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09 May 2026 · 8 min read · updated 13 June 2026

Sigma-Algebras and Measures

Integration needs a prior notion of the size of a set, and not every set can be sized consistently. We define sigma-algebras and measures, prove a measure is continuous along increasing and decreasing limits, prove measurable functions are closed under suprema and pointwise limits, and prove the Caratheodory extension theorem that the sets split additively by an outer measure form a sigma-algebra on which the outer measure is countably additive. Applying it to the length of intervals constructs Lebesgue measure, the ground on which the Lebesgue integral is built.

  • 5 equations
  • 10 results
  • 13 connections
  • measure-theory
  • real-analysis
  • probability
On this page▾
  • Sigma-algebras
  • Measures and continuity
  • Measurable functions and limits
  • The Caratheodory extension theorem
  • Lebesgue measure

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  • Sigma-algebras1m
  • Measures and continuity1m
  • Measurable functions and limits1m
  • The Caratheodory extension theorem2m
  • Lebesgue measure3m

To integrate a function one first measures sets, and on the real line no countably additive notion of length can be assigned to every subset at once. The resolution is to measure only a rich collection of sets, a sigma-algebra, and to build the measure on it by extension from a primitive notion of size. This post supplies the construction the Lebesgue integral takes for granted, the Caratheodory extension theorem and the Lebesgue measure it produces. It uses the limit theory of sequences and completeness [1].

#Sigma-algebras

Definition1

A sigma-algebra on a set Ω\OmegaΩ is a collection A\mathcal AA of subsets containing Ω\OmegaΩ, closed under complement, and closed under countable union. Its members are the measurable sets.

Closure under complement and countable union gives closure under countable intersection by De Morgan, and under set difference. The intersection of any family of sigma-algebras is a sigma-algebra, so for any collection C\mathcal CC of sets there is a smallest sigma-algebra containing it, the one generated by C\mathcal CC. On a metric space the sigma-algebra generated by the open sets is the Borel sigma-algebra, the measurable sets one cannot avoid.

#Measures and continuity

Definition2

A measure on (Ω,A)(\Omega,\mathcal A)(Ω,A) is a function μ:A→[0,∞]\mu:\mathcal A\to[0,\infty]μ:A→[0,∞] with μ(∅)=0\mu(\emptyset)=0μ(∅)=0 that is countably additive, meaning μ(⋃nEn)=∑nμ(En)\mu\big(\bigcup_n E_n\big)=\sum_n\mu(E_n)μ(⋃n​En​)=∑n​μ(En​) for every sequence of pairwise disjoint measurable sets.

Countable additivity forces monotonicity, since E⊆FE\subseteq FE⊆F gives μ(F)=μ(E)+μ(F∖E)≥μ(E)\mu(F)=\mu(E)+\mu(F\setminus E) \ge\mu(E)μ(F)=μ(E)+μ(F∖E)≥μ(E), and it forces continuity along monotone limits.

Proposition3

If E1⊆E2⊆⋯E_1\subseteq E_2\subseteq\cdotsE1​⊆E2​⊆⋯ then μ(⋃nEn)=lim⁡nμ(En)\mu\big(\bigcup_n E_n\big)=\lim_n\mu(E_n)μ(⋃n​En​)=limn​μ(En​). If F1⊇F2⊇⋯F_1\supseteq F_2\supseteq\cdotsF1​⊇F2​⊇⋯ and μ(F1)<∞\mu(F_1)<\inftyμ(F1​)<∞ then μ(⋂nFn)=lim⁡nμ(Fn)\mu\big(\bigcap_n F_n\big)=\lim_n\mu(F_n)μ(⋂n​Fn​)=limn​μ(Fn​).

Proof

For the increasing case write the union as the disjoint union of the differences D1=E1D_1=E_1D1​=E1​ and Dn=En∖En−1D_n=E_n\setminus E_{n-1}Dn​=En​∖En−1​. Countable additivity and then the telescoping finite sums give

μ(⋃nEn)=∑nμ(Dn)=lim⁡N∑n=1Nμ(Dn)=lim⁡Nμ(EN).(1)\mu\Big(\bigcup_n E_n\Big)=\sum_n\mu(D_n)=\lim_N\sum_{n=1}^N\mu(D_n)=\lim_N\mu(E_N). \tag{1}μ(n⋃​En​)=n∑​μ(Dn​)=Nlim​n=1∑N​μ(Dn​)=Nlim​μ(EN​).(1)

For the decreasing case apply the increasing case to En=F1∖FnE_n=F_1\setminus F_nEn​=F1​∖Fn​, which rises to F1∖⋂nFnF_1\setminus\bigcap_n F_nF1​∖⋂n​Fn​. Then μ(F1)−μ(⋂nFn)=μ(F1)−lim⁡nμ(Fn)\mu(F_1)-\mu\big(\bigcap_n F_n\big)=\mu(F_1)-\lim_n\mu(F_n)μ(F1​)−μ(⋂n​Fn​)=μ(F1​)−limn​μ(Fn​), and the finiteness of μ(F1)\mu(F_1)μ(F1​) lets the subtractions cancel, giving the claim.

#Measurable functions and limits

A function f:Ω→Rf:\Omega\to\Rf:Ω→R is measurable when {f>a}={ω:f(ω)>a}∈A\{f>a\}=\{\omega:f(\omega)>a\}\in\mathcal A{f>a}={ω:f(ω)>a}∈A for every a∈Ra\in\Ra∈R, equivalently when the preimage of every Borel set is measurable. The same criterion defines measurability for an extended-real g:Ω→[−∞,+∞]g:\Omega\to[-\infty,+\infty]g:Ω→[−∞,+∞], since it forces {g=+∞}=⋂n{g>n}\{g=+\infty\}= \bigcap_n\{g>n\}{g=+∞}=⋂n​{g>n} and {g=−∞}=(⋃n{g>−n})c\{g=-\infty\}=\big(\bigcup_n\{g>-n\}\big)^c{g=−∞}=(⋃n​{g>−n})c into A\mathcal AA. Measurability survives every limit operation, the fact the convergence theorems rest on, with suprema and limits read as [−∞,+∞][-\infty,+\infty][−∞,+∞]-valued.

Proposition4

If f1,f2,…f_1,f_2,\dotsf1​,f2​,… are measurable then sup⁡nfn\sup_n f_nsupn​fn​, inf⁡nfn\inf_n f_ninfn​fn​, lim sup⁡nfn\limsup_n f_nlimsupn​fn​, and lim inf⁡nfn\liminf_n f_nliminfn​fn​ are measurable, and so is lim⁡nfn\lim_n f_nlimn​fn​ wherever it exists.

Proof

The supremum is measurable because {sup⁡nfn>a}=⋃n{fn>a}\{\sup_n f_n>a\}=\bigcup_n\{f_n>a\}{supn​fn​>a}=⋃n​{fn​>a} is a countable union of measurable sets. The infimum follows from inf⁡nfn=−sup⁡n(−fn)\inf_n f_n=-\sup_n(-f_n)infn​fn​=−supn​(−fn​). Then lim sup⁡nfn=inf⁡Nsup⁡n≥Nfn\limsup_n f_n=\inf_N\sup_{n\ge N}f_nlimsupn​fn​=infN​supn≥N​fn​ and lim inf⁡nfn=sup⁡Ninf⁡n≥Nfn\liminf_n f_n=\sup_N\inf_{n\ge N}f_nliminfn​fn​=supN​infn≥N​fn​ are measurable as countable suprema and infima of measurable functions. Where the limit exists it equals the limit superior, hence is measurable.

#The Caratheodory extension theorem

A measure is hard to define directly, because countable additivity must be checked against all disjoint decompositions. The standard route defines an approximate size on all sets and then carves out those on which it behaves. An outer measure is a function μ∗:2Ω→[0,∞]\om:2^\Omega\to[0,\infty]μ∗:2Ω→[0,∞] with μ∗(∅)=0\om(\emptyset)=0μ∗(∅)=0 that is monotone and countably subadditive, μ∗(⋃nAn)≤∑nμ∗(An)\om\big(\bigcup_n A_n\big)\le\sum_n \om(A_n)μ∗(⋃n​An​)≤∑n​μ∗(An​). A set EEE is Caratheodory measurable when it splits every set additively,

μ∗(A)=μ∗(A∩E)+μ∗(A∖E)for all A⊆Ω.(2)\om(A)=\om(A\cap E)+\om(A\setminus E)\qquad\text{for all }A\subseteq\Omega. \tag{2}μ∗(A)=μ∗(A∩E)+μ∗(A∖E)for all A⊆Ω.(2)

Subadditivity makes ≤\le≤ automatic, so the content of Equation (2) is the reverse inequality.

Theorem5

The Caratheodory measurable sets form a sigma-algebra M\mathcal MM, and the restriction of μ∗\omμ∗ to M\mathcal MM is a measure.

Proof

The empty set is measurable and the definition is symmetric in EEE and its complement, so M\mathcal MM contains ∅\emptyset∅ and is closed under complement. For finite unions, let E,F∈ME,F\in\mathcal ME,F∈M and split an arbitrary AAA first by EEE and then the part outside EEE by FFF,

μ∗(A)=μ∗(A∩E)+μ∗(A∩Ec∩F)+μ∗(A∩Ec∩Fc).(3)\om(A)=\om(A\cap E)+\om(A\cap E^c\cap F)+\om(A\cap E^c\cap F^c). \tag{3}μ∗(A)=μ∗(A∩E)+μ∗(A∩Ec∩F)+μ∗(A∩Ec∩Fc).(3)

The first two pieces cover A∩(E∪F)A\cap(E\cup F)A∩(E∪F), so subadditivity bounds their sum below by μ∗(A∩(E∪F))\om(A\cap(E\cup F))μ∗(A∩(E∪F)), while the third is μ∗(A∩(E∪F)c)\om(A\cap(E\cup F)^c)μ∗(A∩(E∪F)c). Hence μ∗(A)≥μ∗(A∩(E∪F))+μ∗(A∩(E∪F)c)\om(A)\ge\om(A\cap(E\cup F))+\om(A\cap(E\cup F)^c)μ∗(A)≥μ∗(A∩(E∪F))+μ∗(A∩(E∪F)c), and E∪F∈ME\cup F\in\mathcal ME∪F∈M. For disjoint E,F∈ME,F\in\mathcal ME,F∈M, testing the split of E∪FE\cup FE∪F by EEE gives μ∗(A∩(E∪F))=μ∗(A∩E)+μ∗(A∩F)\om(A\cap(E\cup F))=\om(A\cap E)+ \om(A\cap F)μ∗(A∩(E∪F))=μ∗(A∩E)+μ∗(A∩F), which by induction extends to μ∗(A∩⋃k≤nEk)=∑k≤nμ∗(A∩Ek)\om\big(A\cap\bigcup_{k\le n}E_k\big)=\sum_{k\le n} \om(A\cap E_k)μ∗(A∩⋃k≤n​Ek​)=∑k≤n​μ∗(A∩Ek​) for disjoint measurable EkE_kEk​. Now let E=⋃kEkE=\bigcup_k E_kE=⋃k​Ek​ be a countable disjoint union and Fn=⋃k≤nEkF_n=\bigcup_{k\le n}E_kFn​=⋃k≤n​Ek​. Since Fn∈MF_n\in\mathcal MFn​∈M and Fnc⊇EcF_n^c\supseteq E^cFnc​⊇Ec, so monotonicity bounds the second term below by μ∗(A∩Ec)\om(A\cap E^c)μ∗(A∩Ec),

μ∗(A)=μ∗(A∩Fn)+μ∗(A∩Fnc)≥∑k≤nμ∗(A∩Ek)+μ∗(A∩Ec).(4)\om(A)=\om(A\cap F_n)+\om(A\cap F_n^c)\ge\sum_{k\le n}\om(A\cap E_k)+\om(A\cap E^c). \tag{4}μ∗(A)=μ∗(A∩Fn​)+μ∗(A∩Fnc​)≥k≤n∑​μ∗(A∩Ek​)+μ∗(A∩Ec).(4)

Letting n→∞n\to\inftyn→∞ and using subadditivity on the tail, μ∗(A)≥∑kμ∗(A∩Ek)+μ∗(A∩Ec)≥μ∗(A∩E)+μ∗(A∩Ec)≥μ∗(A)\om(A)\ge\sum_k\om(A\cap E_k)+\om(A\cap E^c)\ge\om(A\cap E)+\om(A\cap E^c)\ge\om(A)μ∗(A)≥∑k​μ∗(A∩Ek​)+μ∗(A∩Ec)≥μ∗(A∩E)+μ∗(A∩Ec)≥μ∗(A), so all are equalities. The first equality shows E∈ME\in\mathcal ME∈M, so M\mathcal MM is closed under countable disjoint unions and hence, with complements, under all countable unions. Taking A=EA=EA=E in the chain gives μ∗(E)=∑kμ∗(Ek)\om(E)=\sum_k\om(E_k)μ∗(E)=∑k​μ∗(Ek​), the countable additivity of μ∗\omμ∗ on M\mathcal MM.

The theorem is an engine. Feed it any outer measure and it returns a sigma-algebra carrying a genuine measure, and the measure is complete, since a subset of an μ∗\omμ∗-null set splits every set trivially and so is measurable.

#Lebesgue measure

On the line, length is the primitive size. Define the Lebesgue outer measure of E⊆RE\subseteq\RE⊆R by covering with open intervals and taking the cheapest cover,

μ∗(E)=inf⁡{∑k∣Ik∣:E⊆⋃kIk, Ik open intervals}.(5)\om(E)=\inf\Big\{\sum_k\abs{I_k}:E\subseteq\bigcup_k I_k,\ I_k\text{ open intervals}\Big\}. \tag{5}μ∗(E)=inf{k∑​∣Ik​∣:E⊆k⋃​Ik​, Ik​ open intervals}.(5)

It is monotone because a cover of a larger set covers the smaller, and countably subadditive because the inequality is trivial when some μ∗(An)=∞\om(A_n)=\inftyμ∗(An​)=∞, and otherwise the union of covers of the AnA_nAn​, with the nnn-th cover sharpened to within ε2−n\varepsilon 2^{-n}ε2−n of μ∗(An)\om(A_n)μ∗(An​), covers ⋃nAn\bigcup_n A_n⋃n​An​ at cost within ε\varepsilonε of ∑nμ∗(An)\sum_n\om(A_n)∑n​μ∗(An​). So μ∗\omμ∗ is an outer measure, and Theorem 5 produces the Lebesgue sigma-algebra and Lebesgue measure λ=μ∗\lambda=\omλ=μ∗ on it.

Proposition6

Every interval is Lebesgue measurable, so the Borel sigma-algebra is contained in the Lebesgue sigma-algebra, and λ\lambdaλ assigns each interval its length. Lebesgue measure is translation invariant and is the unique Borel measure giving each interval its length.

Proof

To see a half-line (b,∞)(b,\infty)(b,∞) is measurable, take any AAA and a near-optimal interval cover; splitting each covering interval IkI_kIk​ at bbb replaces it by its part in (b,∞)(b,\infty)(b,∞) and the open interval Ik∩(−∞,b+δk)I_k\cap(-\infty,b+\delta_k)Ik​∩(−∞,b+δk​) covering the rest, with ∑kδk<ε\sum_k\delta_k<\varepsilon∑k​δk​<ε chosen so the endpoint bbb stays covered without raising total length by more than ε\varepsilonε. Thus μ∗(A)≥μ∗(A∩(b,∞))+μ∗(A∖(b,∞))−ε\om(A)\ge\om(A\cap(b,\infty))+\om(A\setminus(b,\infty))-\varepsilonμ∗(A)≥μ∗(A∩(b,∞))+μ∗(A∖(b,∞))−ε, and ε→0\varepsilon\to0ε→0 gives the defining inequality. Half-lines generate the Borel sigma-algebra, so every Borel set is measurable. For μ∗((a,b))=b−a\om((a,b))=b-aμ∗((a,b))=b−a the cover {(a−ε,b+ε)}\{(a-\varepsilon,b+\varepsilon)\}{(a−ε,b+ε)} gives μ∗((a,b))≤b−a\om((a,b))\le b-aμ∗((a,b))≤b−a. For the reverse, fix ε>0\varepsilon>0ε>0 and any open cover {Ik}\{I_k\}{Ik​} of (a,b)(a,b)(a,b). The compact subinterval [a+ε,b−ε][a+\varepsilon, b-\varepsilon][a+ε,b−ε] has a finite subcover by the Heine-Borel theorem, and any finite open cover of an interval of length LLL has total length exceeding LLL, by induction on the cover size starting from an interval that contains the left endpoint and extends past it. Hence ∑k∣Ik∣≥(b−a)−2ε\sum_k\abs{I_k}\ge(b-a)-2\varepsilon∑k​∣Ik​∣≥(b−a)−2ε, and taking the infimum over covers and then ε→0\varepsilon\to0ε→0 gives μ∗((a,b))≥b−a\om((a,b))\ge b-aμ∗((a,b))≥b−a. A singleton has μ∗({b})=0\om(\{b\})=0μ∗({b})=0 by the cover (b−ε,b+ε)(b-\varepsilon,b+\varepsilon)(b−ε,b+ε), so subadditivity and monotonicity give μ∗([a,b])=μ∗([a,b))=μ∗((a,b])=b−a\om([a,b])=\om([a,b))=\om((a,b])=b-aμ∗([a,b])=μ∗([a,b))=μ∗((a,b])=b−a as well, and every bounded interval shares the common length. Translation invariance follows because translating an interval cover preserves its total length. Uniqueness holds because two Borel measures μ,ν\mu,\nuμ,ν agreeing with length on the intervals agree on the algebra of finite interval unions, a generating system closed under intersection, so they agree on every Borel set contained in some (−n,n)(-n,n)(−n,n), the sets of the generating system that have finite measure and increase to R\RR. For a general Borel BBB the sets B∩(−n,n)B\cap(-n,n)B∩(−n,n) increase to BBB with finite measure, so continuity from below (Proposition 3, increasing case) gives μ(B)=lim⁡nμ(B∩(−n,n))=lim⁡nν(B∩(−n,n))=ν(B)\mu(B)=\lim_n\mu(B\cap(-n,n))=\lim_n \nu(B\cap(-n,n))=\nu(B)μ(B)=limn​μ(B∩(−n,n))=limn​ν(B∩(−n,n))=ν(B), agreement on all Borel sets. The exhaustion of R\RR by the finite-measure intervals (−n,n)(-n,n)(−n,n) is what carries finite-measure agreement to the whole sigma-algebra.

With Lebesgue measure in hand the Lebesgue integral is built by integrating simple functions and passing to limits, the program the measures and integration post carries out on the structure produced here. Every measure that later posts use, the probability measures of the probability foundations and the Wiener measure of Brownian motion, is a Caratheodory extension of a primitive size.

[1]
G. B. Folland, Real Analysis: Modern Techniques and Their Applications, 2nd ed. Wiley, 1999.

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cite
@misc{sigma-algebras-and-measures,
  author = {Zac Kienzle},
  title  = {Sigma-Algebras and Measures},
  year   = {2026},
  month  = {05},
  url    = {https://zackienzle.com/blog/sigma-algebras-and-measures}
}