Hahn decomposition theorem

A consequence of the Hahn decomposition theorem is the {{visible anchor|Jordan decomposition theorem}}, which states that every signed measure $\backslash mu$ defined on $\backslash Sigma$ has a unique decomposition into a difference $\backslash mu\; =\; \backslash mu^\{+\}\; -\; \backslash mu^\{-\}$ of two positive measures, $\backslash mu^\{+\}$ and $\backslash mu^\{-\}$, at least one of which is finite, such that $\{\backslash mu^\{+\}\}(E)\; =\; 0$ for every $\backslash Sigma$-measurable subset $E\; \backslash subseteq\; N$ and $\{\backslash mu^\{-\}\}(E)\; =\; 0$ for every $\backslash Sigma$-measurable subset $E\; \backslash subseteq\; P$, for any Hahn decomposition $(P,N)$ of $\backslash mu$. We call $\backslash mu^\{+\}$ and $\backslash mu^\{-\}$ the positive and negative part of $\backslash mu$, respectively. The pair $(\backslash mu^\{+\},\backslash mu^\{-\})$ is called a Jordan decomposition (or sometimes Hahn–Jordan decomposition) of $\backslash mu$. The two measures can be defined as

:$\{\backslash mu^\{+\}\}(E)\; :=\; \backslash mu(E\; \backslash cap\; P)\; \backslash qquad\; \backslash text\{and\}\; \backslash qquad\; \{\backslash mu^\{-\}\}(E)\; :=\; -\; \backslash mu(E\; \backslash cap\; N)$

for every $E\; \backslash in\; \backslash Sigma$ and any Hahn decomposition $(P,N)$ of $\backslash mu$.

Note that the Jordan decomposition is unique, while the Hahn decomposition is only essentially unique.

The Jordan decomposition has the following corollary: Given a Jordan decomposition $(\backslash mu^\{+\},\backslash mu^\{-\})$ of a finite signed measure $\backslash mu$, one has

:$\{\backslash mu^\{+\}\}(E)\; =\; \backslash sup\_\{B\; \backslash in\; \backslash Sigma,\; ~\; B\; \backslash subseteq\; E\}\; \backslash mu(B)\; \backslash quad\; \backslash text\{and\}\; \backslash quad\; \{\backslash mu^\{-\}\}(E)\; =\; -\; \backslash inf\_\{B\; \backslash in\; \backslash Sigma,\; ~\; B\; \backslash subseteq\; E\}\; \backslash mu(B)$

for any $E$ in $\backslash Sigma$. Furthermore, if $\backslash mu\; =\; \backslash nu^\{+\}\; -\; \backslash nu^\{-\}$ for a pair $(\backslash nu^\{+\},\backslash nu^\{-\})$ of finite non-negative measures on $X$, then

:$\backslash nu^\{+\}\; \backslash geq\; \backslash mu^\{+\}\; \backslash quad\; \backslash text\{and\}\; \backslash quad\; \backslash nu^\{-\}\; \backslash geq\; \backslash mu^\{-\}.$

The last expression means that the Jordan decomposition is the minimal decomposition of $\backslash mu$ into a difference of non-negative measures. This is the minimality property of the Jordan decomposition.

Proof of the Jordan decomposition: For an elementary proof of the existence, uniqueness, and minimality of the Jordan measure decomposition see [https://arxiv.org/abs/1206.5449 Fischer (2012)].

Preparation: Assume that $\backslash mu$ does not take the value $-\; \backslash infty$ (otherwise decompose according to $-\; \backslash mu$). As mentioned above, a negative set is a set $A\; \backslash in\; \backslash Sigma$ such that $\backslash mu(B)\; \backslash leq\; 0$ for every $\backslash Sigma$-measurable subset $B\; \backslash subseteq\; A$.

Claim: Suppose that $D\; \backslash in\; \backslash Sigma$ satisfies $\backslash mu(D)\; \backslash leq\; 0$. Then there is a negative set $A\; \backslash subseteq\; D$ such that $\backslash mu(A)\; \backslash leq\; \backslash mu(D)$.

Proof of the claim: Define $A\_\{0\}\; :=\; D$. Inductively assume for $n\; \backslash in\; \backslash mathbb\{N\}\_\{0\}$ that $A\_\{n\}\; \backslash subseteq\; D$ has been constructed. Let

:$t\_\{n\}\; :=\; \backslash sup(\backslash \{\; \backslash mu(B)\; \backslash mid\; B\; \backslash in\; \backslash Sigma\; ~\; \backslash text\{and\}\; ~\; B\; \backslash subseteq\; A\_\{n\}\; \backslash \})$

denote the supremum of $\backslash mu(B)$ over all the $\backslash Sigma$-measurable subsets $B$ of $A\_\{n\}$. This supremum might a priori be infinite. As the empty set $\backslash varnothing$ is a possible candidate for $B$ in the definition of $t\_\{n\}$, and as $\backslash mu(\backslash varnothing)\; =\; 0$, we have $t\_\{n\}\; \backslash geq\; 0$. By the definition of $t\_\{n\}$, there then exists a $\backslash Sigma$-measurable subset $B\_\{n\}\; \backslash subseteq\; A\_\{n\}$ satisfying

:$\backslash mu(B\_\{n\})\; \backslash geq\; \backslash min\; \backslash !\; \backslash left(\; 1,\backslash frac\{t\_\{n\}\}\{2\}\; \backslash right).$

Set $A\_\{n\; +\; 1\}\; :=\; A\_\{n\}\; \backslash setminus\; B\_\{n\}$ to finish the induction step. Finally, define

:$A\; :=\; D\; \backslash Bigg\backslash backslash\; \backslash bigcup\_\{n\; =\; 0\}^\{\backslash infty\}\; B\_\{n\}.$

As the sets $(B\_\{n\})\_\{n\; =\; 0\}^\{\backslash infty\}$ are disjoint subsets of $D$, it follows from the sigma additivity of the signed measure $\backslash mu$ that

:$\backslash mu(D)\; =\; \backslash mu(A)\; +\; \backslash sum\_\{n\; =\; 0\}^\{\backslash infty\}\; \backslash mu(B\_\{n\})\; \backslash geq\; \backslash mu(A)\; +\; \backslash sum\_\{n\; =\; 0\}^\{\backslash infty\}\; \backslash min\; \backslash !\; \backslash left(\; 1,\backslash frac\{t\_\{n\}\}\{2\}\; \backslash right)\backslash geq\; \backslash mu(A).$

This shows that $\backslash mu(A)\; \backslash leq\; \backslash mu(D)$. Assume $A$ were not a negative set. This means that there would exist a $\backslash Sigma$-measurable subset $B\; \backslash subseteq\; A$ that satisfies $\backslash mu(B)\; >\; 0$. Then $t\_\{n\}\; \backslash geq\; \backslash mu(B)$ for every $n\; \backslash in\; \backslash mathbb\{N\}\_\{0\}$, so the series on the right would have to diverge to $+\; \backslash infty$, implying that $\backslash mu(D)\; =\; +\; \backslash infty$, which is a contradiction, since $\backslash mu(D)\; \backslash leq\; 0$. Therefore, $A$ must be a negative set.

Construction of the decomposition: Set $N\_\{0\}\; =\; \backslash varnothing$. Inductively, given $N\_\{n\}$, define

:$s\_\{n\}\; :=\; \backslash inf(\backslash \{\; \backslash mu(D)\; \backslash mid\; D\; \backslash in\; \backslash Sigma\; ~\; \backslash text\{and\}\; ~\; D\; \backslash subseteq\; X\; \backslash setminus\; N\_\{n\}\; \backslash \}).$

as the infimum of $\backslash mu(D)$ over all the $\backslash Sigma$-measurable subsets $D$ of $X\; \backslash setminus\; N\_\{n\}$. This infimum might a priori be $-\; \backslash infty$. As $\backslash varnothing$ is a possible candidate for $D$ in the definition of $s\_\{n\}$, and as $\backslash mu(\backslash varnothing)\; =\; 0$, we have $s\_\{n\}\; \backslash leq\; 0$. Hence, there exists a $\backslash Sigma$-measurable subset $D\_\{n\}\; \backslash subseteq\; X\; \backslash setminus\; N\_\{n\}$ such that

:$\backslash mu(D\_\{n\})\; \backslash leq\; \backslash max\; \backslash !\; \backslash left(\; \backslash frac\{s\_\{n\}\}\{2\},-\; 1\; \backslash right)\; \backslash leq\; 0.$

By the claim above, there is a negative set $A\_\{n\}\; \backslash subseteq\; D\_\{n\}$ such that $\backslash mu(A\_\{n\})\; \backslash leq\; \backslash mu(D\_\{n\})$. Set $N\_\{n\; +\; 1\}\; :=\; N\_\{n\}\; \backslash cup\; A\_\{n\}$ to finish the induction step. Finally, define

:$N\; :=\; \backslash bigcup\_\{n\; =\; 0\}^\{\backslash infty\}\; A\_\{n\}.$

As the sets $(A\_\{n\})\_\{n\; =\; 0\}^\{\backslash infty\}$ are disjoint, we have for every $\backslash Sigma$-measurable subset $B\; \backslash subseteq\; N$ that

:$\backslash mu(B)\; =\; \backslash sum\_\{n\; =\; 0\}^\{\backslash infty\}\; \backslash mu(B\; \backslash cap\; A\_\{n\})$

by the sigma additivity of $\backslash mu$. In particular, this shows that $N$ is a negative set. Next, define $P\; :=\; X\; \backslash setminus\; N$. If $P$ were not a positive set, there would exist a $\backslash Sigma$-measurable subset $D\; \backslash subseteq\; P$ with . Then $s\_\{n\}\; \backslash leq\; \backslash mu(D)$ for all $n\; \backslash in\; \backslash mathbb\{N\}\_\{0\}$ and{{clarify|reason=see talk page|date=July 2023}}

:$\backslash mu(N)\; =\; \backslash sum\_\{n\; =\; 0\}^\{\backslash infty\}\; \backslash mu(A\_\{n\})\; \backslash leq\; \backslash sum\_\{n\; =\; 0\}^\{\backslash infty\}\; \backslash max\; \backslash !\; \backslash left(\; \backslash frac\{s\_\{n\}\}\{2\},-\; 1\; \backslash right)\; =\; -\; \backslash infty,$

which is not allowed for $\backslash mu$. Therefore, $P$ is a positive set.

Proof of the uniqueness statement:

Suppose that $(N\text{'},P\text{'})$ is another Hahn decomposition of $X$. Then $P\; \backslash cap\; N\text{'}$ is a positive set and also a negative set. Therefore, every measurable subset of it has measure zero. The same applies to $N\; \backslash cap\; P\text{'}$. As

:$P\; \backslash triangle\; P\text{'}\; =\; N\; \backslash triangle\; N\text{'}\; =\; (P\; \backslash cap\; N\text{'})\; \backslash cup\; (N\; \backslash cap\; P\text{'}),$

this completes the proof. Q.E.D.

• {{cite book

| last = Billingsley

| first = Patrick

| title = Probability and Measure -- Third Edition

| series = Wiley Series in Probability and Mathematical Statistics

|publisher = John Wiley & Sons

| location = New York

| year = 1995

| isbn = 0-471-00710-2

}}

• {{cite arXiv |last=Fischer |first=Tom |eprint=1206.5449 |class=math.ST |title=Existence, uniqueness, and minimality of the Jordan measure decomposition |year=2012 }}

• [http://planetmath.org/?op=getobj&from=objects&id=4014 Hahn decomposition theorem] at PlanetMath.
• {{springer|title=Hahn decomposition|id=p/h046140}}
• {{springer|title=Jordan decomposition (of a signed measure)|id=p/h046140}}

{{Measure theory}}

Category:Theorems in measure theory

Category:Articles containing proofs