- Hahn decomposition theorem
In
mathematics , the Hahn decomposition theorem, named after theAustria nmathematician Hans Hahn , states that given a measurable space ("X",Σ) and asigned measure "μ" defined on the σ-algebra Σ, there exist two 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, thesymmetric difference s "P" Δ "P"' and "N" Δ "N"' are "μ"-null set s 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".Hahn–Jordan decomposition
A consequence of this theorem is the "Jordan decomposition theorem", which states that every signed measure "μ" can be expressed as a difference of two positive measures "μ"+ and "μ"–, at least one of which is finite; "μ"+ and "μ"– are called the "positive" and "negative part" of "μ", respectively. The two measures can be defined as:and:for every "E" in Σ, and it is an easy task to verify that both "μ"+ and "μ"– are positive measures on the space ("X",Σ), at least one of them is finite (since "μ" cannot take both +∞ and −∞ as values), and satisfy "μ" = "μ"+ − "μ"–. The pair ("μ"+, "μ"–) is called the "Jordan decomposition" (or sometimes "Hahn–Jordan decomposition") of "μ".
Proof of the Hahn decomposition theorem
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 "An" ⊆ "D" has been constructed. Let
:
denote the
supremum of "μ"("B") for all the measurable subsets "B" of "An". This supremum might a priori be infinite. Since the empty set ∅ is a possible "B" in the definition of "tn" and "μ"(∅) = 0, we have "tn" ≥ 0. By definition of "tn" there exists a "Bn" ⊆ "An" in Σ satisfying:
Set "A""n"+1 = "An" "Bn" to finish the induction step. Define
:
Since the sets ("Bn")"n"≥0 are disjoint subsets of "D", it follows from the
sigma additivity of the signed measure "μ" that:
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 "tn" ≥ "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 "Nn", define
:
as the
infimum of "μ"("D") for all the measurable subsets "D" of "X" "Nn". This infimum might a priori be –∞.Since the empty set is a possible "D" and "μ"(∅) = 0, we have "sn" ≤ 0. Hence there exists a "Dn" in Σ with "Dn" ⊆ "X" "Nn" and:
By the claim above, there is a negative set "An" ⊆ "Dn" such that "μ"("An") ≤ "μ"("Dn"). Define "N""n"+1 = "Nn" ∪ "An"to finish the induction step.
Define
:
Since the sets ("An")"n"≥0 are disjoint, we have for every "B" ⊆ "N" in Σ that
:
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 "sn" ≤ "μ"("D") for all "n" and
:
which is not allowed for "μ". Therefore, "P" 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 . Since
:
this completes the proof.
Q.E.D. External links
* [http://planetmath.org/?op=getobj&from=objects&id=4014 Hahn decomposition theorem] at
PlanetMath .
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