13 min read

What You'll Learn

  • State the Carathéodory criterion for an outer measure
  • Explain why the criterion requires equality for every test set
  • Prove that Carathéodory measurable sets form a sigma-algebra
  • Show that outer measure is countably additive on measurable sets
  • Verify that open intervals and outer-measure-zero sets are measurable
  • Apply the splitting test to concrete subsets of the real line

From Outer Measure to Measurability

Outer measure assigns a size to every subset of the real line, but countable subadditivity alone does not guarantee additivity on disjoint sets. Carathéodory measurability identifies the sets that split every test set without losing or creating outer measure. This condition will turn the outer measure constructed in the previous tutorial into a measure on a sigma-algebra.

The test is deliberately required to hold for every set \(A\), not just for the set being tested. A set can have finite outer measure, or even a familiar geometric shape, without those facts alone establishing the required splitting property. We first give the general definition for an outer measure and then work with Lebesgue outer measure \(m^*\) on \(\mathbb{R}\).

Definition: Let \(\mu^*\) be an outer measure on a set \(X\). A set \(E\subseteq X\) is Carathéodory measurable if, for every \(A\subseteq X\),
$$ \mu^*(A)=\mu^*(A\cap E)+\mu^*(A\setminus E). $$
For Lebesgue outer measure on \(\mathbb{R}\), the collection of all such sets is called the Lebesgue measurable sets.

The two sets \(A\cap E\) and \(A\setminus E\) are disjoint and have union \(A\). Countable subadditivity already gives

$$ \mu^*(A)\leq \mu^*(A\cap E)+\mu^*(A\setminus E). $$

Thus the substantive requirement is the reverse inequality: splitting \(A\) along \(E\) must not make the total outer measure larger than \(\mu^*(A)\). Together, the two inequalities give the equality in the definition. The equality is interpreted in the extended nonnegative real numbers, so it also applies when one or more quantities are infinite.

The Measurable Sets Form a Sigma-Algebra

The splitting test is stable under the set operations needed for measure theory. The proof below uses only the outer-measure axioms: monotonicity and countable subadditivity, along with the definition of measurability. In particular, no additivity for arbitrary sets is assumed.

Theorem (Carathéodory Measurable Sets Form a Sigma-Algebra): If \(\mu^*\) is an outer measure on \(X\), then its Carathéodory measurable sets form a sigma-algebra on \(X\).

Proof. First, \(\varnothing\) is measurable: for every \(A\subseteq X\),

$$ \mu^*(A)=\mu^*(A\cap\varnothing)+\mu^*(A\setminus\varnothing) =0+\mu^*(A). $$

If \(E\) is measurable, then its complement is measurable as well. Indeed, \(A\cap E^c=A\setminus E\) and \(A\setminus E^c=A\cap E\), so the defining equality for \(E^c\) is the same equality with its two terms reversed. In particular, \(X\) is measurable because it is the complement of \(\varnothing\).

Next let \(E\) and \(F\) be measurable. For arbitrary \(A\subseteq X\), apply the criterion first to \(E\), then to \(F\) using the test set \(A\setminus E\). This gives

$$ \begin{aligned} \mu^*(A) &=\mu^*(A\cap E)+\mu^*(A\setminus E)\\ &=\mu^*(A\cap E)+\mu^*((A\setminus E)\cap F) +\mu^*(A\setminus(E\cup F)). \end{aligned} $$

The first two sets on the last line have union \(A\cap(E\cup F)\). By subadditivity, the outer measure of that union is at most the sum of their outer measures. Consequently,

$$ \mu^*(A)\geq \mu^*(A\cap(E\cup F))+\mu^*(A\setminus(E\cup F)). $$

Subadditivity applied to the two pieces of \(A\) cut out by \(E\cup F\) gives the reverse inequality. Hence \(E\cup F\) is measurable. Closure under complements and finite unions gives closure under finite intersections and relative differences as well.

Now suppose \(E_1,E_2,\ldots\) are pairwise disjoint measurable sets, and write \(U=\bigcup_{n=1}^{\infty}E_n\) and \(U_N=\bigcup_{n=1}^{N}E_n\). Repeatedly applying the criterion to the finite union \(U_N\) yields

$$ \mu^*(A)=\sum_{n=1}^{N}\mu^*(A\cap E_n)+\mu^*(A\setminus U_N). $$

Since \(A\setminus U\subseteq A\setminus U_N\), monotonicity gives \(\mu^*(A\setminus U)\leq\mu^*(A\setminus U_N)\). If \(\mu^*(A)=+\infty\), then the desired reverse inequality for \(U\) holds automatically. If \(\mu^*(A)<\infty\), the displayed finite-stage equalities imply

$$ \mu^*(A)\geq \sum_{n=1}^{N}\mu^*(A\cap E_n)+\mu^*(A\setminus U) \quad\text{for every }N. $$

Taking the limit of the increasing partial sums gives

$$ \mu^*(A)\geq \sum_{n=1}^{\infty}\mu^*(A\cap E_n)+\mu^*(A\setminus U). $$

By subadditivity, \(\mu^*(A\cap U)\leq\sum_{n=1}^{\infty}\mu^*(A\cap E_n)\). Therefore

$$ \mu^*(A)\geq\mu^*(A\cap U)+\mu^*(A\setminus U). $$

Subadditivity supplies the reverse inequality, so \(U\) is measurable. For an arbitrary sequence of measurable sets, replace it by the disjoint sequence \(F_1=E_1\) and \(F_n=E_n\setminus\bigcup_{k=1}^{n-1}E_k\) for \(n\geq2\). Each \(F_n\) is measurable by the finite closure already proved, and the union of the \(F_n\) is the union of the \(E_n\). Thus arbitrary countable unions are measurable. The measurable sets form a sigma-algebra. \(\square\)

Outer Measure Becomes a Measure

Once the domain is restricted to the Carathéodory measurable sets, the outer measure is countably additive. This is the key reason for using the splitting criterion: the test does not merely identify a useful collection of sets; it ensures that outer measure behaves as a measure on that collection.

Theorem (Countable Additivity on the Carathéodory Measurable Sets): Let \(\mathcal{M}\) be the sigma-algebra of Carathéodory measurable sets for an outer measure \(\mu^*\). If \(E_1,E_2,\ldots\in\mathcal{M}\) are pairwise disjoint, then
$$ \mu^*\left(\bigcup_{n=1}^{\infty}E_n\right) =\sum_{n=1}^{\infty}\mu^*(E_n). $$

Proof. Put \(U=\bigcup_{n=1}^{\infty}E_n\), which is measurable by the sigma-algebra theorem. Countable subadditivity gives \(\mu^*(U)\leq\sum_{n=1}^{\infty}\mu^*(E_n)\). For each \(N\), apply the finite splitting equality successively to the measurable sets \(E_1,\ldots,E_N\), with test set \(U\). Pairwise disjointness gives

$$ \mu^*(U)=\sum_{n=1}^{N}\mu^*(E_n)+\mu^*(U\setminus U_N) \geq\sum_{n=1}^{N}\mu^*(E_n). $$

This holds for every \(N\), so \(\mu^*(U)\) is at least the limit of the partial sums, namely \(\sum_{n=1}^{\infty}\mu^*(E_n)\). Combining the two inequalities proves countable additivity, including the case in which the sum is infinite. \(\square\)

Outer-Measure-Zero Sets Pass the Test

A useful first class of measurable sets consists of sets whose outer measure is zero. This fact does not require completeness or any prior assumption that the set is measurable. It follows directly from monotonicity and subadditivity.

Theorem (Every Outer-Measure-Zero Set Is Carathéodory Measurable): If \(N\subseteq X\) and \(\mu^*(N)=0\), then \(N\) is Carathéodory measurable.

Proof. Fix an arbitrary \(A\subseteq X\). By monotonicity, \(\mu^*(A\cap N)=0\), since \(A\cap N\subseteq N\). Also \(A\setminus N\subseteq A\), so \(\mu^*(A\setminus N)\leq\mu^*(A)\). On the other hand, subadditivity applied to \(A=(A\cap N)\cup(A\setminus N)\) gives

$$ \mu^*(A)\leq\mu^*(A\cap N)+\mu^*(A\setminus N) =\mu^*(A\setminus N). $$

Thus \(\mu^*(A)=\mu^*(A\cap N)+\mu^*(A\setminus N)\) for every \(A\), as required. \(\square\)

Worked Example: The Rational Numbers

The rational numbers \(\mathbb{Q}\) are countable, so the earlier theorem that every countable set has Lebesgue outer measure zero gives \(m^*(\mathbb{Q})=0\). The zero-outer-measure theorem therefore shows that \(\mathbb{Q}\) is Carathéodory measurable. For any \(A\subseteq\mathbb{R}\), the calculation behind the criterion is

$$ m^*(A\cap\mathbb{Q})=0, \qquad m^*(A\setminus\mathbb{Q})=m^*(A). $$

The second equality follows because subadditivity gives \(m^*(A)\leq m^*(A\setminus\mathbb{Q})\), while monotonicity gives the reverse inequality. In particular, the rational numbers have measure zero when outer measure is restricted to the measurable sets.

Intervals Are Measurable

The splitting condition is compatible with the geometry of intervals. We first use the interval calculation from the previous tutorial, together with monotonicity, to handle the pieces made by cutting a covering interval at fixed endpoints.

For any bounded interval \(J\) with endpoints \(a\leq b\), regardless of whether either endpoint is included, \(m^*(J)=b-a\). Indeed, \(J\subseteq[a,b]\), so monotonicity and the closed-interval theorem give \(m^*(J)\leq b-a\). If \(a<b\), then for every sufficiently small \(\delta>0\), the closed interval \([a+\delta,b-\delta]\) lies in \(J\), so

$$ m^*(J)\geq m^*([a+\delta,b-\delta]) =(b-a)-2\delta. $$

Letting \(\delta\) decrease to zero gives the reverse bound. If \(a=b\), the interval is empty or a singleton, and its outer measure is zero by monotonicity and the countable-set theorem. Thus endpoint inclusion does not affect the outer measure of a bounded interval.

Theorem (Open Intervals Are Carathéodory Measurable): Every open interval \(E=(a,b)\), with \(a<b\), is Carathéodory measurable for Lebesgue outer measure.

Proof. Fix \(A\subseteq\mathbb{R}\). If \(m^*(A)=+\infty\), subadditivity implies that \(m^*(A\cap E)+m^*(A\setminus E)=+\infty\), and the criterion holds. Suppose \(m^*(A)<\infty\), and let \(\varepsilon>0\). By the definition of outer measure, there is a finite or countable collection of open intervals \(I_j\) covering \(A\) such that

$$ \sum_j |I_j|<m^*(A)+\varepsilon. $$

For each \(j\), split \(I_j\) into its portions to the left of or at \(a\), between \(a\) and \(b\), and to the right of or at \(b\): \(L_j=I_j\cap(-\infty,a]\), \(M_j=I_j\cap(a,b)\), and \(R_j=I_j\cap[b,\infty)\). Empty portions are discarded. These are bounded intervals, possibly with endpoints included, and they are pairwise disjoint with union \(I_j\). Their lengths add to the length of \(I_j\), so the interval calculation just established gives

$$ m^*(L_j)+m^*(M_j)+m^*(R_j)=|I_j|. $$

The intervals \(M_j\) cover \(A\cap E\), while all the \(L_j\) and \(R_j\) together cover \(A\setminus E\). Countable subadditivity therefore yields

$$ \begin{aligned} m^*(A\cap E)+m^*(A\setminus E) &\leq \sum_j\bigl(m^*(M_j)+m^*(L_j)+m^*(R_j)\bigr)\\ &=\sum_j|I_j|\\ &<m^*(A)+\varepsilon. \end{aligned} $$

Since this holds for every \(\varepsilon>0\), the sum on the left is at most \(m^*(A)\). Subadditivity gives the reverse inequality, proving the criterion for \(E\). \(\square\)

Worked Example: Splitting a Closed Interval at Two Endpoints

Take \(A=[-4,4]\) and \(E=(-2,3)\). The pieces are exactly

$$ A\cap E=(-2,3), \qquad A\setminus E=[-4,-2]\cup[3,4]. $$

They are disjoint: the point \(-2\) belongs to the outside piece, not the open interval, and \(3\) also belongs to the outside piece. Their union is exactly \([-4,4]\). The interval theorem gives \(m^*(A)=8\) and \(m^*((-2,3))=5\). The two closed intervals in the outside piece are disjoint measurable sets, so countable additivity on measurable sets gives

$$ m^*(A\setminus E)=((-2)-(-4))+(4-3)=2+1=3. $$

Thus \(m^*(A\cap E)+m^*(A\setminus E)=5+3=8=m^*(A)\), verifying the splitting equality for this particular test set. The theorem proves that the equality holds for every test set \(A\), not only for this example.

Worked Example: A Measurable Set of Measure One

The open interval \((0,1)\) is measurable and has outer measure \(1\). The set \(\mathbb{Q}\cap(0,1)\) is countable and has outer measure zero, so it is measurable as well. Closure of the measurable sets under relative differences shows that the irrational numbers in \((0,1)\),

$$ (0,1)\setminus\mathbb{Q}, $$

are measurable. The two disjoint measurable sets \(\mathbb{Q}\cap(0,1)\) and \((0,1)\setminus\mathbb{Q}\) have union \((0,1)\). Countable additivity on measurable sets therefore gives

$$ 1=m^*((0,1)) =m^*(\mathbb{Q}\cap(0,1))+m^*((0,1)\setminus\mathbb{Q}) =0+m^*((0,1)\setminus\mathbb{Q}). $$

Hence the irrational numbers in \((0,1)\) have measure \(1\), even though the rational numbers in that interval are dense and have measure zero.

Why the Universal Test Matters

Measurability is not established by checking the splitting equality for just one convenient set. The definition quantifies over every \(A\subseteq X\), including sets that may themselves be nonmeasurable. For example, calculating the two pieces of a particular interval, as in the first worked example, illustrates the criterion but does not replace the proof for all possible test sets.

The resulting sigma-algebra is also not the power set automatically. The outer measure is defined on every subset, but Carathéodory measurability imposes an additional condition. What the criterion guarantees is that the sets passing the test form a sigma-algebra and that outer measure is countably additive on it. In the next tutorial, this framework is used to construct Lebesgue measure by restricting Lebesgue outer measure to the Carathéodory measurable sets.

Key takeaway: A set is Carathéodory measurable when it splits every test set additively in outer measure. The measurable sets form a sigma-algebra, outer measure is a measure on that sigma-algebra, and both intervals and all outer-measure-zero sets are measurable.

Check Your Understanding

Use the splitting criterion and the proofs in this tutorial to answer the following questions.

  1. Why does countable subadditivity always give one direction of the Carathéodory equality?
  2. How does applying the criterion successively to disjoint measurable sets help prove that their union is measurable?
  3. Why is every set of outer measure zero measurable, even if no measurability information is known in advance?
  4. In the split of \([-4,4]\) by \((-2,3)\), which piece contains the points \(-2\) and \(3\), and why?
  5. Why does verifying the splitting equality for one test set \(A\) not establish that \(E\) is measurable?

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