Hahn decomposition theorem

Short description: Measurability theorem

In mathematics, the Hahn decomposition theorem, named after the Austrian mathematician Hans Hahn, states that for any measurable space [math]\displaystyle{ (X,\Sigma) }[/math] and any signed measure [math]\displaystyle{ \mu }[/math] defined on the [math]\displaystyle{ \sigma }[/math]-algebra [math]\displaystyle{ \Sigma }[/math], there exist two [math]\displaystyle{ \Sigma }[/math]-measurable sets, [math]\displaystyle{ P }[/math] and [math]\displaystyle{ N }[/math], of [math]\displaystyle{ X }[/math] such that:

1. [math]\displaystyle{ P \cup N = X }[/math] and [math]\displaystyle{ P \cap N = \varnothing }[/math].
2. For every [math]\displaystyle{ E \in \Sigma }[/math] such that [math]\displaystyle{ E \subseteq P }[/math], one has [math]\displaystyle{ \mu(E) \geq 0 }[/math], i.e., [math]\displaystyle{ P }[/math] is a positive set for [math]\displaystyle{ \mu }[/math].
3. For every [math]\displaystyle{ E \in \Sigma }[/math] such that [math]\displaystyle{ E \subseteq N }[/math], one has [math]\displaystyle{ \mu(E) \leq 0 }[/math], i.e., [math]\displaystyle{ N }[/math] is a negative set for [math]\displaystyle{ \mu }[/math].

Moreover, this decomposition is essentially unique, meaning that for any other pair [math]\displaystyle{ (P',N') }[/math] of [math]\displaystyle{ \Sigma }[/math]-measurable subsets of [math]\displaystyle{ X }[/math] fulfilling the three conditions above, the symmetric differences [math]\displaystyle{ P \triangle P' }[/math] and [math]\displaystyle{ N \triangle N' }[/math] are [math]\displaystyle{ \mu }[/math]-null sets in the strong sense that every [math]\displaystyle{ \Sigma }[/math]-measurable subset of them has zero measure. The pair [math]\displaystyle{ (P,N) }[/math] is then called a Hahn decomposition of the signed measure [math]\displaystyle{ \mu }[/math].

Jordan measure decomposition

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

[math]\displaystyle{ {\mu^{+}}(E) := \mu(E \cap P) \qquad \text{and} \qquad {\mu^{-}}(E) := - \mu(E \cap N) }[/math]

for every [math]\displaystyle{ E \in \Sigma }[/math] and any Hahn decomposition [math]\displaystyle{ (P,N) }[/math] of [math]\displaystyle{ \mu }[/math].

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 [math]\displaystyle{ (\mu^{+},\mu^{-}) }[/math] of a finite signed measure [math]\displaystyle{ \mu }[/math], one has

[math]\displaystyle{ {\mu^{+}}(E) = \sup_{B \in \Sigma, ~ B \subseteq E} \mu(B) \quad \text{and} \quad {\mu^{-}}(E) = - \inf_{B \in \Sigma, ~ B \subseteq E} \mu(B) }[/math]

for any [math]\displaystyle{ E }[/math] in [math]\displaystyle{ \Sigma }[/math]. Furthermore, if [math]\displaystyle{ \mu = \nu^{+} - \nu^{-} }[/math] for a pair [math]\displaystyle{ (\nu^{+},\nu^{-}) }[/math] of finite non-negative measures on [math]\displaystyle{ X }[/math], then

[math]\displaystyle{ \nu^{+} \geq \mu^{+} \quad \text{and} \quad \nu^{-} \geq \mu^{-}. }[/math]

The last expression means that the Jordan decomposition is the minimal decomposition of [math]\displaystyle{ \mu }[/math] 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 Fischer (2012).

Proof of the Hahn decomposition theorem

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

Claim: Suppose that [math]\displaystyle{ D \in \Sigma }[/math] satisfies [math]\displaystyle{ \mu(D) \leq 0 }[/math]. Then there is a negative set [math]\displaystyle{ A \subseteq D }[/math] such that [math]\displaystyle{ \mu(A) \leq \mu(D) }[/math].

Proof of the claim: Define [math]\displaystyle{ A_{0} := D }[/math]. Inductively assume for [math]\displaystyle{ n \in \mathbb{N}_{0} }[/math] that [math]\displaystyle{ A_{n} \subseteq D }[/math] has been constructed. Let

[math]\displaystyle{ t_{n} := \sup(\{ \mu(B) \mid B \in \Sigma ~ \text{and} ~ B \subseteq A_{n} \}) }[/math]

denote the supremum of [math]\displaystyle{ \mu(B) }[/math] over all the [math]\displaystyle{ \Sigma }[/math]-measurable subsets [math]\displaystyle{ B }[/math] of [math]\displaystyle{ A_{n} }[/math]. This supremum might a priori be infinite. As the empty set [math]\displaystyle{ \varnothing }[/math] is a possible candidate for [math]\displaystyle{ B }[/math] in the definition of [math]\displaystyle{ t_{n} }[/math], and as [math]\displaystyle{ \mu(\varnothing) = 0 }[/math], we have [math]\displaystyle{ t_{n} \geq 0 }[/math]. By the definition of [math]\displaystyle{ t_{n} }[/math], there then exists a [math]\displaystyle{ \Sigma }[/math]-measurable subset [math]\displaystyle{ B_{n} \subseteq A_{n} }[/math] satisfying

[math]\displaystyle{ \mu(B_{n}) \geq \min \! \left( 1,\frac{t_{n}}{2} \right). }[/math]

Set [math]\displaystyle{ A_{n + 1} := A_{n} \setminus B_{n} }[/math] to finish the induction step. Finally, define

[math]\displaystyle{ A := D \Bigg\backslash \bigcup_{n = 0}^{\infty} B_{n}. }[/math]

As the sets [math]\displaystyle{ (B_{n})_{n = 0}^{\infty} }[/math] are disjoint subsets of [math]\displaystyle{ D }[/math], it follows from the sigma additivity of the signed measure [math]\displaystyle{ \mu }[/math] that

[math]\displaystyle{ \mu(D) = \mu(A) + \sum_{n = 0}^{\infty} \mu(B_{n}) \geq \mu(A) + \sum_{n = 0}^{\infty} \min \! \left( 1,\frac{t_{n}}{2} \right)\geq \mu(A). }[/math]

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

Construction of the decomposition: Set [math]\displaystyle{ N_{0} = \varnothing }[/math]. Inductively, given [math]\displaystyle{ N_{n} }[/math], define

[math]\displaystyle{ s_{n} := \inf(\{ \mu(D) \mid D \in \Sigma ~ \text{and} ~ D \subseteq X \setminus N_{n} \}). }[/math]

as the infimum of [math]\displaystyle{ \mu(D) }[/math] over all the [math]\displaystyle{ \Sigma }[/math]-measurable subsets [math]\displaystyle{ D }[/math] of [math]\displaystyle{ X \setminus N_{n} }[/math]. This infimum might a priori be [math]\displaystyle{ - \infty }[/math]. As [math]\displaystyle{ \varnothing }[/math] is a possible candidate for [math]\displaystyle{ D }[/math] in the definition of [math]\displaystyle{ s_{n} }[/math], and as [math]\displaystyle{ \mu(\varnothing) = 0 }[/math], we have [math]\displaystyle{ s_{n} \leq 0 }[/math]. Hence, there exists a [math]\displaystyle{ \Sigma }[/math]-measurable subset [math]\displaystyle{ D_{n} \subseteq X \setminus N_{n} }[/math] such that

[math]\displaystyle{ \mu(D_{n}) \leq \max \! \left( \frac{s_{n}}{2},- 1 \right) \leq 0. }[/math]

By the claim above, there is a negative set [math]\displaystyle{ A_{n} \subseteq D_{n} }[/math] such that [math]\displaystyle{ \mu(A_{n}) \leq \mu(D_{n}) }[/math]. Set [math]\displaystyle{ N_{n + 1} := N_{n} \cup A_{n} }[/math] to finish the induction step. Finally, define

[math]\displaystyle{ N := \bigcup_{n = 0}^{\infty} A_{n}. }[/math]

As the sets [math]\displaystyle{ (A_{n})_{n = 0}^{\infty} }[/math] are disjoint, we have for every [math]\displaystyle{ \Sigma }[/math]-measurable subset [math]\displaystyle{ B \subseteq N }[/math] that

[math]\displaystyle{ \mu(B) = \sum_{n = 0}^{\infty} \mu(B \cap A_{n}) }[/math]

by the sigma additivity of [math]\displaystyle{ \mu }[/math]. In particular, this shows that [math]\displaystyle{ N }[/math] is a negative set. Next, define [math]\displaystyle{ P := X \setminus N }[/math]. If [math]\displaystyle{ P }[/math] were not a positive set, there would exist a [math]\displaystyle{ \Sigma }[/math]-measurable subset [math]\displaystyle{ D \subseteq P }[/math] with [math]\displaystyle{ \mu(D) \lt 0 }[/math]. Then [math]\displaystyle{ s_{n} \leq \mu(D) }[/math] for all [math]\displaystyle{ n \in \mathbb{N}_{0} }[/math] and

[math]\displaystyle{ \mu(N) = \sum_{n = 0}^{\infty} \mu(A_{n}) \leq \sum_{n = 0}^{\infty} \max \! \left( \frac{s_{n}}{2},- 1 \right) = - \infty, }[/math]

which is not allowed for [math]\displaystyle{ \mu }[/math]. Therefore, [math]\displaystyle{ P }[/math] is a positive set.

Proof of the uniqueness statement: Suppose that [math]\displaystyle{ (N',P') }[/math] is another Hahn decomposition of [math]\displaystyle{ X }[/math]. Then [math]\displaystyle{ P \cap N' }[/math] is a positive set and also a negative set. Therefore, every measurable subset of it has measure zero. The same applies to [math]\displaystyle{ N \cap P' }[/math]. As

[math]\displaystyle{ P \triangle P' = N \triangle N' = (P \cap N') \cup (N \cap P'), }[/math]

this completes the proof. Q.E.D.

References

• Billingsley, Patrick (1995). Probability and Measure -- Third Edition. Wiley Series in Probability and Mathematical Statistics. New York: John Wiley & Sons. ISBN 0-471-00710-2. 
• Fischer, Tom (2012). "Existence, uniqueness, and minimality of the Jordan measure decomposition". arXiv:1206.5449 [math.ST].

External links

• Hahn decomposition theorem at PlanetMath.
• Hazewinkel, Michiel, ed. (2001), "Hahn decomposition", Encyclopedia of Mathematics, Springer Science+Business Media B.V. / Kluwer Academic Publishers, ISBN 978-1-55608-010-4, https://www.encyclopediaofmath.org/index.php?title=p/h046140 
• Hazewinkel, Michiel, ed. (2001), "Jordan decomposition (of a signed measure)", Encyclopedia of Mathematics, Springer Science+Business Media B.V. / Kluwer Academic Publishers, ISBN 978-1-55608-010-4, https://www.encyclopediaofmath.org/index.php?title=p/h046140