Hahn decomposition theorem

In mathematics, the Hahn decomposition theorem, named after the Austrian mathematician Hans Hahn, states that for any measurable space $(X, Σ)$ and any signed measure $μ$ defined on the $σ$ -algebra $Σ$ , there exist two $Σ$ -measurable sets, $P$ and $N$ , of $X$ such that:

$P \cup N = X$ and $P \cap N = \emptyset$ .
For every $E \in Σ$ such that $E \subseteq P$ , one has $μ (E) \geq 0$ , i.e., $P$ is a positive set for $μ$ .
For every $E \in Σ$ such that $E \subseteq N$ , one has $μ (E) \leq 0$ , i.e., $N$ is a negative set for $μ$ .

Moreover, this decomposition is essentially unique, meaning that for any other pair $(P^{'}, N^{'})$ of $Σ$ -measurable subsets of $X$ fulfilling the three conditions above, the symmetric differences $P △ P^{'}$ and $N △ N^{'}$ are $μ$ -null sets in the strong sense that every $Σ$ -measurable subset of them has zero measure. The pair $(P, N)$ is then called a Hahn decomposition of the signed measure $μ$ .

Jordan measure decomposition

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

μ^{+} (E) := μ (E \cap P) and μ^{-} (E) := - μ (E \cap N)

for every $E \in Σ$ and any Hahn decomposition $(P, N)$ of $μ$ .

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 $(μ^{+}, μ^{-})$ of a finite signed measure $μ$ , one has

μ^{+} (E) = sup_{B \in Σ, B \subseteq E} μ (B) and μ^{-} (E) = - inf_{B \in Σ, B \subseteq E} μ (B)

for any $E$ in $Σ$ . Furthermore, if $μ = ν^{+} - ν^{-}$ for a pair $(ν^{+}, ν^{-})$ of finite non-negative measures on $X$ , then

ν^{+} \geq μ^{+} and ν^{-} \geq μ^{-} .

The last expression means that the Jordan decomposition is the minimal decomposition of $μ$ into the 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 $μ$ does not take the value $- \infty$ (otherwise decompose according to $- μ$ ). As mentioned above, a negative set is a set $A \in Σ$ such that $μ (B) \leq 0$ for every $Σ$ -measurable subset $B \subseteq A$ .

Claim: Suppose that $D \in Σ$ satisfies $μ (D) \leq 0$ . Then there is a negative set $A \subseteq D$ such that $μ (A) \leq μ (D)$ .

Proof of the claim: Define $A_{0} := D$ . Inductively assume for $n \in ℕ_{0}$ that $A_{n} \subseteq D$ has been constructed. Let

t_{n} := sup ({μ (B) ∣ B \in Σ and B \subseteq A_{n}})

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

μ (B_{n}) \geq min (1, \frac{t_{n}}{2}) .

Set $A_{n + 1} := A_{n} ∖ B_{n}$ to finish the induction step. Finally, define

A := D \ ⋃_{n = 0}^{\infty} B_{n} .

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

μ (D) = μ (A) + \sum_{n = 0}^{\infty} μ (B_{n}) \geq μ (A) + \sum_{n = 0}^{\infty} min (1, \frac{t_{n}}{2}) \geq μ (A) .

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

Construction of the decomposition: Set $N_{0} = \emptyset$ . Inductively, given $N_{n}$ , define

s_{n} := inf ({μ (D) ∣ D \in Σ and D \subseteq X ∖ N_{n}}) .

as the infimum of $μ (D)$ over all the $Σ$ -measurable subsets $D$ of $X ∖ N_{n}$ . This infimum might a priori be $- \infty$ . As $\emptyset$ is a possible candidate for $D$ in the definition of $s_{n}$ , and as $μ (\emptyset) = 0$ , we have $s_{n} \leq 0$ . Hence, there exists a $Σ$ -measurable subset $D_{n} \subseteq X ∖ N_{n}$ such that

μ (D_{n}) \leq max (\frac{s_{n}}{2}, - 1) \leq 0.

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

N := ⋃_{n = 0}^{\infty} A_{n} .

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

μ (B) = \sum_{n = 0}^{\infty} μ (B \cap A_{n})

by the sigma additivity of $μ$ . In particular, this shows that $N$ is a negative set. Next, define $P := X ∖ N$ . If $P$ were not a positive set, there would exist a $Σ$ -measurable subset $D \subseteq P$ with $μ (D) < 0$ . Then $s_{n} \leq μ (D)$ for all $n \in ℕ_{0}$ and^{[clarification needed]}

μ (N) = \sum_{n = 0}^{\infty} μ (A_{n}) \leq \sum_{n = 0}^{\infty} max (\frac{s_{n}}{2}, - 1) = - \infty,

which is not allowed for $μ$ . Therefore, $P$ is a positive set.

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

P △ P^{'} = N △ N^{'} = (P \cap N^{'}) \cup (N \cap P^{'}),

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.
"Hahn decomposition", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
"Jordan decomposition (of a signed measure)", Encyclopedia of Mathematics, EMS Press, 2001 [1994]

Jordan measure decomposition

Proof of the Hahn decomposition theorem

See also

References

External links