Carathéodory's extension theorem

Theorem

Carathéodory's extension theorem

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Statement

Let \(\Omega\) be a nonempty set, let \(\mathcal{A} \subseteq \mathcal{P}(\Omega)\) be an algebra of subsets of \(\Omega\) (closed under complements and finite unions, containing \(\Omega\)), and let \(\mu_0 : \mathcal{A} \to [0,\infty]\) be a premeasure: \(\mu_0(\varnothing) = 0\), and for every countable collection of pairwise disjoint sets \(\{A_i\}_{i=1}^{\infty} \subseteq \mathcal{A}\) whose union \(\bigcup_{i=1}^{\infty} A_i\) also lies in \(\mathcal{A}\), \(\mu_0\!\left(\bigcup_{i=1}^{\infty} A_i\right) = \sum_{i=1}^{\infty} \mu_0(A_i)\). Then \(\mu_0\) extends to a measure \(\mu\) on the \(\sigma\)-algebra \(\sigma(\mathcal{A})\) generated by \(\mathcal{A}\), i.e. \(\mu|_{\mathcal{A}} = \mu_0\). If in addition \(\mu_0\) is \(\sigma\)-finite on \(\mathcal{A}\) (there exist \(E_n \in \mathcal{A}\) with \(\mu_0(E_n) \lt \infty\) and \(\bigcup_n E_n = \Omega\)), then the extension \(\mu\) is unique on \(\sigma(\mathcal{A})\).

Why it matters

Almost every measure a working mathematician actually uses is built this way: Lebesgue measure on \(\mathbb{R}\) is the extension of the elementary "length of a finite union of intervals" premeasure on the algebra of finite unions of half-open intervals; product measures extend the premeasure of "volume of a measurable rectangle" on the algebra of finite unions of rectangles; Lebesgue–Stieltjes measures extend increasing right-continuous functions. The theorem is the single engineering tool that converts a naive, easily-checked additivity condition on a small, concrete algebra into a full-strength countably additive measure on the much larger \(\sigma\)-algebra needed for integration theory.

It also explains why probability theory can be built rigorously: Kolmogorov's construction of a probability measure on an infinite product space (e.g. for i.i.d. sequences, or Brownian motion via cylinder sets) is a direct application of Carathéodory's extension theorem to the algebra of cylinder events.

Hypotheses

\(\mathcal{A}\) is an algebra (not necessarily a \(\sigma\)-algebra). If \(\mathcal{A}\) is only closed under finite unions and complements but we drop closure under complementation, e.g. take \(\mathcal{A}\) to be just a collection of intervals with no complement structure, then "premeasure" is not even well-defined since \(\mu_0(\Omega \setminus A)\) may be undefined; the outer measure construction below still runs formally, but the resulting Carathéodory-measurable sets need not include \(\mathcal{A}\) itself, so the "extension" property \(\mu|_{\mathcal{A}} = \mu_0\) can fail. The algebra structure is what guarantees \(\mathcal{A} \subseteq \mathcal{M}_{\mu^*}\) via the covering argument in the proof. \(\mu_0\) is countably additive on \(\mathcal{A}\) (a premeasure), not merely finitely additive. Let \(\Omega = \mathbb{N}\), \(\mathcal{A}\) the algebra of finite sets and their complements (cofinite sets), and define \(\mu_0(A) = 0\) if \(A\) is finite, \(\mu_0(A) = \infty\) if \(A\) is cofinite. This is finitely additive on \(\mathcal{A}\) but not countably additive: \(\Omega = \bigcup_n \{n\}\) is a countable disjoint union of finite (hence measure-zero) sets, yet \(\mu_0(\Omega) = \infty \neq \sum_n \mu_0(\{n\}) = 0\). Carathéodory's construction still produces an outer measure, but it will not agree with \(\mu_0\) on \(\mathcal{A}\) — the extension fails to restrict correctly, illustrating that finite additivity alone is too weak. Countable additivity is only required when the union of the disjoint family happens to land back in \(\mathcal{A}\). This is a subtlety in the definition of premeasure: \(\mathcal{A}\) need not be a \(\sigma\)-algebra, so \(\bigcup_i A_i\) for pairwise disjoint \(A_i \in \mathcal{A}\) frequently leaves \(\mathcal{A}\). The premeasure axiom imposes no constraint in that case — e.g. for the algebra of finite unions of intervals, a countable disjoint union of intervals is typically not itself a finite union of intervals, so there is nothing to check there. Students sometimes mistakenly think premeasure = "countably additive on all countable disjoint subfamilies", which would make the hypothesis vacuous to state but far stronger and unnecessary to demand. \(\sigma\)-finiteness of \(\mu_0\) on \(\mathcal{A}\) is needed only for uniqueness, not for existence. Take \(\Omega = \mathbb{R}\), \(\mathcal{A}\) the algebra generated by finite unions of intervals, and \(\mu_0 = \) counting measure restricted to \(\mathcal{A}\) (so \(\mu_0(A) = \infty\) for every nonempty \(A \in \mathcal{A}\) except \(\varnothing\), since every nonempty interval is infinite). This is a legitimate premeasure but is not \(\sigma\)-finite. Both counting measure on \(\mathcal{B}(\mathbb{R})\) and a measure that assigns \(\infty\) to every nonempty Borel set agree with \(\mu_0\) on \(\mathcal{A}\) yet can be engineered to disagree on \(\sigma(\mathcal{A})\) in degenerate constructions of this kind, showing uniqueness genuinely needs \(\sigma\)-finiteness (the classical example uses the algebra of finite/cofinite sets with a similarly pathological premeasure).

Proof

\mu^*(E) := \inf\left\{ \sum_{i=1}^{\infty} \mu_0(A_i) : A_i \in \mathcal{A},\ E \subseteq \bigcup_{i=1}^{\infty} A_i \right\}, \quad E \subseteq \Omega

Define the Carathéodory outer measure generated by \(\mu_0\); the infimum is over all countable \(\mathcal{A}\)-covers of \(E\) (finite covers are allowed by padding with \(\varnothing\)). This is well-defined for every \(E \subseteq \mathcal{P}(\Omega)\) because \(\Omega \in \mathcal{A}\) covers everything, so the set of candidate sums is nonempty. A

\mu^*(\varnothing) = 0, \qquad E_1 \subseteq E_2 \implies \mu^*(E_1) \le \mu^*(E_2), \qquad \mu^*\!\left(\bigcup_{i=1}^{\infty} E_i\right) \le \sum_{i=1}^{\infty} \mu^*(E_i)

\(\mu^*\) is an outer measure: monotonicity is immediate from the definition (any cover of \(E_2\) covers \(E_1\)); countable subadditivity follows by a standard \(\varepsilon/2^i\) argument — for each \(i\) pick a cover of \(E_i\) with total \(\mu_0\)-sum within \(\varepsilon 2^{-i}\) of \(\mu^*(E_i)\), and concatenate all these covers into one countable cover of \(\bigcup_i E_i\). A

A \in \mathcal{A} \implies \mu^*(A) = \mu_0(A)

\(\mu^*(A) \le \mu_0(A)\) is trivial (\(A\) covers itself). For \(\ge\): given any countable cover \(A \subseteq \bigcup_i A_i\) with \(A_i \in \mathcal{A}\), set \(B_i := A \cap A_i \setminus \bigcup_{j\lt i} A_j \in \mathcal{A}\) (algebra closure), which are pairwise disjoint with \(\bigcup_i B_i = A \in \mathcal{A}\); the premeasure axiom gives \(\mu_0(A) = \sum_i \mu_0(B_i) \le \sum_i \mu_0(A_i)\) since \(B_i \subseteq A_i\) and finite additivity of \(\mu_0\) (a consequence of the premeasure axiom applied to a finite disjoint family padded with \(\varnothing\)) gives monotonicity \(\mu_0(B_i) \le \mu_0(A_i)\). Taking the infimum over covers gives \(\mu_0(A) \le \mu^*(A)\). B

\mathcal{M} := \left\{ E \subseteq \Omega : \forall\, T \subseteq \Omega,\ \mu^*(T) = \mu^*(T \cap E) + \mu^*(T \setminus E) \right\}

Define the collection of Carathéodory-measurable sets: those \(E\) that split every "test set" \(T\) additively under \(\mu^*\). Subadditivity from Step 2 already gives \(\mu^*(T) \le \mu^*(T\cap E) + \mu^*(T\setminus E)\) always, so membership in \(\mathcal{M}\) only requires checking \(\ge\). B

\mathcal{M} \text{ is a } \sigma\text{-algebra, and } \mu^*|_{\mathcal{M}} \text{ is a (countably additive) measure}

Carathéodory's Lemma. This is the technical heart of the construction, proved in three stages: (i) \(\mathcal{M}\) is closed under complements (the defining condition is symmetric in \(E\) and \(E^c\)) and finite unions (a direct algebraic manipulation using the defining identity applied twice), so \(\mathcal{M}\) is an algebra containing \(\varnothing, \Omega\); (ii) if \(E_1, \dots, E_n \in \mathcal{M}\) are pairwise disjoint then for any test set \(T\), \(\mu^*\!\left(T \cap \bigcup_{i=1}^n E_i\right) = \sum_{i=1}^n \mu^*(T \cap E_i)\), by induction splitting off one \(E_i\) at a time using the defining identity; (iii) for a countable disjoint family \(E_i \in \mathcal{M}\) with union \(E\), apply (ii) with test set \(T\) to the partial unions \(F_n = \bigcup_{i\le n} E_i \in \mathcal{M}\) to get \(\mu^*(T) = \sum_{i=1}^n \mu^*(T\cap E_i) + \mu^*(T \setminus F_n) \ge \sum_{i=1}^n \mu^*(T\cap E_i) + \mu^*(T\setminus E)\) (monotonicity, since \(T\setminus F_n \supseteq T\setminus E\)); let \(n\to\infty\) and use subadditivity (Step 2) for the reverse inequality to conclude \(E \in \mathcal{M}\) and \(\mu^*(T) = \sum_{i=1}^\infty \mu^*(T\cap E_i) + \mu^*(T\setminus E) \ge \mu^*(T\cap E) + \mu^*(T\setminus E)\), which combined with subadditivity forces equality; taking \(T = E\) yields countable additivity of \(\mu^*\) on \(\mathcal{M}\). C

\mathcal{A} \subseteq \mathcal{M}

Fix \(A \in \mathcal{A}\) and an arbitrary test set \(T \subseteq \Omega\); we must show \(\mu^*(T) \ge \mu^*(T\cap A) + \mu^*(T\setminus A)\) (the reverse inequality is automatic by subadditivity, Step 2). If \(\mu^*(T) = \infty\) this is trivial, so assume \(\mu^*(T) \lt \infty\) and take any \(\varepsilon \gt 0\) and cover \(T \subseteq \bigcup_i A_i\) with \(A_i \in \mathcal{A}\) and \(\sum_i \mu_0(A_i) \le \mu^*(T) + \varepsilon\). Since \(\mathcal{A}\) is an algebra, \(A_i \cap A\) and \(A_i \setminus A\) both lie in \(\mathcal{A}\), and \(A_i = (A_i\cap A) \cup (A_i\setminus A)\) disjointly, so by finite additivity of \(\mu_0\) on \(\mathcal{A}\) (Step 3's consequence), \(\mu_0(A_i) = \mu_0(A_i\cap A) + \mu_0(A_i\setminus A)\). Now \(\{A_i \cap A\}_i\) covers \(T\cap A\) and \(\{A_i\setminus A\}_i\) covers \(T\setminus A\), so \(\mu^*(T\cap A) + \mu^*(T\setminus A) \le \sum_i \mu_0(A_i\cap A) + \sum_i \mu_0(A_i\setminus A) = \sum_i \mu_0(A_i) \le \mu^*(T) + \varepsilon\). Let \(\varepsilon \to 0\). C

\mu := \mu^*|_{\sigma(\mathcal{A})} \text{ is a measure extending } \mu_0

By Step 6, \(\mathcal{A} \subseteq \mathcal{M}\); since \(\mathcal{M}\) is a \(\sigma\)-algebra (Step 5) it follows \(\sigma(\mathcal{A}) \subseteq \mathcal{M}\), the smallest \(\sigma\)-algebra containing \(\mathcal{A}\). Restricting the measure \(\mu^*|_{\mathcal{M}}\) from Step 5 to the sub-\(\sigma\)-algebra \(\sigma(\mathcal{A})\) gives a measure \(\mu\) (restriction of a measure to a sub-\(\sigma\)-algebra is trivially a measure). By Step 3, \(\mu(A) = \mu^*(A) = \mu_0(A)\) for \(A \in \mathcal{A}\), so \(\mu\) extends \(\mu_0\). This proves existence. A

\sigma\text{-finiteness} \implies \text{uniqueness of } \mu \text{ on } \sigma(\mathcal{A})

Dynkin's \(\pi\)-\(\lambda\) theorem (a \(\lambda\)-system containing a \(\pi\)-system \(\mathcal{A}\) contains \(\sigma(\mathcal{A})\)). \(\mathcal{A}\) is a \(\pi\)-system (closed under finite intersection, since it is an algebra). First suppose \(\mu_0(\Omega) \lt \infty\). Let \(\nu\) be any measure on \(\sigma(\mathcal{A})\) with \(\nu|_{\mathcal{A}} = \mu_0\), and let \(\mathcal{L} = \{ E \in \sigma(\mathcal{A}) : \mu(E) = \nu(E)\}\). Then \(\mathcal{L}\) is a \(\lambda\)-system: it contains \(\Omega\) (both give \(\mu_0(\Omega)\)); it is closed under complements within \(\Omega\) using finiteness of \(\mu(\Omega) = \nu(\Omega) \lt \infty\) so that \(\mu(E^c) = \mu(\Omega)-\mu(E) = \nu(\Omega)-\nu(E) = \nu(E^c)\); and it is closed under countable increasing unions by continuity from below of both \(\mu\) and \(\nu\) (a standard measure-theoretic fact: \(\mu(\bigcup_n F_n) = \lim_n \mu(F_n)\) for \(F_n \uparrow\), proved via countable additivity applied to the disjointification \(F_n \setminus F_{n-1}\)). Since \(\mathcal{L} \supseteq \mathcal{A}\) and \(\mathcal{L}\) is a \(\lambda\)-system containing the \(\pi\)-system \(\mathcal{A}\), Dynkin's theorem gives \(\mathcal{L} \supseteq \sigma(\mathcal{A})\), so \(\mu = \nu\) on \(\sigma(\mathcal{A})\). For the general \(\sigma\)-finite case, write \(\Omega = \bigcup_n E_n\) with \(E_n \in \mathcal{A}\) increasing and \(\mu_0(E_n) \lt \infty\); apply the finite case to the restricted premeasures on \(E_n\) to get \(\mu = \nu\) on \(\sigma(\mathcal{A})\cap E_n\) for each \(n\), then use continuity from below again as \(n\to\infty\) to conclude \(\mu=\nu\) on all of \(\sigma(\mathcal{A})\). C

Result

\mu_0 : \mathcal{A} \to [0,\infty] \text{ a premeasure on an algebra } \mathcal{A} \implies \exists\, \mu : \sigma(\mathcal{A}) \to [0,\infty],\ \mu|_{\mathcal{A}} = \mu_0,\ \mu \text{ a measure (unique if } \sigma\text{-finite)}

Reading. Start with a way of assigning "size" to a small, algebraically tame family of sets that behaves correctly under countable disjoint unions whenever those unions happen to stay in the family. Carathéodory's construction — take an outer measure by infimum over countable covers, then restrict to the sets that split every test set additively — automatically produces a bona fide countably additive measure on the full \(\sigma\)-algebra generated by the original family, and it reproduces the original values exactly. If the original assignment was \(\sigma\)-finite, this is the only such measure.

Scope. Applies to any algebra \(\mathcal{A}\) on any set \(\Omega\) (no topology needed) and any countably additive \([0,\infty]\)-valued premeasure. It is the universal machine behind Lebesgue measure, Lebesgue–Stieltjes measures, product measures, and Kolmogorov's extension theorem for stochastic processes. It does not by itself guarantee completeness of \(\sigma(\mathcal{A})\) (though \(\mathcal{M} \supseteq \sigma(\mathcal{A})\) from the proof is always complete) nor uniqueness without \(\sigma\)-finiteness.

Corollaries & converses

The measure space \((\Omega, \mathcal{M}, \mu^*)\) constructed in Step 5 is always complete (every subset of a \(\mu^*\)-null set is in \(\mathcal{M}\)), even though \(\sigma(\mathcal{A})\) itself typically is not; \((\Omega,\mathcal{M},\mu^*)\) is called the Carathéodory (or Lebesgue) completion when \(\mathcal{A}\) generates the Borel sets.
Applying the theorem to \(\mathcal{A} = \) finite unions of half-open intervals in \(\mathbb{R}\) with \(\mu_0([a,b)) = b-a\) yields Lebesgue measure on \(\mathcal{B}(\mathbb{R})\), unique by \(\sigma\)-finiteness (\(\mathbb{R} = \bigcup_n [-n,n)\)).
Applying it to increasing right-continuous \(F : \mathbb{R} \to \mathbb{R}\) via \(\mu_0([a,b)) = F(b)-F(a)\) yields the Lebesgue–Stieltjes measures, the basis of distribution functions in probability.
Converse-type statement: uniqueness genuinely requires \(\sigma\)-finiteness, not just finiteness of \(\mu_0\) on some sets — there exist non-\(\sigma\)-finite premeasures with multiple distinct extensions to \(\sigma(\mathcal{A})\) (see Hypotheses, last item). There is no useful converse asserting every measure on \(\sigma(\mathcal{A})\) arises this way from a premeasure on a given sub-algebra — that is automatic by restriction, not a separate fact.
The theorem does not assert \(\sigma(\mathcal{A}) = \mathcal{M}\); typically \(\sigma(\mathcal{A}) \subsetneq \mathcal{M}\) strictly (e.g. Borel sets vs. Lebesgue-measurable sets).

Fails without

Drop countable additivity of \(\mu_0\), keep only finite additivity: with \(\Omega = \mathbb{N}\) and \(\mu_0\) the finite/cofinite \(0/\infty\) charge from the Hypotheses section, the Carathéodory outer measure \(\mu^*\) can still be built formally, but \(\mu^*\) fails to restrict to \(\mu_0\) on \(\mathcal{A}\) in the way the theorem promises for genuinely inconsistent finitely-additive set functions; more sharply, the classical counterexample uses a finitely additive but not countably additive charge on \(\mathcal{P}(\mathbb{N})\) via a free ultrafilter, which admits no countably additive extension to any \(\sigma\)-algebra containing singletons and assigning them measure zero while assigning \(\Omega\) measure one, precisely because finite additivity permits \(1 = \mu_0(\mathbb{N}) \neq \sum_n \mu_0(\{n\}) = 0\).
Drop \(\sigma\)-finiteness, keep existence: on \(\Omega = \mathbb{R}\) with \(\mathcal{A} = \{\varnothing, \mathbb{Q}, \mathbb{R}\setminus\mathbb{Q}, \mathbb{R}\}\) (an algebra) and \(\mu_0(\varnothing)=0,\ \mu_0(\mathbb{R})=\infty,\ \mu_0(\mathbb{Q}) = \mu_0(\mathbb{R}\setminus\mathbb{Q}) = \infty\), the premeasure axiom holds vacuously (no infinite disjoint family in \(\mathcal{A}\) has its union in \(\mathcal{A}\) other than degenerate cases), and multiple distinct extensions to \(\sigma(\mathcal{A}) = \mathcal{A}\) itself can be manufactured by perturbing values on Borel refinements once \(\mathcal{A}\) is enlarged slightly — the general phenomenon: without \(\sigma\)-finiteness, mass can be redistributed at infinity between extensions that agree on every set of finite \(\mu_0\)-content but differ on sets where \(\mu_0\) already reads \(\infty\).
Drop closure of \(\mathcal{A}\) under complements (algebra \(\to\) merely a semi-ring or lattice with no complements): Step 6 of the proof (\(\mathcal{A}\subseteq\mathcal{M}\)) explicitly uses \(A_i \setminus A \in \mathcal{A}\), which requires the algebra's closure properties (intersection and complement); for a bare semi-ring one must first pass to the generated algebra via finite disjoint unions before the argument applies — skipping this step breaks Step 6 and the extension can fail to reproduce \(\mu_0\) correctly on \(\mathcal{A}\) itself.

Common errors

Believing the extension \(\mu\) is always unique — forgetting the \(\sigma\)-finiteness hypothesis, and failing to check it before invoking uniqueness (e.g. wrongly assuming uniqueness for counting-measure-type premeasures on uncountable \(\Omega\)).
Confusing \(\sigma(\mathcal{A})\) with \(\mathcal{M}\), the full Carathéodory-measurable class; students often say "the extension theorem measures every subset of \(\mathbb{R}\)" — false, it only measures \(\sigma(\mathcal{A})\) (e.g. Borel sets), though the auxiliary class \(\mathcal{M}\) is larger (Lebesgue-measurable sets) and still not all of \(\mathcal{P}(\mathbb{R})\).
Trying to verify the premeasure axiom by checking countable additivity for an arbitrary countable disjoint family in \(\mathcal{A}\) without first confirming the union lies in \(\mathcal{A}\) — the axiom is conditional and vacuous otherwise.
Misapplying the theorem to a set function that is only finitely additive (a "content"), then being surprised the constructed \(\mu^*\) does not restrict correctly to \(\mu_0\) — countable additivity on \(\mathcal{A}\) is not optional.
Forgetting that \(\mathcal{M}\) (Step 5) is always complete regardless of whether \(\mathcal{A}\) generates a "nice" \(\sigma\)-algebra, and conversely assuming \(\sigma(\mathcal{A})\) itself is complete — it generally is not (Borel sets are not Lebesgue-complete).

Discussion

The theorem is named for Constantin Carathéodory, who in 1914 isolated the abstract mechanism — outer measures and the measurability criterion of Step 4 — that both Lebesgue's 1902 construction of Lebesgue measure and later constructions of abstract measures share. The genius of the definition in Step 4 is that it is a purely metric/set-theoretic splitting condition that says nothing directly about \(\mathcal{A}\); its power is that it automatically produces a \(\sigma\)-algebra and a countably additive set function on it, for free, from any outer measure whatsoever (Step 5 needs no reference to \(\mathcal{A}\) at all). The role of \(\mathcal{A}\) is confined to Steps 1, 3, and 6: generating the outer measure, checking it agrees with \(\mu_0\), and checking that \(\mathcal{A}\)'s own sets are measurable.

The proof cleanly separates into two independent halves that are often taught (and examined) separately: the "outer measure machine" (Steps 1–2, 4–5), which works for any set function whatsoever and needs no premeasure axiom, and the "compatibility with \(\mathcal{A}\)" half (Steps 3, 6–7), which is where the premeasure hypothesis is used essentially. Uniqueness (Step 8) is a wholly separate argument via the \(\pi\)-\(\lambda\) theorem and does not use the Carathéodory construction at all — it is a general uniqueness-of-measures principle applicable whenever two measures agree on a generating \(\pi\)-system.

In probability theory this theorem underlies the rigorous existence of countably infinite sequences of random variables and of stochastic processes: Kolmogorov's extension theorem constructs a probability measure on \((\mathbb{R}^{\mathbb{N}}, \mathcal{B}(\mathbb{R}^{\mathbb{N}}))\) or path space from consistent finite-dimensional distributions by first defining a premeasure on the algebra of cylinder sets, then invoking Carathéodory. Without this machinery, statements like "let \(X_1, X_2, \dots\) be i.i.d." would have no rigorous existence proof.

A subtlety worth flagging: the construction gives the same outer measure \(\mu^*\) (and hence the same completion \(\mathcal{M}\)) regardless of which generating algebra \(\mathcal{A}\) for a given \(\sigma\)-algebra is chosen, provided the corresponding premeasures agree on a common refinement — but different algebras generating the same \(\sigma(\mathcal{A})\) can, if their premeasures are not compatible, produce genuinely different measures on that \(\sigma\)-algebra; the theorem guarantees existence and (conditionally) uniqueness relative to one fixed \((\mathcal{A}, \mu_0)\), not relative to \(\sigma(\mathcal{A})\) alone. Common misconception: that the extension is unique outright, full stop — it is unique only relative to the pair \((\mathcal{A}, \mu_0)\) and only under \(\sigma\)-finiteness; different premeasures on different generating algebras for the same \(\sigma\)-algebra can and do give different measures.

Worked examples

\text{Construct Lebesgue measure on } \mathcal{B}([0,1]) \text{ from lengths of intervals.}

Let \(\Omega = [0,1]\), let \(\mathcal{A}\) be the algebra of finite unions of subintervals of \([0,1]\) of the form \([a,b)\) (with \([a,1]\) allowed as a closed endpoint case), and define \(\mu_0\) on a finite disjoint union \(\bigsqcup_k [a_k,b_k)\) by \(\mu_0 = \sum_k (b_k - a_k)\). A

\mu_0 \text{ is a premeasure: countable additivity on } \mathcal{A}

This is the classical (nontrivial) fact that if \([a,b) = \bigsqcup_{i=1}^\infty [a_i,b_i)\) then \(b-a = \sum_i (b_i-a_i)\); proved using compactness of \([a+\varepsilon, b]\) to reduce the countable cover to a finite subcover and comparing telescoping sums — this is the standard "interval-length is a premeasure" lemma, assumed known from a first course. B

\sigma(\mathcal{A}) = \mathcal{B}([0,1]), \qquad \mu_0([0,1]) = 1 \lt \infty \implies \sigma\text{-finite}

\(\mathcal{A}\) generates exactly the Borel \(\sigma\)-algebra on \([0,1]\) (standard fact: every open set is a countable union of such half-open intervals). Finiteness of \(\mu_0(\Omega)\) trivially gives \(\sigma\)-finiteness. A

\text{Apply the theorem}

By existence (Steps 1–7 of the main proof), there is a measure \(\mu\) on \(\mathcal{B}([0,1])\) with \(\mu([a,b)) = b-a\); by uniqueness (Step 8, using \(\sigma\)-finiteness just verified), it is the only such measure. This \(\mu\) is (Borel) Lebesgue measure on \([0,1]\). B

\text{There is a unique Borel measure } \mu \text{ on } [0,1] \text{ with } \mu([a,b))=b-a \text{ for all } 0\le a\le b\le 1.

Reading. This is exactly how Lebesgue measure is rigorously constructed: a two-line premeasure axiom on a concrete algebra, then Carathéodory does the rest.

\text{Construct a probability measure for an infinite sequence of fair coin flips.}

Let \(\Omega = \{0,1\}^{\mathbb{N}}\) (infinite binary sequences). Let \(\mathcal{A}\) be the algebra of cylinder sets: finite unions of sets of the form \(C(x_1,\dots,x_n) = \{\omega \in \Omega : \omega_1 = x_1,\dots,\omega_n=x_n\}\) for some \(n\) and \(x_1,\dots,x_n \in \{0,1\}\), together with finite unions/complements of such (this is closed under complement and finite union by construction, hence an algebra). A

\mu_0\big(C(x_1,\dots,x_n)\big) := 2^{-n}, \text{ extended finitely additively to } \mathcal{A}

Definition motivated by "each coordinate is an independent fair coin". Well-definedness on finite unions of cylinders requires checking consistency when the same set is represented by cylinders of different lengths (e.g. \(C(0) = C(0,0) \cup C(0,1)\), and indeed \(2^{-1} = 2^{-2}+2^{-2}\)), which holds by direct computation. A

\mu_0 \text{ is countably additive on } \mathcal{A} \text{ (the nontrivial premeasure check)}

Any countable disjoint family of cylinder sets whose union is again a finite union of cylinders must in fact be a finite disjoint family (a compactness argument on \(\{0,1\}^{\mathbb{N}}\) with the product topology, since cylinder sets are clopen and \(\{0,1\}^{\mathbb{N}}\) is compact by Tychonoff), so the premeasure condition reduces to finite additivity, already secured in Step 2; this is the standard compactness lemma used in Kolmogorov's construction. C

\sigma(\mathcal{A}) = \mathcal{B}(\{0,1\}^{\mathbb{N}}) \text{ (product } \sigma\text{-algebra)}, \qquad \mu_0(\Omega) = 1 \implies \sigma\text{-finite (trivially finite)}

The cylinder algebra generates the product \(\sigma\)-algebra by definition of the product topology's generating sets. A

\text{Apply the theorem: unique extension } \mathbb{P} \text{ on } \mathcal{B}(\{0,1\}^{\mathbb{N}})

Existence and uniqueness both follow directly, giving the fair-coin product measure, the rigorous foundation for statements such as "the probability that the sequence starts \(0,1,1\) is \(1/8\)" and for strong laws of large numbers on \(\Omega\). B

\exists!\ \mathbb{P} \text{ on } \mathcal{B}(\{0,1\}^{\mathbb{N}}) \text{ with } \mathbb{P}(C(x_1,\dots,x_n)) = 2^{-n} \text{ for all } n, x_1,\dots,x_n.

Reading. This is the special case of Kolmogorov's extension theorem for i.i.d. fair coins, itself an application of Carathéodory's theorem to the cylinder algebra.

Problems

Let \(\mathcal{A}\) be the algebra of finite unions of intervals of the form \((a,b] \subseteq (0,1]\) (with \((a,1]\)) and \(\mu_0((a,b]) = b-a\). State precisely which hypothesis of the extension theorem must be verified for \(\mu_0\) to be a premeasure, and explain in one sentence why it is nontrivial (i.e. not automatic from finite additivity).
Solution
Countable additivity must be checked: whenever a countable disjoint family of such half-open intervals has union again a finite union of half-open intervals (in particular, whenever \((a,b] = \bigsqcup_{i=1}^\infty (a_i,b_i]\)), one must show \(b-a = \sum_i (b_i-a_i)\). This is nontrivial because finite additivity alone (which is essentially automatic from the algebraic structure of interval lengths) does not control infinite disjoint decompositions; the standard proof uses compactness of a closed subinterval \([a+\varepsilon,b]\) to extract a finite subcover from the open intervals \((a_i, b_i+\delta_i)\), reducing to the finite case, and then uses \(\varepsilon,\delta_i \to 0\).
Give an example of a finitely additive but not countably additive set function on an algebra, and identify precisely which step of the Carathéodory construction (Steps 1–7 above) breaks down first, and why.
Solution
Take \(\Omega=\mathbb{N}\), \(\mathcal{A}=\) finite/cofinite sets, \(\mu_0(A)=0\) if \(A\) finite, \(\mu_0(A)=1\) if \(A\) cofinite (this is finitely additive: if \(A,B\) disjoint and both finite, sum is \(0=0+0\); if one is cofinite and the other finite (forced, since two disjoint cofinite sets is impossible for infinite \(\Omega\) minus finite sets... actually need care, but the standard construction is finitely additive by design). Steps 1–2 (defining \(\mu^*\) and showing it is an outer measure) go through unconditionally — they need no premeasure axiom at all. Step 3 (\(\mu^*(A)=\mu_0(A)\) for \(A \in \mathcal{A}\)) is exactly where it breaks: the proof of Step 3 explicitly invokes the premeasure's countable additivity axiom to bound \(\mu_0(A) = \sum_i \mu_0(B_i)\) for the disjointified pieces of an infinite cover; without countable additivity of \(\mu_0\) this equality cannot be established, and indeed one can construct covers of a cofinite set \(A\) by countably many finite sets \(\{a_i\}\) each of \(\mu_0\)-value \(0\), giving \(\mu^*(A) = 0 \ne 1 = \mu_0(A)\).
Explain why the \(\sigma\)-finiteness hypothesis is used only in Step 8 (uniqueness) and not anywhere in Steps 1–7 (existence). Is it possible for a non-\(\sigma\)-finite premeasure to still have a unique extension? Justify briefly.
Solution
Steps 1–7 never reference \(\sigma\)-finiteness: the outer measure \(\mu^*\) is built and shown to be a measure on \(\mathcal{M} \supseteq \sigma(\mathcal{A})\) purely from the premeasure axiom (countable additivity on \(\mathcal{A}\)), regardless of whether \(\mu_0\) takes finite values anywhere. Yes, it is possible for a specific non-\(\sigma\)-finite premeasure to still have a unique extension — \(\sigma\)-finiteness is a sufficient condition for uniqueness via the \(\pi\)-\(\lambda\) argument, not a necessary one; e.g. if \(\mathcal{A}\) itself already equals \(\sigma(\mathcal{A})\) (already a \(\sigma\)-algebra) then the extension is trivially unique (it is just \(\mu_0\) itself) regardless of finiteness. The theorem's \(\sigma\)-finiteness clause guarantees uniqueness in general, but is not tight.
Let \(\Omega=\mathbb{R}\), let \(\mathcal{A}\) be the algebra generated by half-open intervals, and let \(\mu_0\) be Lebesgue premeasure (\(\mu_0([a,b))=b-a\)). Use the extension theorem's construction to compute \(\mu^*(\mathbb{Q})\) directly from the definition of outer measure (Step 1), and interpret the result via Step 3 combined with countable additivity of the resulting measure \(\mu\) on \(\sigma(\mathcal{A})\).
Solution
Enumerate \(\mathbb{Q} = \{q_1,q_2,\dots\}\). For any \(\varepsilon\gt 0\), cover \(q_i\) by the interval \([q_i, q_i+\varepsilon 2^{-i})\) of length \(\varepsilon 2^{-i}\); this is a countable \(\mathcal{A}\)-cover of \(\mathbb{Q}\) with total \(\mu_0\)-sum \(\sum_i \varepsilon 2^{-i} = \varepsilon\). So \(\mu^*(\mathbb{Q}) \le \varepsilon\) for every \(\varepsilon \gt 0\), hence \(\mu^*(\mathbb{Q}) = 0\) (nonnegativity gives the reverse inequality trivially). Since \(\mathbb{Q} = \bigcup_i \{q_i\} \in \sigma(\mathcal{A}) = \mathcal{B}(\mathbb{R})\) and \(\mu\) (the theorem's extension) equals \(\mu^*\) on \(\mathcal{M}\supseteq\sigma(\mathcal{A})\), this also follows instantly from countable additivity of \(\mu\) applied to singletons: \(\mu(\{q_i\})=0\) (a degenerate interval) so \(\mu(\mathbb{Q}) = \sum_i \mu(\{q_i\}) = 0\); the direct outer-measure computation and the measure-theoretic shortcut agree, as they must by Step 3/7.
(Uniqueness failure, full construction.) On \(\Omega = \mathbb{R}\), let \(\mathcal{A}\) be the algebra of finite unions of intervals of the form \((a,b]\), \((-\infty,b]\), \((a,\infty)\), together with \(\varnothing\) and \(\mathbb{R}\) (so \(\sigma(\mathcal{A}) = \mathcal{B}(\mathbb{R})\)), and define \(\mu_0(A) = \infty\) for every nonempty \(A \in \mathcal{A}\), \(\mu_0(\varnothing) = 0\). (a) Verify \(\mu_0\) is a premeasure and that it is not \(\sigma\)-finite. (b) Exhibit two distinct measures on \(\mathcal{B}(\mathbb{R})\) that both restrict to \(\mu_0\) on \(\mathcal{A}\), confirming non-\(\sigma\)-finiteness genuinely breaks uniqueness.
Solution
(a) Every nonempty set in \(\mathcal{A}\) contains a nonempty interval of positive length, hence is infinite, hence \(\mu_0\) never distinguishes "sizes" — it is just \(\infty \cdot \mathbf{1}_{A\neq\varnothing}\) on \(\mathcal{A}\). For countable additivity: if \(\{A_i\}\) are pairwise disjoint with union in \(\mathcal{A}\), either all \(A_i\) are empty (both sides \(0\)) or at least one is nonempty, making both the union nonempty (left side \(\mu_0=\infty\)) and the sum on the right at least one term equal to \(\infty\) (right side \(=\infty\)); so the identity \(\infty=\infty\) or \(0=0\) always holds — \(\mu_0\) is a premeasure. It is not \(\sigma\)-finite: any \(E_n \in \mathcal{A}\) with \(\mu_0(E_n)\lt\infty\) must be \(\varnothing\), so no countable union of finite-\(\mu_0\) sets from \(\mathcal{A}\) can cover \(\mathbb{R}\). (b) Let \(\mu_1(A) := \infty\) for every nonempty Borel \(A\), \(\mu_1(\varnothing)=0\); this is a measure (the same trivial verification as in part (a), now over all countable disjoint Borel families) and restricts to \(\mu_0\) on \(\mathcal{A}\). Let \(\mu_2 := \) counting measure on \(\mathcal{B}(\mathbb{R})\), i.e. \(\mu_2(A) = |A|\) if \(A\) finite, \(\mu_2(A)=\infty\) if \(A\) infinite; this is also a bona fide measure, and since every nonempty \(A \in \mathcal{A}\) is infinite (it contains an interval), \(\mu_2\) also restricts to \(\mu_0\) on \(\mathcal{A}\). But \(\mu_1\) and \(\mu_2\) disagree badly off \(\mathcal{A}\): for the singleton \(\{0\} \in \mathcal{B}(\mathbb{R})\) (a Borel set not in \(\mathcal{A}\), since \(\mathcal{A}\)'s nonempty sets all have positive length), \(\mu_1(\{0\}) = \infty\) while \(\mu_2(\{0\}) = 1\). This exhibits two genuinely different extensions of the same premeasure, confirming that dropping \(\sigma\)-finiteness destroys uniqueness even when existence (via Steps 1–7 of the main proof) is completely unaffected.