Proof of Uniqueness of Measures Agreeing on a Generating π-System

Throughout, $\mathcal{F}\subseteq\mathcal{P}(X)$ by σ-Algebra and Measurable Space (i), and $\mathcal{E}\subseteq\sigma(\mathcal{E})=\mathcal{F}$ by Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (iii). The union $\bigcup_{n\in\mathbb{N}}A_{n}$ of a sequence $(A_{n})$ is the union of the family, whose elements are the $x$ with $x\in A_{n}$ for some $n\in\mathbb{N}$.

Agreement on traces. Let $G\in\mathcal{F}$ with $\mu(G)=\nu(G)$ and $\mu(G)<\infty$, and with $\mu(E\cap G)=\nu(E\cap G)$ for every $E\in\mathcal{E}$. We show that $\mu(A\cap G)=\nu(A\cap G)$ for every $A\in\mathcal{F}$. Note that $A\cap G\in\mathcal{F}$ for every $A\in\mathcal{F}$, by Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (ix). Let $\mathcal{L}_{G}=\{A\in\mathcal{F}:\mu(A\cap G)=\nu(A\cap G)\}$, a set by Typed Terms under a Predicative Guard: Variables, Parameters, Values of Maps and Ordered Pairs, and Predicative Equations and Memberships between Them (vii), as $\mu(A\cap G)$ and $\nu(A\cap G)$ are the values at $A$ of the maps $\mu\circ g_{G}$ and $\nu\circ g_{G}$ into $[0,\infty]$ (Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (ii)), with $g_{G}$ the trace map $A\mapsto A\cap G$ of Sequences and Maps of Sets: Complements, Traces, Two-Term Sequences, the Disjointing Sequence, and Sequences in a Subset (ii), a map $\mathcal{F}\to\mathcal{F}$ by Functions, Values of a Function, and Functions from One Class to Another (iii) as its values lie in $\mathcal{F}$, so typed terms 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); by that clause it is a subset of $\mathcal{F}$, hence of $\mathcal{P}(X)$.

First, $X\in\mathcal{L}_{G}$: $X\in\mathcal{F}$ by σ-Algebra and Measurable Space (i), and $X\cap G=G$ as $G\subseteq X$, while $\mu(G)=\nu(G)$.

Next, let $A\in\mathcal{L}_{G}$. Then $X\setminus A\in\mathcal{F}$ by σ-Algebra and Measurable Space (i), and $(X\setminus A)\cap G=G\setminus(A\cap G)$ as $G\subseteq X$, where $A\cap G\in\mathcal{F}$ and $A\cap G\subseteq G$. As $\mu(G)<\infty$ and $\nu(G)=\mu(G)$, also $\nu(G)<\infty$. As $(X,\mathcal{F},\mu)$ and $(X,\mathcal{F},\nu)$ are measure spaces by Measure, Measure Space, and Probability Measure (ii), Basic Properties of a Measure (iv) applies to $\mu$ and to $\nu$: the values $\mu(G)$, $\nu(G)$, $\mu(A\cap G)$ and $\nu(A\cap G)$ are real, and

$$\mu\bigl(G\setminus(A\cap G)\bigr)=\mu(G)-\mu(A\cap G)=\nu(G)-\nu(A\cap G)=\nu\bigl(G\setminus(A\cap G)\bigr).$$

Hence $X\setminus A\in\mathcal{L}_{G}$.

Last, let $(A_{n})$ be a sequence in $\mathcal{L}_{G}$ whose terms are pairwise disjoint, and let $U=\bigcup_{n\in\mathbb{N}}A_{n}$. As $\mathcal{L}_{G}\subseteq\mathcal{F}$, $(A_{n})$ is a sequence in $\mathcal{F}$ and $U\in\mathcal{F}$ by σ-Algebra and Measurable Space (i). By Sequences and Maps of Sets: Complements, Traces, Two-Term Sequences, the Disjointing Sequence, and Sequences in a Subset (ii) with $Y=G$, the map $n\mapsto A_{n}\cap G$ is a sequence in $\mathcal{P}(G)$, hence in $\mathcal{P}(X)$ as $G\subseteq X$, and as each $A_{n}\cap G$ lies in $\mathcal{F}$ it is a sequence in $\mathcal{F}$ by Sequences and Maps of Sets: Complements, Traces, Two-Term Sequences, the Disjointing Sequence, and Sequences in a Subset (v). Its terms are pairwise disjoint in the sense of Measure, Measure Space, and Probability Measure (i): for $m\neq n$, the disjoint sets $A_{m}$ and $A_{n}$ satisfy $A_{m}\cap A_{n}=\emptyset$ (Monotone Sequences of Sets, Unions of Traces, and Disjointness of Smaller Sets (iv)), and as $A_{n}\cap G\subseteq A_{n}$ for every $n$, $(A_{m}\cap G)\cap(A_{n}\cap G)=\emptyset$ (Monotone Sequences of Sets, Unions of Traces, and Disjointness of Smaller Sets (v)). Its union is $U\cap G$ (Monotone Sequences of Sets, Unions of Traces, and Disjointness of Smaller Sets (iii)). By the countable additivity of Measure, Measure Space, and Probability Measure (i), applied to $\mu$ and to $\nu$,

$$\mu(U\cap G)=\sum_{n\in\mathbb{N}}\mu(A_{n}\cap G),\qquad \nu(U\cap G)=\sum_{n\in\mathbb{N}}\nu(A_{n}\cap G).$$

The sequences $n\mapsto\mu(A_{n}\cap G)$ and $n\mapsto\nu(A_{n}\cap G)$ in $[0,\infty]$ have equal terms, as every $A_{n}$ lies in $\mathcal{L}_{G}$; so they are the same sequence, and their sums coincide. Hence $\mu(U\cap G)=\nu(U\cap G)$, and $U\in\mathcal{L}_{G}$.

By λ-System (i), $\mathcal{L}_{G}$ is therefore a $\lambda$-system on $X$. It contains $\mathcal{E}$, as every $E\in\mathcal{E}$ lies in $\mathcal{F}$ and satisfies $\mu(E\cap G)=\nu(E\cap G)$ by assumption. By Dynkin's Pi-Lambda Theorem (ii), $\mathcal{F}=\sigma(\mathcal{E})\subseteq\mathcal{L}_{G}$, that is, $\mu(A\cap G)=\nu(A\cap G)$ for every $A\in\mathcal{F}$.

Proof of (i). Take $G=X$ in the paragraphs on agreement on traces: $X\in\mathcal{F}$, $\mu(X)=\nu(X)$ and $\mu(X)<\infty$ by assumption, and for $E\in\mathcal{E}$ we have $E\cap X=E$, as $E\subseteq X$, so $\mu(E\cap X)=\nu(E\cap X)$ by the hypothesis on $\mathcal{E}$. Hence, for every $A\in\mathcal{F}$, $\mu(A)=\mu(A\cap X)=\nu(A\cap X)=\nu(A)$, as $A\subseteq X$. Since $\mu$ and $\nu$ are maps on $\mathcal{F}$ with the same values, $\mu=\nu$ (Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (iv)).

Proof of (ii). Let $(E_{m})$ be a sequence in $\mathcal{E}$ with $\bigcup_{m\in\mathbb{N}}E_{m}=X$ and $\mu(E_{m})<\infty$ for every $m\in\mathbb{N}$. Fix $n\in\mathbb{N}$ and take $G=E_{n}$ in the paragraphs on agreement on traces: $E_{n}\in\mathcal{E}\subseteq\mathcal{F}$, $\mu(E_{n})=\nu(E_{n})$ by the hypothesis on $\mathcal{E}$, $\mu(E_{n})<\infty$, and for $E\in\mathcal{E}$ we have $E\cap E_{n}\in\mathcal{E}$ by π-System (i), so $\mu(E\cap E_{n})=\nu(E\cap E_{n})$. Hence

$$\mu(A\cap E_{n})=\nu(A\cap E_{n})\quad\text{for every }A\in\mathcal{F}\text{ and every }n\in\mathbb{N}.\tag{$*$}$$

As $\mathcal{E}\subseteq\mathcal{P}(X)$, $(E_{m})$ is a sequence in $\mathcal{P}(X)$, and by Sequences and Maps of Sets: Complements, Traces, Two-Term Sequences, the Disjointing Sequence, and Sequences in a Subset (iv) there are exactly one sequence $(C_{n})$ in $\mathcal{P}(X)$ with $C_{1}=X$ and $C_{n+1}=C_{n}\cap(X\setminus E_{n})$ for every $n\in\mathbb{N}$, and exactly one sequence $(B_{n})$ in $\mathcal{P}(X)$ with $B_{n}=E_{n}\cap C_{n}$ for every $n\in\mathbb{N}$. As every $E_{n}$ lies in $\mathcal{E}\subseteq\mathcal{F}$, The Disjointing Sequence of a Sequence of Sets (v) gives $C_{n}\in\mathcal{F}$ for every $n\in\mathbb{N}$. By The Disjointing Sequence of a Sequence of Sets (ii), $B_{n}\subseteq E_{n}$ for every $n\in\mathbb{N}$; by The Disjointing Sequence of a Sequence of Sets (iii), $B_{m}$ and $B_{n}$ are disjoint for all $m,n\in\mathbb{N}$ with $m\neq n$; and by The Disjointing Sequence of a Sequence of Sets (iv), $\bigcup_{n\in\mathbb{N}}B_{n}=\bigcup_{m\in\mathbb{N}}E_{m}=X$.

Finally let $A\in\mathcal{F}$. By Sequences and Maps of Sets: Complements, Traces, Two-Term Sequences, the Disjointing Sequence, and Sequences in a Subset (ii) with $Y=A$, applied to the sequence $(B_{n})$ in $\mathcal{P}(X)$, the map $n\mapsto B_{n}\cap A$ is a sequence in $\mathcal{P}(A)$, hence in $\mathcal{P}(X)$ as $A\subseteq X$, and as each $B_{n}\cap A=(A\cap C_{n})\cap E_{n}$ lies in $\mathcal{F}$ by Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (ix), $A$, $C_{n}$ and $E_{n}$ lying in $\mathcal{F}$, it is a sequence in $\mathcal{F}$ by Sequences and Maps of Sets: Complements, Traces, Two-Term Sequences, the Disjointing Sequence, and Sequences in a Subset (v). Its terms are pairwise disjoint in the sense of Measure, Measure Space, and Probability Measure (i): $B_{m}\cap B_{n}=\emptyset$ for $m\neq n$ by The Disjointing Sequence of a Sequence of Sets (iii), and as $B_{n}\cap A\subseteq B_{n}$ for every $n$, $(B_{m}\cap A)\cap(B_{n}\cap A)=\emptyset$ (Monotone Sequences of Sets, Unions of Traces, and Disjointness of Smaller Sets (v)). Its union is $\bigl(\bigcup_{n\in\mathbb{N}}B_{n}\bigr)\cap A$ (Monotone Sequences of Sets, Unions of Traces, and Disjointness of Smaller Sets (iii)), that is, $X\cap A=A$. For every $n\in\mathbb{N}$, $B_{n}\cap A=(A\cap C_{n})\cap E_{n}$ with $A\cap C_{n}\in\mathcal{F}$, so $\mu(B_{n}\cap A)=\nu(B_{n}\cap A)$ by $(*)$; thus the sequences $n\mapsto\mu(B_{n}\cap A)$ and $n\mapsto\nu(B_{n}\cap A)$ in $[0,\infty]$ are the same sequence. By the countable additivity of Measure, Measure Space, and Probability Measure (i), applied to $\mu$ and to $\nu$, and the sum of that sequence,

$$\mu(A)=\sum_{n\in\mathbb{N}}\mu(B_{n}\cap A)=\sum_{n\in\mathbb{N}}\nu(B_{n}\cap A)=\nu(A).$$

Since $\mu$ and $\nu$ are maps on $\mathcal{F}$ with the same values, $\mu=\nu$ (Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (iv)).