Outer Measure and the Splitting Condition
Lebesgue outer measure \(m^*\) assigns a size to every subset of \(\mathbb{R}\), whether or not that set is measurable. Its defining covering construction uses countable covers by open intervals: \(m^*(A)\) is the infimum of the total lengths of such covers of \(A\). In particular, \(m^*(A)\) may be infinite. Outer measure is monotone and countably subadditive, but it is not assumed to be additive on arbitrary sets.
The Carathéodory criterion identifies the sets across which outer measure does split additively. In this tutorial, all sets being tested are arbitrary subsets of \(\mathbb{R}\); measurability means Lebesgue measurability, not merely Borel measurability. The resulting Lebesgue measure \(m\) is the restriction of \(m^*\) to the sets satisfying the criterion.
Countable subadditivity always gives one direction: since \(A=(A\cap E)\cup(A\cap E^c)\), we have \(m^*(A)\le m^*(A\cap E)+m^*(A\cap E^c)\). Thus the criterion asks for the reverse inequality as well, for every test set \(A\). It is not enough to check the split for \(A=E\), or for a few convenient test sets.
The Carathéodory Theorem
Proof. We first prove closure under complements. The criterion for \(E^c\) is exactly the criterion for \(E\), with the two terms on the right interchanged. Thus \(E\in\mathcal{L}\) implies \(E^c\in\mathcal{L}\). Also, \(\varnothing\in\mathcal{L}\), since \(m^*(A\cap\varnothing)+m^*(A\cap\mathbb{R})=0+m^*(A)=m^*(A)\).
Next suppose \(E,F\in\mathcal{L}\). Fix any \(A\subseteq\mathbb{R}\). Apply the criterion for \(E\), and then the criterion for \(F\) to \(A\cap E\): $$ m^*(A)=m^*(A\cap E)+m^*(A\cap E^c) \ge m^*(A\cap E\cap F)+m^*((A\cap E)\cap F^c)+m^*(A\cap E^c). $$ The last two sets have union \(A\setminus(E\cap F)\). By subadditivity, their outer measures sum to at least \(m^*(A\setminus(E\cap F))\). Therefore \(m^*(A)\ge m^*(A\cap(E\cap F))+m^*(A\setminus(E\cap F))\). Subadditivity gives the opposite inequality. Since \(A\) was arbitrary, \(E\cap F\in\mathcal{L}\). Complement closure now also gives closure under finite unions and differences.
For countable unions, first let \(E_1,E_2,\ldots\) be pairwise disjoint members of \(\mathcal{L}\), and write \(U=\bigcup_{n=1}^{\infty}E_n\) and \(U_N=\bigcup_{n=1}^{N}E_n\). Finite unions are measurable by the finite-operations result. For every \(A\subseteq\mathbb{R}\), applying the criterion to \(U_N\) and repeatedly splitting the part inside \(U_N\) gives $$ m^*(A)=\sum_{n=1}^{N}m^*(A\cap E_n)+m^*(A\setminus U_N) \ge \sum_{n=1}^{N}m^*(A\cap E_n)+m^*(A\setminus U). $$ Taking the supremum over \(N\) yields \(m^*(A)\ge\sum_{n=1}^{\infty}m^*(A\cap E_n)+m^*(A\setminus U)\). Subadditivity gives \(m^*(A\cap U)\le\sum_n m^*(A\cap E_n)\), so \(m^*(A)\ge m^*(A\cap U)+m^*(A\setminus U)\). The reverse inequality follows from subadditivity. This proves \(U\in\mathcal{L}\).
For arbitrary \(E_n\in\mathcal{L}\), define \(F_1=E_1\) and \(F_n=E_n\setminus\bigcup_{k=1}^{n-1}E_k\) for \(n\ge2\). Finite operations show \(F_n\in\mathcal{L}\); the \(F_n\) are pairwise disjoint and \(\bigcup_nF_n=\bigcup_nE_n\). The disjoint-union result therefore proves closure under arbitrary countable unions. Hence \(\mathcal{L}\) is a sigma-algebra.
Finally, let \(E_n\in\mathcal{L}\) be pairwise disjoint and \(E=\bigcup_nE_n\). The disjoint-union inequality just proved, applied with \(A=E\), gives \(m^*(E)\ge\sum_n m^*(E_n)\), since \(E\setminus E=\varnothing\). Countable subadditivity gives \(m^*(E)\le\sum_n m^*(E_n)\). Thus \(m^*(E)=\sum_n m^*(E_n)\). This is countable additivity on \(\mathcal{L}\), so the restriction \(m\) is a measure. \(\square\)
Testing Sets with Interval Covers
For Lebesgue outer measure, the criterion can often be checked by splitting each interval in a cover of \(A\). The needed interval facts are that an interval has outer measure equal to its length, endpoint changes do not change that outer measure, and a singleton has outer measure zero. These are the interval outer-measure results established earlier in the course. For an interval \(I\) cut at finitely many points, the outer measures of its resulting interval pieces sum to the length of \(I\).
Worked Example: The Half-Line \((-\infty,0]\)
Let \(E=(-\infty,0]\). We verify the criterion for every \(A\subseteq\mathbb{R}\). If \(m^*(A)=\infty\), subadditivity gives \(\infty=m^*(A)\le m^*(A\cap E)+m^*(A\cap E^c)\); consequently the sum on the right is infinite and equality holds.
Now suppose \(m^*(A)<\infty\), and fix \(\varepsilon>0\). By the definition of outer measure, cover \(A\) by open intervals \(I_j\) whose total length is less than \(m^*(A)+\varepsilon\). The intervals \(I_j\cap E\) cover \(A\cap E\), and the sets \(I_j\cap E^c\) cover \(A\cap E^c\). Splitting an interval at \(0\) gives \(m^*(I_j\cap E)+m^*(I_j\cap E^c)=|I_j|\); whether the cut point belongs to one side or the other does not change the sum, because a singleton has outer measure zero. By countable subadditivity applied to the two covers,
This holds for every \(\varepsilon>0\), so the left side is at most \(m^*(A)\). Subadditivity gives the reverse inequality, proving equality. Thus \((-\infty,0]\) satisfies the criterion. Check: taking \(A=[-2,3]\), the two pieces have outer measures \(2\) and \(3\), and \(2+3=5=m^*([-2,3])\).
Worked Example: The Bounded Interval \([0,1]\)
Let \(E=[0,1]\). If \(m^*(A)=\infty\), subadditivity again forces the sum of the two pieces to be infinite. For finite \(m^*(A)\), choose an open-interval cover \((I_j)\) with total length less than \(m^*(A)+\varepsilon\). Each \(I_j\) is split by \(0\) and \(1\) into its part in \(E\) and at most two parts outside \(E\). The interval lengths of all these pieces sum to \(|I_j|\); endpoints have outer measure zero. Therefore $$ \begin{aligned} m^*(A\cap E)+m^*(A\cap E^c) &\le \sum_j\bigl(m^*(I_j\cap E)+m^*(I_j\cap E^c)\bigr)\\ &=\sum_j |I_j|<m^*(A)+\varepsilon. \end{aligned} $$ Letting \(\varepsilon\) decrease to zero gives the reverse of the subadditivity inequality, hence the criterion. Check: for \(A=[-1,2]\), the pieces are \([0,1]\) of outer measure \(1\) and \([-1,0)\cup(1,2]\) of outer measure \(2\); their sum \(1+2=3\) is \(m^*([-1,2])\).
Worked Example: The Rationals \(\mathbb{Q}\)
We verify the criterion for \(E=\mathbb{Q}\). First, enumerate \(\mathbb{Q}=\{q_1,q_2,\ldots\}\). Given \(\varepsilon>0\), cover \(q_n\) by an open interval of length \(\varepsilon/2^n\). The total length is \(\sum_{n=1}^{\infty}\varepsilon/2^n=\varepsilon\), so \(m^*(\mathbb{Q})=0\). By monotonicity, \(m^*(A\cap\mathbb{Q})=0\) for every \(A\subseteq\mathbb{R}\).
Since \(A\setminus\mathbb{Q}\subseteq A\), monotonicity gives \(m^*(A\setminus\mathbb{Q})\le m^*(A)\). In the other direction, \(A=(A\cap\mathbb{Q})\cup(A\setminus\mathbb{Q})\), so subadditivity gives \(m^*(A)\le 0+m^*(A\setminus\mathbb{Q})\). These inequalities imply \(m^*(A)=m^*(A\cap\mathbb{Q})+m^*(A\setminus\mathbb{Q})\), including when \(m^*(A)=\infty\). Thus \(\mathbb{Q}\) is Lebesgue measurable. Check: for \(A=[0,1]\), the rational part has outer measure \(0\), and the irrational part has outer measure \(1\), so the sum is \(1=m^*([0,1])\).
A Common Pitfall: Subadditivity Is Only One Direction
A frequent error is to write \(A=(A\cap E)\cup(A\cap E^c)\), invoke subadditivity, and conclude that the criterion holds. That computation proves only \(m^*(A)\le m^*(A\cap E)+m^*(A\cap E^c)\). To establish measurability, one must also prove the reverse inequality for every \(A\). For the interval examples, the reverse inequality came from splitting each covering interval and taking covers with total length arbitrarily close to \(m^*(A)\); for \(\mathbb{Q}\), it followed from the zero outer measure of \(A\cap\mathbb{Q}\) and monotonicity.
Do not assume the criterion merely because \(E\) has a familiar description, or because a single test set gives equality. The definition quantifies over all \(A\subseteq\mathbb{R}\). Once the criterion is proved, the Carathéodory theorem supplies the structural consequences: measurable sets form a sigma-algebra, and outer measure restricted to that sigma-algebra is countably additive.
Check Your Understanding
For each question, identify which direction follows from subadditivity and what additional argument is needed.
- Write the Carathéodory criterion for a set \(E\subseteq\mathbb{R}\), including its quantifier over test sets.
- Why does \(m^*(A)\le m^*(A\cap E)+m^*(A\cap E^c)\) hold for every \(A,E\subseteq\mathbb{R}\)?
- In the half-line example, what role does choosing an interval cover with total length less than \(m^*(A)+\varepsilon\) play?
- Give the covering lengths used to prove \(m^*(\mathbb{Q})=0\), and evaluate their sum.
- In the proof that the Carathéodory sets form a sigma-algebra, why is it useful to first handle pairwise disjoint countable unions?