Hahn decomposition theorem

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


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

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

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


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

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

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


 * $$ {\mu^{+}}(E) = \sup_{B \in \Sigma, ~ B \subseteq E} \mu(B) \quad \text{and} \quad {\mu^{-}}(E) = - \inf_{B \in \Sigma, ~ B \subseteq E} \mu(B) $$

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


 * $$ \nu^{+} \geq \mu^{+} \quad \text{and} \quad \nu^{-} \geq \mu^{-}. $$

The last expression means that the Jordan decomposition is the minimal decomposition of $$ \mu $$ 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 Fischer (2012).

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

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

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


 * $$ t_{n} := \sup(\{ \mu(B) \mid B \in \Sigma ~ \text{and} ~ B \subseteq A_{n} \}) $$

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


 * $$ \mu(B_{n}) \geq \min \! \left( 1,\frac{t_{n}}{2} \right). $$

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


 * $$ A := D \Bigg\backslash \bigcup_{n = 0}^{\infty} B_{n}. $$

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


 * $$ \mu(D) = \mu(A) + \sum_{n = 0}^{\infty} \mu(B_{n}) \geq \mu(A) + \sum_{n = 0}^{\infty} \min \! \left( 1,\frac{t_{n}}{2} \right)\geq \mu(A). $$

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

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


 * $$ s_{n} := \inf(\{ \mu(D) \mid D \in \Sigma ~ \text{and} ~ D \subseteq X \setminus N_{n} \}). $$

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


 * $$ \mu(D_{n}) \leq \max \! \left( \frac{s_{n}}{2},- 1 \right) \leq 0. $$

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


 * $$ N := \bigcup_{n = 0}^{\infty} A_{n}. $$

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


 * $$ \mu(B) = \sum_{n = 0}^{\infty} \mu(B \cap A_{n}) $$

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


 * $$ \mu(N) = \sum_{n = 0}^{\infty} \mu(A_{n}) \leq \sum_{n = 0}^{\infty} \max \! \left( \frac{s_{n}}{2},- 1 \right) = - \infty, $$

which is not allowed for $$ \mu $$. 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


 * $$ P \triangle P' = N \triangle N' = (P \cap N') \cup (N \cap P'), $$

this completes the proof. Q.E.D.