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:

# $P \cup N = X$ and $P \cap N = \varnothing$.
# For every $E \in \Sigma$ such that $E \subseteq P$, one has $\mu(E) \geq 0$, i.e., $P$ is a positive set for $\mu$.
# 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 $(E) = 0$ for every $\Sigma$-measurable subset $E \subseteq N$ and $(E) = 0$ 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

:$(E) := \mu(E \cap P) \qquad \text \qquad (E) := - \mu(E \cap N)$

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

:$(E) = \sup_ \mu(B) \quad \text \quad (E) = - \inf_ \mu(B)$

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

:$\nu^ \geq \mu^ \quad \text \quad \nu^ \geq \mu^.$

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 se
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 $A_ := D$. Inductively assume for $n \in \mathbb_$ that $A_ \subseteq D$ has been constructed. Let

:$t_ := \sup(\)$

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

:$\mu(B_) \geq \min \! \left( 1,\frac \right).$

Set $A_ := A_ \setminus B_$ to finish the induction step. Finally, define

:$A := D \Bigg\backslash \bigcup_^ B_.$

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

:$\mu(D) = \mu(A) + \sum_^ \mu(B_) \geq \mu(A) + \sum_^ \min \! \left( 1,\frac \right)\geq \mu(A).$

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 $t_ \geq \mu(B)$ for every $n \in \mathbb_$, 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 $N_ = \varnothing$. Inductively, given $N_$, define

:$s_ := \inf(\).$

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

:$\mu(D_) \leq \max \! \left( \frac,- 1 \right) \leq 0.$

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

:$N := \bigcup_^ A_.$

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

:$\mu(B) = \sum_^ \mu(B \cap A_)$

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 $s_ \leq \mu(D)$ for all $n \in \mathbb_$ and

:$\mu(N) = \sum_^ \mu(A_) \leq \sum_^ \max \! \left( \frac,- 1 \right) = - \infty,$

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.

References

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External links

Hahn decomposition theorem
at PlanetMath.
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Jordan decomposition of a signed measure
a
Encyclopedia of Mathematics

{{Measure theory

Theorems in measure theory
Articles containing proofs