Proof of Carathéodory's Theorem: the Measurable Sets of an Outer Measure Form a σ-Algebra on Which It Is a Measure

Throughout, $X$, $\mathcal{P}(X)$, $\mu^{*}$ and $\mathcal{M}$ are as in the statement; $\mu^{*}(\emptyset)=0$, monotonicity and countable subadditivity are those of Outer Measure (i), and $\emptyset\in\mathcal{P}(X)$ as noted there. A set lies in $\mathcal{P}(X)$ exactly when it is a subset of $X$ (The Union Set and the Power Set of a Set (ii)). Subclasses of sets are sets (Subclasses of Sets Are Sets, the Union and Power Set of a Set Exist Uniquely, Binary Unions of Sets Are Sets, and the Universal Class Is Proper (i)) and the union of two sets is a set (Subclasses of Sets Are Sets, the Union and Power Set of a Set Exist Uniquely, Binary Unions of Sets Are Sets, and the Universal Class Is Proper (iv)), so every subclass of $X$, and the intersection, difference and union of two elements of $\mathcal{P}(X)$, lie in $\mathcal{P}(X)$. Identities between such sets are checked elementwise from The Boolean Operations on Classes, Disjointness, and the Universal Class (i). By Carathéodory Measurable Sets of an Outer Measure (ii) and Carathéodory Measurable Sets of an Outer Measure (i), $E\in\mathcal{M}$ if and only if $E\in\mathcal{P}(X)$ and $\mu^{*}(A)=\mu^{*}(A\cap E)+\mu^{*}(A\setminus E)$ for every $A\in\mathcal{P}(X)$; and $\mathcal{M}\subseteq\mathcal{P}(X)$, so a sequence in $\mathcal{M}$ is a sequence in $\mathcal{P}(X)$ (Functions, Values of a Function, and Functions from One Class to Another (iii)).

In $[0,\infty]$, $\le$ is a total order, in particular reflexive, antisymmetric and transitive (Order and Suprema in the Extended Real Line (i)), with greatest element $\infty$ (Order and Suprema in the Extended Real Line (ii)); and $+$ is associative and commutative with neutral element $0$ (Arithmetic in [0,∞]: Monoid Laws, Distributivity, Monotonicity and Cancellation (i)). (O) If $a,b,c,d\in[0,\infty]$, $a\le b$ and $c\le d$, then $a+c\le b+d$: indeed $a+c\le b+c$ and $c+b\le d+b$ by Arithmetic in [0,∞]: Monoid Laws, Distributivity, Monotonicity and Cancellation (iv), and $b+c=c+b$, $d+b=b+d$.

Step 1 (subadditivity for two sets). Let $U,V\in\mathcal{P}(X)$; we show $\mu^{*}(U\cup V)\le\mu^{*}(U)+\mu^{*}(V)$. By Sequences and Maps of Sets: Complements, Traces, Two-Term Sequences, the Disjointing Sequence, and Sequences in a Subset (iii) there is a sequence $(E_{n})$ in $\mathcal{P}(X)$ with $E_{1}=U$, $E_{2}=V$ and $E_{n}=\emptyset$ for every $n\in\mathbb{N}$ with $n\neq1$ and $n\neq2$. Its union is $U\cup V$: an element of some $E_{n}$ shows $E_{n}\neq\emptyset$, so $n=1$ or $n=2$; and $U=E_{1}$, $V=E_{2}$. By countable subadditivity, applied to $(E_{n})$, $\mu^{*}(U\cup V)\le\sum_{n\in\mathbb{N}}\mu^{*}(E_{n})$, where $n\mapsto\mu^{*}(E_{n})$ is the composite of $(E_{n})$ with $\mu^{*}$, a sequence in $[0,\infty]$ (Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (ii)). Since $2=1+1$ by Arithmetic and Order of the Natural Numbers (xiv), $1<2$ by Arithmetic and Order of the Natural Numbers (xi). If $k\in\mathbb{N}$ and $2<k$, then $k\neq2$ by Arithmetic and Order of the Natural Numbers (vii), and $1<k$ by Arithmetic and Order of the Natural Numbers (vi), so $k\neq1$; thus $\mu^{*}(E_{k})=\mu^{*}(\emptyset)=0$. By Sums of Sequences in [0,∞]: Real Series, Linearity, Comparison, Rearrangement, Double Series and Monotone Convergence (vii) with $n=2$, the partial sum $\sum_{k=1}^{2}\mu^{*}(E_{k})$ being also the sum over the interval $\{1,\dots,2\}$ (Sums and Products over a Finite Set and over an Interval (iv)), and by Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals (vii) with $m=n=1$ and Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals (i), as $\{1,\dots,1\}=[1]=\{1\}$ (Intervals of Natural Numbers: Initial Segments, Adding One Element, Splitting and Shifting (i)) and $+$ is associative and commutative with neutral element $0$ (Arithmetic in [0,∞]: Monoid Laws, Distributivity, Monotonicity and Cancellation (i)),

$$\sum_{n\in\mathbb{N}}\mu^{*}(E_{n})=\sum_{k=1}^{2}\mu^{*}(E_{k})=\mu^{*}(E_{1})+\mu^{*}(E_{2})=\mu^{*}(U)+\mu^{*}(V).$$

Step 2 (one inequality suffices). Let $E,A\in\mathcal{P}(X)$. As $A=(A\cap E)\cup(A\setminus E)$, Step 1 gives $\mu^{*}(A)\le\mu^{*}(A\cap E)+\mu^{*}(A\setminus E)$. By antisymmetry of $\le$, therefore:

(2) for $E\in\mathcal{P}(X)$, $E\in\mathcal{M}$ if and only if $\mu^{*}(A\cap E)+\mu^{*}(A\setminus E)\le\mu^{*}(A)$ for every $A\in\mathcal{P}(X)$.

Proof of (i). Let $E$ be a subset of $X$, so $E\in\mathcal{P}(X)$. If $E\in\mathcal{M}$, then $\mu^{*}(A\cap E)+\mu^{*}(A\setminus E)=\mu^{*}(A)$ for every $A\in\mathcal{P}(X)$, and the inequality holds by reflexivity, in particular when $\mu^{*}(A)<\infty$. Conversely, assume the inequality for every $A\in\mathcal{P}(X)$ with $\mu^{*}(A)<\infty$, and let $A\in\mathcal{P}(X)$. If $\mu^{*}(A)<\infty$, the inequality holds by assumption. Otherwise, as $\mu^{*}(A)\le\infty$ and $x<y$ means $x\le y$ and $x\neq y$ (The Extended Real Line, Its Order, and [0,∞] (iii)), $\mu^{*}(A)=\infty$, and $\mu^{*}(A\cap E)+\mu^{*}(A\setminus E)\le\infty=\mu^{*}(A)$ because $\infty$ is the greatest element. So the inequality holds for every $A\in\mathcal{P}(X)$, and $E\in\mathcal{M}$ by (2).

Step 3 (the whole set, complements, unions and intersections of two sets). (a) $X\in\mathcal{M}$: $X\in\mathcal{P}(X)$, and for $A\in\mathcal{P}(X)$, $A\cap X=A$ and $A\setminus X=\emptyset$, so $\mu^{*}(A\cap X)+\mu^{*}(A\setminus X)=\mu^{*}(A)+0=\mu^{*}(A)$.

(b) If $E\in\mathcal{M}$, then $X\setminus E\in\mathcal{M}$: $X\setminus E\in\mathcal{P}(X)$, and for $A\in\mathcal{P}(X)$, as $A\subseteq X$, $A\cap(X\setminus E)=A\setminus E$ and $A\setminus(X\setminus E)=A\cap E$, so by commutativity $\mu^{*}(A\cap(X\setminus E))+\mu^{*}(A\setminus(X\setminus E))=\mu^{*}(A\setminus E)+\mu^{*}(A\cap E)=\mu^{*}(A\cap E)+\mu^{*}(A\setminus E)=\mu^{*}(A)$.

(c) If $E,F\in\mathcal{M}$, then $E\cup F\in\mathcal{M}$. Let $A\in\mathcal{P}(X)$. As $E\in\mathcal{M}$, $\mu^{*}(A)=\mu^{*}(A\cap E)+\mu^{*}(A\setminus E)$; as $F\in\mathcal{M}$ and $A\setminus E\in\mathcal{P}(X)$, $\mu^{*}(A\setminus E)=\mu^{*}((A\setminus E)\cap F)+\mu^{*}((A\setminus E)\setminus F)$. Now $(A\setminus E)\setminus F=A\setminus(E\cup F)$ and $(A\cap E)\cup((A\setminus E)\cap F)=A\cap(E\cup F)$, elementwise by The Boolean Operations on Classes, Disjointness, and the Universal Class (i), so Step 1 gives $\mu^{*}(A\cap(E\cup F))\le\mu^{*}(A\cap E)+\mu^{*}((A\setminus E)\cap F)$. By (O) and associativity,

$$\mu^{*}(A\cap(E\cup F))+\mu^{*}(A\setminus(E\cup F))\le\mu^{*}(A\cap E)+\bigl(\mu^{*}((A\setminus E)\cap F)+\mu^{*}((A\setminus E)\setminus F)\bigr)=\mu^{*}(A\cap E)+\mu^{*}(A\setminus E)=\mu^{*}(A).$$

As $E\cup F\in\mathcal{P}(X)$, $E\cup F\in\mathcal{M}$ by (2).

(d) If $E,F\in\mathcal{M}$, then $E\cap F\in\mathcal{M}$: as $E,F\subseteq X$, $E\cap F=X\setminus\bigl((X\setminus E)\cup(X\setminus F)\bigr)$, which lies in $\mathcal{M}$ by (b), (c) and (b).

Step 4 (splitting off a measurable set). Let $E\in\mathcal{M}$ and $F\in\mathcal{P}(X)$ with $E\cap F=\emptyset$, and let $A\in\mathcal{P}(X)$. Then $\mu^{*}(A\cap(E\cup F))=\mu^{*}(A\cap E)+\mu^{*}(A\cap F)$. Indeed, $E\in\mathcal{M}$ applied to the set $A\cap(E\cup F)\in\mathcal{P}(X)$ gives the claim, since $(A\cap(E\cup F))\cap E=A\cap E$, and $(A\cap(E\cup F))\setminus E=A\cap F$ because no set is an element of both $E$ and $F$ (Monotone Sequences of Sets, Unions of Traces, and Disjointness of Smaller Sets (iv)).

Step 5 (finite unions of a disjoint sequence). Let $(E_{m})$ be a sequence in $\mathcal{M}$ that is pairwise disjoint: $E_{m}\cap E_{k}=\emptyset$ for all $m,k\in\mathbb{N}$ with $m\neq k$. For $n\in\mathbb{N}$, $[n]=\{k\in\mathbb{N}:k\le n\}\subseteq\mathbb{N}$ (Intervals of Natural Numbers: Initial Segments, Adding One Element, Splitting and Shifting (i)), so the restriction of $(E_{m})$ to $[n]$ is a function with domain $[n]$ and the same values (Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (ix)), a family $(E_{i})_{i\in[n]}$ indexed by $[n]$. Let $G_{n}=\bigcup_{i\in[n]}E_{i}$ be its union (Indexed Families of Sets and Their Union, Intersection and Product (ii)), a set by The Union and Product of a Family of Sets Indexed by a Set, and the Intersection of a Family with an Inhabited Index Class, Are Sets (i): $x\in G_{n}$ if and only if $x\in E_{i}$ for some $i\in[n]$, so $G_{n}\subseteq X$ and $G_{n}\in\mathcal{P}(X)$. The expression $t(n)=\bigcup_{i\in[n]}E_{i}$, with the sequence $(E_{m})$, $\mathbb{N}$ and the order $\le$ of $\mathbb{N}_{0}$ as parameters ($i\in[n]$ meaning $i\in\mathbb{N}\wedge i\le n$), is used properly and lies in $\mathcal{P}(X)$ for every $n\in\mathbb{N}$; the condition $z=t(n)$ is $\forall x\,(x\in z\Leftrightarrow\exists i\,(i\in[n]\wedge x\in E_{i}))$, written with abbreviations and with no quantifier applied to a class variable, hence predicative by Predicativity of Expressions with Abbreviations, and Symbol-Free Forms of the Basic Defined Set Symbols (i). So Maps and Relations Given by Formulas (i), applied to the sets $\mathbb{N}$ and $\mathcal{P}(X)$ and to $t$, gives exactly one map $n\mapsto G_{n}$ from $\mathbb{N}$ to $\mathcal{P}(X)$ with value $t(n)$ at each $n$, and $(G_{n})$ is a sequence in $\mathcal{P}(X)$. As $[1]=\{1\}$ (Intervals of Natural Numbers: Initial Segments, Adding One Element, Splitting and Shifting (i)), $G_{1}=E_{1}$. For $n\in\mathbb{N}$, $[n+1]=[n]\cup\{n+1\}$ (Intervals of Natural Numbers: Initial Segments, Adding One Element, Splitting and Shifting (ii)), so $x\in G_{n+1}$ if and only if $x\in G_{n}$ or $x\in E_{n+1}$; that is, $G_{n+1}=G_{n}\cup E_{n+1}$. Moreover no set is an element of both $E_{n+1}$ and $G_{n}$: if $x\in G_{n}$, then $x\in E_{i}$ for some $i\in[n]$, and $i\neq n+1$ as $n+1\notin[n]$ (Intervals of Natural Numbers: Initial Segments, Adding One Element, Splitting and Shifting (ii)), so $E_{i}\cap E_{n+1}=\emptyset$ and $x\notin E_{n+1}$ (Monotone Sequences of Sets, Unions of Traces, and Disjointness of Smaller Sets (iv)). Hence $E_{n+1}\cap G_{n}=\emptyset$ (Monotone Sequences of Sets, Unions of Traces, and Disjointness of Smaller Sets (iv)).

Fix $A\in\mathcal{P}(X)$. By Sequences and Maps of Sets: Complements, Traces, Two-Term Sequences, the Disjointing Sequence, and Sequences in a Subset (ii), with $Y=A$, there is a sequence $(T_{m})$ in $\mathcal{P}(A)$ with $T_{m}=E_{m}\cap A=A\cap E_{m}$; every subset of $A$ is a subset of $X$, so $\mathcal{P}(A)\subseteq\mathcal{P}(X)$ and $(T_{m})$ is a sequence in $\mathcal{P}(X)$ (Functions, Values of a Function, and Functions from One Class to Another (iii)). Let $a_{m}=\mu^{*}(A\cap E_{m})$; $(a_{m})$ is a sequence in $[0,\infty]$, the composite of $(T_{m})$ with $\mu^{*}$ (Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (ii)). Let $s_{n}=\sum_{i=1}^{n}a_{i}$ be its partial sums, a sequence $(s_{n})$ in $[0,\infty]$. As $+$ is associative and commutative with neutral element $0$ (Arithmetic in [0,∞]: Monoid Laws, Distributivity, Monotonicity and Cancellation (i)), $s_{n}$ is also the sum of $(a_{i})$ over the interval $\{1,\dots,n\}$ (Sums and Products over a Finite Set and over an Interval (iv)). So $s_{1}=a_{1}$ by Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals (i), as $\{1,\dots,1\}=[1]=\{1\}$ (Intervals of Natural Numbers: Initial Segments, Adding One Element, Splitting and Shifting (i)); and $s_{n+1}=s_{n}+a_{n+1}$ for every $n\in\mathbb{N}$ by Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals (vii) with $m=1$.

We show by Induction for Predicative Formulas, and the Order of ℕ below 1 and below n+1 (i) that, for every $n\in\mathbb{N}$, $G_{n}\in\mathcal{M}$ and $\mu^{*}(A\cap G_{n})=s_{n}$. (This condition on $n$, with the guard $n\in\mathbb{N}$, is expressed by a predicative formula: $n$ is a term of type $\mathbb{N}$ by Typed Terms under a Predicative Guard: Variables, Parameters, Values of Maps and Ordered Pairs, and Predicative Equations and Memberships between Them (i); $G_{n}$ and $s_{n}$, the values of the sequences $(G_{n})$ and $(s_{n})$ at $n$, are terms of type $\mathcal{P}(X)$ and $[0,\infty]$ by Typed Terms under a Predicative Guard: Variables, Parameters, Values of Maps and Ordered Pairs, and Predicative Equations and Memberships between Them (iii); $A\cap G_{n}$ is a term of type $\mathcal{P}(X)$, by the predicative form of $z=u\cap v$ in Predicativity of Expressions with Abbreviations, and Symbol-Free Forms of the Basic Defined Set Symbols (ix) with $G_{n}$ eliminated by Predicativity of Expressions with Abbreviations, and Symbol-Free Forms of the Basic Defined Set Symbols (ii), and $\mu^{*}(A\cap G_{n})$ a term of type $[0,\infty]$ by Typed Terms under a Predicative Guard: Variables, Parameters, Values of Maps and Ordered Pairs, and Predicative Equations and Memberships between Them (iii); and $G_{n}\in\mathcal{M}$ and $\mu^{*}(A\cap G_{n})=s_{n}$ are expressed by predicative formulas by Typed Terms under a Predicative Guard: Variables, Parameters, Values of Maps and Ordered Pairs, and Predicative Equations and Memberships between Them (vi) and Typed Terms under a Predicative Guard: Variables, Parameters, Values of Maps and Ordered Pairs, and Predicative Equations and Memberships between Them (v), whose conjunction expresses the condition.) For $n=1$: $G_{1}=E_{1}\in\mathcal{M}$ and $\mu^{*}(A\cap G_{1})=a_{1}=s_{1}$. If the claim holds for $n$, then $G_{n+1}=G_{n}\cup E_{n+1}\in\mathcal{M}$ by Step 3(c), and Step 4, applied to $E_{n+1}\in\mathcal{M}$ and $G_{n}$, gives

$$\mu^{*}(A\cap G_{n+1})=\mu^{*}(A\cap(E_{n+1}\cup G_{n}))=\mu^{*}(A\cap E_{n+1})+\mu^{*}(A\cap G_{n})=a_{n+1}+s_{n}=s_{n}+a_{n+1}=s_{n+1}.$$

Step 6 (countable unions of a disjoint sequence). Let $(E_{m})$, $(G_{n})$, $A$, $(T_{m})$, $(a_{m})$ and $(s_{n})$ be as in Step 5, and let $E=\bigcup_{m\in\mathbb{N}}E_{m}$, the union of $(E_{m})$ as a family indexed by $\mathbb{N}$, a set by The Union and Product of a Family of Sets Indexed by a Set, and the Intersection of a Family with an Inhabited Index Class, Are Sets (i) and a subset of $X$ by Indexed Families of Sets and Their Union, Intersection and Product (ii), so that $E\in\mathcal{P}(X)$. We show

(6) $\sum_{m\in\mathbb{N}}a_{m}+\mu^{*}(A\setminus E)\le\mu^{*}(A)$ and $\mu^{*}(A\cap E)\le\sum_{m\in\mathbb{N}}a_{m}$.

Let $n\in\mathbb{N}$. By Indexed Families of Sets and Their Union, Intersection and Product (ii), $G_{n}\subseteq E$, so $A\setminus E\subseteq A\setminus G_{n}$ and $\mu^{*}(A\setminus E)\le\mu^{*}(A\setminus G_{n})$ by monotonicity. As $G_{n}\in\mathcal{M}$, Step 5 and (O) give

$$\mu^{*}(A)=\mu^{*}(A\cap G_{n})+\mu^{*}(A\setminus G_{n})=s_{n}+\mu^{*}(A\setminus G_{n})\ge s_{n}+\mu^{*}(A\setminus E).$$

Let $c=\mu^{*}(A\setminus E)$ and $S=\{s_{n}:n\in\mathbb{N}\}$, the range of $(s_{n})$; as $(s_{n})$ is a sequence in $[0,\infty]$, $S$ is a subclass of $[0,\infty]$, hence a set by Subclasses of Sets Are Sets, the Union and Power Set of a Set Exist Uniquely, Binary Unions of Sets Are Sets, and the Universal Class Is Proper (i) and a subset of $[0,\infty]$, with $s_{1}\in S$. The map $n\mapsto s_{n}+c$ on $\mathbb{N}$ is given by an expression, as in Routine Facts Used Without Citation: Arithmetic, Maps, Predicativity and Indices (iv), and is a sequence in $[0,\infty]$; its range $\{s_{n}+c:n\in\mathbb{N}\}$ is a subclass of $[0,\infty]$, hence a set by Subclasses of Sets Are Sets, the Union and Power Set of a Set Exist Uniquely, Binary Unions of Sets Are Sets, and the Universal Class Is Proper (i), and it equals $\{x+c:x\in S\}$, the elements of $S$ being the $s_{n}$. Thus $\mu^{*}(A)$ is an upper bound of $\{x+c:x\in S\}=\{s_{n}+c:n\in\mathbb{N}\}$, whose supremum exists by Order and Suprema in the Extended Real Line (iii) and, being the least upper bound (Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iv)), is at most $\mu^{*}(A)$. By Suprema and Finite Sums in [0,∞] (i) and Sum of a Sequence in [0,∞] (ii), that supremum is $\sup S+c=\sum_{m\in\mathbb{N}}a_{m}+\mu^{*}(A\setminus E)$, which proves the first inequality. By Monotone Sequences of Sets, Unions of Traces, and Disjointness of Smaller Sets (iii), applied to $(E_{m})$ as a sequence in $\mathcal{P}(X)$ with $Y=A$, $\bigcup_{m\in\mathbb{N}}T_{m}=E\cap A=A\cap E$; countable subadditivity, applied to $(T_{m})$, gives $\mu^{*}(A\cap E)\le\sum_{m\in\mathbb{N}}\mu^{*}(T_{m})=\sum_{m\in\mathbb{N}}a_{m}$, the second inequality.

By (6), (O) and transitivity, $\mu^{*}(A\cap E)+\mu^{*}(A\setminus E)\le\sum_{m\in\mathbb{N}}a_{m}+\mu^{*}(A\setminus E)\le\mu^{*}(A)$. As $A\in\mathcal{P}(X)$ was arbitrary, $E\in\mathcal{M}$ by (2).

Step 7 (countable unions). Let $(D_{n})$ be a sequence in $\mathcal{M}$, hence in $\mathcal{P}(X)$. By Sequences and Maps of Sets: Complements, Traces, Two-Term Sequences, the Disjointing Sequence, and Sequences in a Subset (iv) there is exactly one sequence $(C_{n})$ in $\mathcal{P}(X)$ with $C_{1}=X$ and $C_{n+1}=C_{n}\cap(X\setminus D_{n})$ for every $n\in\mathbb{N}$, and exactly one sequence $(B_{n})$ in $\mathcal{P}(X)$ with $B_{n}=D_{n}\cap C_{n}$ for every $n\in\mathbb{N}$. By Induction for Predicative Formulas, and the Order of ℕ below 1 and below n+1 (i), $C_{n}\in\mathcal{M}$ for every $n\in\mathbb{N}$ (this condition on $n$, with the guard $n\in\mathbb{N}$, is expressed by a predicative formula: $C_{n}$, the value of the sequence $(C_{n})$ at the term $n$ of type $\mathbb{N}$, is a term of type $\mathcal{P}(X)$ by Typed Terms under a Predicative Guard: Variables, Parameters, Values of Maps and Ordered Pairs, and Predicative Equations and Memberships between Them (i) and Typed Terms under a Predicative Guard: Variables, Parameters, Values of Maps and Ordered Pairs, and Predicative Equations and Memberships between Them (iii), and $C_{n}\in\mathcal{M}$ is expressed by a predicative formula by Typed Terms under a Predicative Guard: Variables, Parameters, Values of Maps and Ordered Pairs, and Predicative Equations and Memberships between Them (vi)): $C_{1}=X\in\mathcal{M}$ by Step 3(a), and if $C_{n}\in\mathcal{M}$, then $X\setminus D_{n}\in\mathcal{M}$ by Step 3(b) and $C_{n+1}\in\mathcal{M}$ by Step 3(d). Hence $B_{n}=D_{n}\cap C_{n}\in\mathcal{M}$ for every $n$ by Step 3(d), and $(B_{n})$ is a sequence in $\mathcal{M}$ by Sequences and Maps of Sets: Complements, Traces, Two-Term Sequences, the Disjointing Sequence, and Sequences in a Subset (v), $\mathcal{M}$ being a subset of $\mathcal{P}(X)$. It is pairwise disjoint, $B_{m}\cap B_{k}=\emptyset$ for $m\neq k$, by The Disjointing Sequence of a Sequence of Sets (iii), and $\bigcup_{n\in\mathbb{N}}B_{n}=\bigcup_{n\in\mathbb{N}}D_{n}$ by The Disjointing Sequence of a Sequence of Sets (iv). By Step 6, $\bigcup_{n\in\mathbb{N}}D_{n}=\bigcup_{n\in\mathbb{N}}B_{n}\in\mathcal{M}$.

Proof of (ii). $\mathcal{M}$ is a set with $\mathcal{M}\subseteq\mathcal{P}(X)$ (Carathéodory Measurable Sets of an Outer Measure (ii)); $X\in\mathcal{M}$ by Step 3(a); $X\setminus E\in\mathcal{M}$ for every $E\in\mathcal{M}$ by Step 3(b); and $\bigcup_{n\in\mathbb{N}}D_{n}\in\mathcal{M}$ for every sequence $(D_{n})$ in $\mathcal{M}$ by Step 7. So $\mathcal{M}$ is a $\sigma$-algebra on $X$ by σ-Algebra and Measurable Space (i), and $(X,\mathcal{M})$ is a measurable space by σ-Algebra and Measurable Space (ii).

Proof of (iii). Let $N$ be a subset of $X$ with $\mu^{*}(N)=0$, so $N\in\mathcal{P}(X)$, and let $A\in\mathcal{P}(X)$. As $A\cap N\subseteq N$, monotonicity gives $\mu^{*}(A\cap N)\le\mu^{*}(N)=0$, so $\mu^{*}(A\cap N)=0$ by Small Facts about [0,∞]: Zero, Finiteness and Sums of Zeros (i). As $A\setminus N\subseteq A$, $\mu^{*}(A\setminus N)\le\mu^{*}(A)$. Hence $\mu^{*}(A\cap N)+\mu^{*}(A\setminus N)=0+\mu^{*}(A\setminus N)=\mu^{*}(A\setminus N)\le\mu^{*}(A)$, and $N\in\mathcal{M}$ by (2).

Proof of (iv). Let $\nu=\mu^{*}|_{\mathcal{M}}$, the map $\nu:\mathcal{M}\to[0,\infty]$ with $\nu(E)=\mu^{*}(E)$ for $E\in\mathcal{M}$, as stated in (iv). By (ii) and Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (vi), $\emptyset\in\mathcal{M}$, and $\nu(\emptyset)=\mu^{*}(\emptyset)=0$. Let $(E_{m})$ be a pairwise disjoint sequence in $\mathcal{M}$ and $E=\bigcup_{m\in\mathbb{N}}E_{m}$. Apply (6) of Step 6 with $A=E$: by Indexed Families of Sets and Their Union, Intersection and Product (ii), $E_{m}\subseteq E$, so $A\cap E_{m}=E_{m}$ and $a_{m}=\mu^{*}(E_{m})$ for every $m\in\mathbb{N}$, and $A\setminus E=\emptyset$. Thus (6) gives $\sum_{m\in\mathbb{N}}\mu^{*}(E_{m})=\sum_{m\in\mathbb{N}}\mu^{*}(E_{m})+0\le\mu^{*}(E)$, while countable subadditivity, applied to $(E_{m})$ as a sequence in $\mathcal{P}(X)$, gives $\mu^{*}(E)\le\sum_{m\in\mathbb{N}}\mu^{*}(E_{m})$, the sequence $m\mapsto\mu^{*}(E_{m})$ being the composite of $(E_{m})$ with $\mu^{*}$ (Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (ii)). By antisymmetry, $\mu^{*}(E)=\sum_{m\in\mathbb{N}}\mu^{*}(E_{m})$. As $\nu(E)=\mu^{*}(E)$ and $\nu(E_{m})=\mu^{*}(E_{m})$ for every $m$, the sequences $m\mapsto\nu(E_{m})$ and $m\mapsto\mu^{*}(E_{m})$ coincide and

$$\nu\Bigl(\bigcup_{m\in\mathbb{N}}E_{m}\Bigr)=\sum_{m\in\mathbb{N}}\nu(E_{m}).$$

So $\nu$ is a measure on $(X,\mathcal{M})$ by Measure, Measure Space, and Probability Measure (i).