Proof of Hahn and Jordan Decompositions of a Finite Signed Measure

Step 1: elementary properties of $\nu$. Throughout, $\nu$ is the finite signed measure of the statement, and the field and order laws of $\mathbb{R}$ are used freely (The Structure of ℝ as a Field and an Ordered Ring (i), The Structure of ℝ as a Field and an Ordered Ring (ii), Rules of Arithmetic and Order in an Ordered Field (iv), Rules of Arithmetic and Order in an Ordered Field (v), Rules of Arithmetic and Order in an Ordered Field (viii)).

(a) Finite additivity. Let $J$ be a finite set and $(B_{j})_{j\in J}$ a family in $\mathcal{F}$ with $B_{i}\cap B_{j}=\emptyset$ for $i\neq j$. Then, with the finite sum over $J$,

$$\nu\Bigl(\bigcup_{j\in J}B_{j}\Bigr)=\sum_{j\in J}\nu(B_{j}).$$

For $J=\emptyset$ both sides are $0$. Otherwise list $J$ as $j_{1},\dots,j_{r}$ and apply countable additivity to the pairwise disjoint sequence $B_{j_{1}},\dots,B_{j_{r}},\emptyset,\emptyset,\dots$: since $\nu(\emptyset)=0$, its partial sums equal $\sum_{j\in J}\nu(B_{j})$ from the $r$-th on, so this is the sum of the series (Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (i), Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (v)). In particular, for $A,B\in\mathcal{F}$ with $B\subseteq A$, $A$ is the disjoint union of $B$ and $A\setminus B$, so

$$\nu(A\setminus B)=\nu(A)-\nu(B).$$

(b) Continuity from below. Let $(B_{m})$ be a sequence in $\mathcal{F}$ with $B_{m}\subseteq B_{m+1}$ for every $m$, and $B=\bigcup_{m\in\mathbb{N}}B_{m}$. Its disjointing sequence $(D_{m})$ is a pairwise disjoint sequence in $\mathcal{F}$ with union $B$ (The Disjointing Sequence of a Sequence of Sets (v), The Disjointing Sequence of a Sequence of Sets (iii), The Disjointing Sequence of a Sequence of Sets (iv)), and since $(B_{m})$ increases, $D_{1}\cup\dots\cup D_{m}=B_{m}$ (The Disjointing Sequence of a Sequence of Sets (i), Monotone Sequences of Sets, Unions of Traces, and Disjointness of Smaller Sets (i)). By (a) the $m$-th partial sum of $\sum_{k}\nu(D_{k})$ is $\nu(B_{m})$, so countable additivity gives

$$\nu(B_{m})\to\nu(B).$$

(c) Continuity from above. Let $(E_{m})$ be a sequence in $\mathcal{F}$ with $E_{m+1}\subseteq E_{m}$ for every $m$, and $E=\bigcap_{m\in\mathbb{N}}E_{m}$. The sets $E_{1}\setminus E_{m}$ increase and have union $E_{1}\setminus E$, so by (b) and (a), $\nu(E_{1})-\nu(E_{m})\to\nu(E_{1})-\nu(E)$, that is (Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (vi))

$$\nu(E_{m})\to\nu(E).$$

Step 2: a maximizing sequence. Assume the axiom of dependent choice. Let

$$\alpha=\sup\{\nu(A):A\in\mathcal{F}\}\in[-\infty,\infty];$$

$\alpha\ge\nu(\emptyset)=0$. For $n\in\mathbb{N}$ put $t_{n}=\alpha-1/n$ if $\alpha\in\mathbb{R}$, and $t_{n}=n$ if $\alpha=\infty$. Since $t_{n}<\alpha$, the set $\{A\in\mathcal{F}:\nu(A)>t_{n}\}$ is nonempty (Order and Suprema in the Extended Real Line (x)). By countable choice, which follows from dependent choice by Countable Choice Follows from Dependent Choice (i), there is a sequence $(A_{n})$ in $\mathcal{F}$ with

$$\nu(A_{n})>t_{n}\qquad(n\in\mathbb{N}).$$

This is the only use of the axiom of dependent choice.

Step 3: the atoms generated by $(A_{n})$. For $n\in\mathbb{N}$ let $\mathcal{E}_{n}$ be the finite set of maps $\varepsilon:[n]\to\{0,1\}$, and for $\varepsilon\in\mathcal{E}_{n}$ let

$$B_{n,\varepsilon}=\bigcap_{k=1}^{n}A_{k}^{\varepsilon(k)},\qquad A_{k}^{1}=A_{k},\quad A_{k}^{0}=X\setminus A_{k}.$$

For fixed $n$, these atoms of generation $n$ lie in $\mathcal{F}$, are pairwise disjoint and cover $X$: a point $x\in X$ lies in $B_{n,\varepsilon}$ exactly for the $\varepsilon$ with $\varepsilon(k)=1\Leftrightarrow x\in A_{k}$ ($k\in[n]$). Consequently $A_{n}$ is the disjoint union of the atoms $B_{n,\varepsilon}$ with $\varepsilon(n)=1$. Let $C_{n}$ be the union of the atoms of generation $n$ of positive measure:

$$C_{n}=\bigcup\{B_{n,\varepsilon}:\varepsilon\in\mathcal{E}_{n},\ \nu(B_{n,\varepsilon})>0\}.$$

By Step 1(a), dropping the terms that are not positive and then adding positive ones,

$$\nu(A_{n})=\sum_{\varepsilon\in\mathcal{E}_{n},\,\varepsilon(n)=1}\nu(B_{n,\varepsilon})\le\sum_{\varepsilon\in\mathcal{E}_{n},\,\varepsilon(n)=1,\,\nu(B_{n,\varepsilon})>0}\nu(B_{n,\varepsilon})\le\sum_{\varepsilon\in\mathcal{E}_{n},\,\nu(B_{n,\varepsilon})>0}\nu(B_{n,\varepsilon})=\nu(C_{n}).$$

Step 4: the unions $C_{n}\cup\dots\cup C_{m}$. Let $j\le m$. An atom $B_{m,\delta}$ of generation $m$ lies in the single atom $B_{j,\delta|_{[j]}}$ of generation $j$ and is disjoint from all other atoms of generation $j$; as $C_{j}$ is a union of atoms of generation $j$, $B_{m,\delta}$ is therefore contained in $C_{j}$ or disjoint from it. Now fix $n$ and put $F_{n,m}=C_{n}\cup C_{n+1}\cup\dots\cup C_{m}$ for $m\ge n$. Let $m>n$. By what was just shown, every atom of generation $m$ lies in $F_{n,m-1}$ or is disjoint from it, so $C_{m}\setminus F_{n,m-1}$ is the union of the atoms $B_{m,\varepsilon}$ with $\nu(B_{m,\varepsilon})>0$ that are disjoint from $F_{n,m-1}$. By Step 1(a), $\nu(C_{m}\setminus F_{n,m-1})\ge0$, and since $F_{n,m}$ is the disjoint union of $F_{n,m-1}$ and $C_{m}\setminus F_{n,m-1}$,

$$\nu(F_{n,m})=\nu(F_{n,m-1})+\nu(C_{m}\setminus F_{n,m-1})\ge\nu(F_{n,m-1}).$$

By induction on $m$ and Step 3,

$$\nu(F_{n,m})\ge\nu(F_{n,n})=\nu(C_{n})\ge\nu(A_{n})\qquad(m\ge n).$$

Step 5: a set of maximal measure. Let $E_{n}=\bigcup_{m\ge n}C_{m}$, the union of the increasing sequence $(F_{n,n+k-1})_{k\in\mathbb{N}}$. By Step 1(b) and Step 4 (Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (x)),

$$\nu(E_{n})\ge\nu(A_{n})>t_{n}\qquad(n\in\mathbb{N}).$$

The sets $E_{n}$ decrease; let $D=\bigcap_{n\in\mathbb{N}}E_{n}\in\mathcal{F}$. By Step 1(c), $\nu(E_{n})\to\nu(D)$. If $\alpha=\infty$, then $\nu(E_{n})>n$ for every $n$, which is impossible for a convergent, hence bounded, sequence (Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (iii)). So $\alpha\in\mathbb{R}$, and for every $N\in\mathbb{N}$ we have $\nu(E_{n})>\alpha-1/N$ for all $n\ge N$, hence $\nu(D)\ge\alpha-1/N$ (Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (x)). Thus $\nu(D)\ge\alpha$, and $\nu(D)\le\alpha$ by the definition of $\alpha$:

$$\nu(D)=\alpha=\sup\{\nu(A):A\in\mathcal{F}\}.$$

Step 6: $D$ is a Hahn set. Let $A\in\mathcal{F}$. If $A\subseteq D$ and $\nu(A)<0$, then $\nu(D\setminus A)=\nu(D)-\nu(A)>\alpha$ by Step 1(a), contradicting the definition of $\alpha$. If $A\subseteq X\setminus D$ and $\nu(A)>0$, then $\nu(D\cup A)=\nu(D)+\nu(A)>\alpha$, again impossible. So $D$ is a Hahn set for $\nu$. This proves (i).

Step 7: uniqueness. Let $P,P'$ be Hahn sets for $\nu$ and $A\in\mathcal{F}$ with $A\subseteq(P\setminus P')\cup(P'\setminus P)$. The set $A\cap(P\setminus P')$ lies in $P$ and in $X\setminus P'$, so its measure is both $\ge0$ and $\le0$, hence $0$; likewise $\nu(A\cap(P'\setminus P))=0$. These two sets are disjoint with union $A$, so $\nu(A)=0$ by Step 1(a).

Now let $A\in\mathcal{F}$ be arbitrary. The set $A\cap P$ is the disjoint union of $A\cap P\cap P'$ and $A\cap(P\setminus P')$, and $A\cap P'$ is the disjoint union of $A\cap P\cap P'$ and $A\cap(P'\setminus P)$; the second sets have measure $0$ by what was just shown. By Step 1(a),

$$\nu(A\cap P)=\nu(A\cap P\cap P')=\nu(A\cap P'),\qquad\nu(A\setminus P)=\nu(A)-\nu(A\cap P)=\nu(A)-\nu(A\cap P')=\nu(A\setminus P').$$

This proves (iii).

Step 8: the Jordan decomposition. Let $P$ be a Hahn set for $\nu$. For $A\in\mathcal{F}$ we have $A\cap P\subseteq P$ and $A\setminus P=A\cap(X\setminus P)\subseteq X\setminus P$, so $\nu^{+}(A)=\nu(A\cap P)$ and $\nu^{-}(A)=-\nu(A\setminus P)$ are real and nonnegative, hence elements of $[0,\infty]$; and $\nu^{\pm}(\emptyset)=0$. Let $(A_{m})$ be a pairwise disjoint sequence in $\mathcal{F}$ with union $A$. The sequences $(A_{m}\cap P)$ and $(A_{m}\cap(X\setminus P))$ are pairwise disjoint, with unions $A\cap P$ and $A\setminus P$ (Monotone Sequences of Sets, Unions of Traces, and Disjointness of Smaller Sets (v), Monotone Sequences of Sets, Unions of Traces, and Disjointness of Smaller Sets (iii)). Countable additivity of $\nu$, together with Elementary Properties of Series of Real Numbers: Linearity, Null Terms, the Cauchy Criterion, Index Shifts, Tails, Order and Telescoping (i) for $\nu^{-}$, gives convergent real series

$$\sum_{m=1}^{\infty}\nu^{+}(A_{m})=\nu^{+}(A),\qquad\sum_{m=1}^{\infty}\nu^{-}(A_{m})=\nu^{-}(A),$$

and these are also the sums in $[0,\infty]$ (Sums of Sequences in [0,∞]: Real Series, Linearity, Comparison, Rearrangement, Double Series and Monotone Convergence (vi)). So $\nu^{+}$ and $\nu^{-}$ are measures; they are finite because $\nu^{+}(X)=\nu(P)$ and $\nu^{-}(X)=-\nu(X\setminus P)$ are real (Small Facts about [0,∞]: Zero, Finiteness and Sums of Zeros (ii)). By Step 1(a),

$$\nu(A)=\nu(A\cap P)+\nu(A\setminus P)=\nu^{+}(A)-\nu^{-}(A)\qquad(A\in\mathcal{F}).$$

By (iii), $\nu^{+}$ and $\nu^{-}$ do not change when $P$ is replaced by another Hahn set $P'$. Finally, let $\mu$ be a measure on $(X,\mathcal{F})$ with $\nu\ll\mu$ in the sense of Absolute Continuity of Measures (ii), and let $A\in\mathcal{F}$ with $\mu(A)=0$. By monotonicity of $\mu$ (Basic Properties of a Measure (iii)), $\mu(A\cap P)=0$ and $\mu(A\setminus P)=0$, so $\nu^{+}(A)=\nu(A\cap P)=0$ and $\nu^{-}(A)=-\nu(A\setminus P)=0$. Thus $\nu^{+}\ll\mu$ and $\nu^{-}\ll\mu$ in the sense of Absolute Continuity of Measures (i). This proves (ii).