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

*P* ∪ *N* = *X* and *P* ∩ *N* = ∅.
- For each
*E* in Σ such that *E* ⊆ *P* one has *μ*(*E*) ≥ 0; that is, *P* is a positive set for *μ*.
- For each
*E* in Σ such that *E* ⊆ *N* one has *μ*(*E*) ≤ 0; that is, *N* is a negative set for *μ*.

Moreover, this decomposition is essentially unique, in the sense that for any other pair (*P*', *N*') of measurable sets fulfilling the above three conditions, the symmetric differences *P* Δ *P*' and *N* Δ *N*' are *μ*-null sets in the strong sense that every measurable subset of them has zero measure. The pair (*P*,*N*) is called a *Hahn decomposition* of the signed measure *μ*.

A consequence of the Hahn decomposition theorem is the *Jordan decomposition theorem*, which states that every signed measure *μ* has a *unique* decomposition into a difference μ = μ^{+} − μ^{−} of two positive measures *μ*^{+} and *μ*^{−}, at least one of which is finite, such that μ^{+}(E) = 0 if E ⊆ N and μ^{−}(E) = 0 if E ⊆ P for any Hahn decomposition (P,N) of μ. *μ*^{+} and *μ*^{−} are called 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
∩
P
)
and

μ
−
(
E
)
:=
−
μ
(
E
∩
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,

μ
+
(
E
)
=
sup
B
∈
Σ
,
B
⊂
E
μ
(
B
)
and

μ
−
(
E
)
=
−
inf
B
∈
Σ
,
B
⊂
E
μ
(
B
)
for any E in Σ. Also, if μ = ν^{+} − ν^{−} for a pair of finite non-negative measures (ν^{+}, ν^{−}), then

ν
+
≥
μ
+
and
ν
−
≥
μ
−
.
The last expression means that the Jordan decomposition is the *minimal* decomposition of μ 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).

**Preparation:** Assume that *μ* does not take the value −∞ (otherwise decompose according to −*μ*). As mentioned above, a negative set is a set *A* in Σ such that *μ*(*B*) ≤ 0 for every *B* in Σ which is a subset of *A*.

**Claim:** Suppose that a set *D* in Σ satisfies *μ*(*D*) ≤ 0. Then there is a negative set *A* ⊆ *D* such that *μ*(*A*) ≤ *μ*(*D*).

**Proof of the claim:** Define *A*_{0} = *D*. Inductively assume for a natural number *n* that *A*_{n} ⊆ *D* has been constructed. Let

t
n
=
sup
{
μ
(
B
)
:
B
∈
Σ
,
B
⊂
A
n
}
denote the supremum of *μ*(*B*) for all the measurable subsets *B* of *A*_{n}. This supremum might a priori be infinite. Since the empty set ∅ is a possible *B* in the definition of *t*_{n} and *μ*(∅) = 0, we have *t*_{n} ≥ 0. By definition of *t*_{n} there exists a *B*_{n} ⊆ *A*_{n} in Σ satisfying

μ
(
B
n
)
≥
min
{
1
,
t
n
/
2
}
.
Set *A*_{n+1} = *A*_{n} *B*_{n} to finish the induction step. Define

A
=
D
∖
⋃
n
=
0
∞
B
n
.
Since the sets (*B*_{n})_{n≥0} are disjoint subsets of *D*, it follows from the sigma additivity of the signed measure *μ* that

μ
(
A
)
=
μ
(
D
)
−
∑
n
=
0
∞
μ
(
B
n
)
≤
μ
(
D
)
−
∑
n
=
0
∞
min
{
1
,
t
n
/
2
}
.
This shows that *μ*(*A*) ≤ *μ*(*D*). Assume *A* were not a negative set. That means there exists a *B* in Σ which is a subset of *A* and satisfies *μ*(*B*) > 0. Then *t*_{n} ≥ *μ*(*B*) for every *n*, hence the series on the right has to diverge to +∞, which means *μ*(*A*) = –∞, which is not allowed. Therefore, *A* must be a negative set.

**Construction of the decomposition:** Set *N*_{0} = ∅. Inductively, given *N*_{n}, define

s
n
:=
inf
{
μ
(
D
)
:
D
∈
Σ
,
D
⊂
X
∖
N
n
}
.
as the infimum of *μ*(*D*) for all the measurable subsets *D* of *X* *N*_{n}. This infimum might a priori be –∞. Since the empty set is a possible *D* and *μ*(∅) = 0, we have *s*_{n} ≤ 0. Hence there exists a *D*_{n} in Σ with *D*_{n} ⊆ *X* *N*_{n} and

μ
(
D
n
)
≤
max
{
s
n
/
2
,
−
1
}
≤
0.
By the claim above, there is a negative set *A*_{n} ⊆ *D*_{n} such that *μ*(*A*_{n}) ≤ *μ*(*D*_{n}). Define *N*_{n+1} = *N*_{n} ∪ *A*_{n} to finish the induction step.

Define

N
=
⋃
n
=
0
∞
A
n
.
Since the sets (*A*_{n})_{n≥0} are disjoint, we have for every *B* ⊆ *N* in Σ that

μ
(
B
)
=
∑
n
=
0
∞
μ
(
B
∩
A
n
)
by the sigma additivity of *μ*. In particular, this shows that *N* is a negative set. Define *P* = *X* *N*. If *P* were not a positive set, there exists a *D* ⊆ *P* in Σ with *μ*(*D*) < 0. Then *s*_{n} ≤ *μ*(*D*) for all *n* and

μ
(
N
)
=
∑
n
=
0
∞
μ
(
A
n
)
≤
∑
n
=
0
∞
max
{
s
n
/
2
,
−
1
}
=
−
∞
,
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
∩
N
′
is a positive set and also a negative set. Therefore, every measurable subset of it has measure zero. The same applies to
N
∩
P
′
. Since

P
△
P
′
=
N
△
N
′
=
(
P
∩
N
′
)
∪
(
N
∩
P
′
)
,
this completes the proof. Q.E.D.