Measurability theorem
In mathematics, the Hahn decomposition theorem, named after the Austrian mathematician Hans Hahn, states that for any measurable space
and any signed measure
defined on the
-algebra
, there exist two
-measurable sets,
and
, of
such that:
and
. - For every
such that
, one has
, i.e.,
is a positive set for
. - For every
such that
, one has
, i.e.,
is a negative set for
.
Moreover, this decomposition is essentially unique, meaning that for any other pair
of
-measurable subsets of
fulfilling the three conditions above, the symmetric differences
and
are
-null sets in the strong sense that every
-measurable subset of them has zero measure. The pair
is then called a Hahn decomposition of the signed measure
.
Jordan measure decomposition
A consequence of the Hahn decomposition theorem is the Jordan decomposition theorem, which states that every signed measure
defined on
has a unique decomposition into a difference
of two positive measures,
and
, at least one of which is finite, such that
for every
-measurable subset
and
for every
-measurable subset
, for any Hahn decomposition
of
. We call
and
the positive and negative part of
, respectively. The pair
is called a Jordan decomposition (or sometimes Hahn–Jordan decomposition) of
. The two measures can be defined as
![{\displaystyle {\mu ^{+}}(E):=\mu (E\cap P)\qquad {\text{and}}\qquad {\mu ^{-}}(E):=-\mu (E\cap N)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5abefc6545bc60b9070dcd82c3eca34b069a34d2)
for every
and any Hahn decomposition
of
.
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
of a finite signed measure
, one has
![{\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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/44ed48f00d4afc5456dc61badab418e54b2e7508)
for any
in
. Furthermore, if
for a pair
of finite non-negative measures on
, then
![{\displaystyle \nu ^{+}\geq \mu ^{+}\quad {\text{and}}\quad \nu ^{-}\geq \mu ^{-}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5e66317be010af8d4defbb4a7dfd872c4940e0e3)
The last expression means that the Jordan decomposition is the minimal decomposition of
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
does not take the value
(otherwise decompose according to
). As mentioned above, a negative set is a set
such that
for every
-measurable subset
.
Claim: Suppose that
satisfies
. Then there is a negative set
such that
.
Proof of the claim: Define
. Inductively assume for
that
has been constructed. Let
![{\displaystyle t_{n}:=\sup(\{\mu (B)\mid B\in \Sigma ~{\text{and}}~B\subseteq A_{n}\})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/03f5aab9e523a4ad69346f5c74a4ad7a5d13179d)
denote the supremum of
over all the
-measurable subsets
of
. This supremum might a priori be infinite. As the empty set
is a possible candidate for
in the definition of
, and as
, we have
. By the definition of
, there then exists a
-measurable subset
satisfying
![{\displaystyle \mu (B_{n})\geq \min \!\left(1,{\frac {t_{n}}{2}}\right).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/64147327eb03a0f973bfbfe5a5e78962f155d59e)
Set
to finish the induction step. Finally, define
![{\displaystyle A:=D{\Bigg \backslash }\bigcup _{n=0}^{\infty }B_{n}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8b055839821f6a9551a04d4c2179ccfbad557547)
As the sets
are disjoint subsets of
, it follows from the sigma additivity of the signed measure
that
![{\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).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fea4bf81741494e40ad6eb427514aae7a4abafeb)
This shows that
. Assume
were not a negative set. This means that there would exist a
-measurable subset
that satisfies
. Then
for every
, so the series on the right would have to diverge to
, implying that
, which is a contradiction, since
. Therefore,
must be a negative set.
Construction of the decomposition: Set
. Inductively, given
, define
![{\displaystyle s_{n}:=\inf(\{\mu (D)\mid D\in \Sigma ~{\text{and}}~D\subseteq X\setminus N_{n}\}).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3dc7ae5b534d6761d274073cc30838fa4388877d)
as the infimum of
over all the
-measurable subsets
of
. This infimum might a priori be
. As
is a possible candidate for
in the definition of
, and as
, we have
. Hence, there exists a
-measurable subset
such that
![{\displaystyle \mu (D_{n})\leq \max \!\left({\frac {s_{n}}{2}},-1\right)\leq 0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6e48ad3a61f68de0a4d13d08a5c55d77f34acbb2)
By the claim above, there is a negative set
such that
. Set
to finish the induction step. Finally, define
![{\displaystyle N:=\bigcup _{n=0}^{\infty }A_{n}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c61168e2bcbde8996134f03cdfe0156398738e1)
As the sets
are disjoint, we have for every
-measurable subset
that
![{\displaystyle \mu (B)=\sum _{n=0}^{\infty }\mu (B\cap A_{n})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9abf05e33b2af484b624a12839aa418e6ee54843)
by the sigma additivity of
. In particular, this shows that
is a negative set. Next, define
. If
were not a positive set, there would exist a
-measurable subset
with
. Then
for all
and[clarification needed]
![{\displaystyle \mu (N)=\sum _{n=0}^{\infty }\mu (A_{n})\leq \sum _{n=0}^{\infty }\max \!\left({\frac {s_{n}}{2}},-1\right)=-\infty ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dedd1764d7ae1a6ab6f6312af69a0c7f67cba427)
which is not allowed for
. Therefore,
is a positive set.
Proof of the uniqueness statement: Suppose that
is another Hahn decomposition of
. Then
is a positive set and also a negative set. Therefore, every measurable subset of it has measure zero. The same applies to
. As
![{\displaystyle P\triangle P'=N\triangle N'=(P\cap N')\cup (N\cap P'),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a83d3c608cd952e56f3aef08e8b9a4d04b6ef464)
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]
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