Signed Measure: Hahn and Jordan Decomposition

Signed Measure: Hahn and Jordan decomposition

Yongheng Zhang

Recall that an abstract measure is a nonnegative extended real valued function defined on a measurable space (X, B). In contrast, a signed measure may also take negative values. A signed measure ν by definition satisfies the following properties: (1) ν assumes at most one of the values −∞ and ∞. (2) ν(∅) = 0. S∞ P∞ (3) ν( i=1 Ei) = i=1 ν(Ei) for any sequence hEii of disjoint measurable sets. And we require S∞ that if ν( i=1 Ei) is finite then the series converges absolutely.

Property (1) suﬃces to avoid ∞ − ∞. Property (2) makes sure we have at least one positive set (this notion will be explained shortly). Property (3) is crucial for the existence of a positive set with positive measure contained in a measurable set with ﬁnite positive measure.

We say a set A is positive w.r.t. a signed measure ν if A is measurable and all measurable subsets of A have nonnegative measure. Similarly, B is called negative if it is measurable and every measurable subset of it has nonpositive measure. A set C is called a null set if it is both positive and negative. Notice that positivity, negativity and nullity are notions involving not just the single set itself but a spectrum of its subsets.

Obviously, every measurable subset of a positive set is positive and a union of a countable col- lection of disjoint positive sets is also positive (by property (3)). If we have a countable collec- tion of positive sets hPii which are not necessarily disjoint, to show their union is still positive, let S Si−1 E be an arbitrary measurable set of Pi, and deﬁne Ei = E ∪ Pi\ n=1 Pn. Then hEii is a sequence of disjoint measurable sets whose union is E and each Ei is contained in positive Pi. Thus, S P ν(E) = ν( Ei) = ν(Ei) ≥ 0 by (3).

It will be eventually proved that given a signed measure space (X, B, ν), there is a positive set A and a negative set B such that A and B partition X. This is called the Hahn Decomposition Theorem. Before proving that, we justify an important lemma. Lemma 1. Let E be a measurable set having finite positive measure. Then there is a positive set P having positive measure contained in E. Proof. If E itself is positive, then we are done. Suppose E is not postive, then it has a measurable subset of negative measure. Let n1 ∈ Z+ be the smallest such that there is a measurable set E1 1 with ν(E1) < − n (If we make n smaller, there is no E1 satisfying the condition.). Then consider E\E1. First, ν(E\E1) = ν(E) − ν(E1) > ν(E) > 0. If E\E1 is positive, then we turn off the proof. Sk−1 Inductively, if E\ i=1 Ei is not positive, let nk ∈ Z+ be the smallest such that there is a measurable 1 S∞ S set Ek with ν(Ek) < − n . If we never stop, let P = E\ i=1 Ei. Then E = P ∪ ( Ei). Notice k P that P and hEii are disjoint, so by (3) ν(E) = ν(P ) + ν(Ei). But ν(E) < ∞, by definition (3), P 1 ν(Ei) is absolutely convergent. Hence, limi→∞ ni = 0. Let > 0. So there is k such that < . nk−1 1 Sk−1 Hence, − < − . Notice that P is contained in E\ Ei. If A contains a measurable set nk−1 i=1 Sk−1 1 which is henceforth contained in E\ Ei with measure less than − and thus less than − i=1 nk−1 1 (and then less than − :this is useless), then nk is no longer the smallest positive integer making the nk existence of a measurable set having measure less than the negative of its reciprocal (but nk − 1 is!). 1 2

Therefore, P contains no measurable set having measure less than −. Since is arbitrary, P contains no measurable set having negative measure. This shows P is positive.

P We also know ν(P ) = ν(E) − ν(Ei) > ν(E) > 0. This completes the proof.

We don’t need to consider the case when nk = 1 so that there is no way to make nk − 1 ∈ Z+ since we only concern the case when is very small.

Now we prove the Hahn Decomposition Theorem. Theorem 2 (Hahn Decomposition Theorem). Let (x, B, ν) be a signed measure space. Then there is a positive set P and a negative set N such that X = P ∪ N and P ∩ N = ∅.

Proof. We may assume ν never takes ∞. Let p = supP is positiveν(P ). Because ∅ is positive, p ≥ 0. By the definition of sup, there is a sequence of positive sets hPiisuch that p = limi→∞ ν(Pi). Let S P = Pi. So P is positive. Hence, p ≥ ν(P ). But for any i, ν(A) = ν(Ai) + ν(A\Ai) ≥ ν(Ai). Therefore, ν(A) ≥ p. Thus, 0 ≤ ν(A) = p < ∞. Let N = X\P . We claim that N is negative and then we are done. Suppose E is a positive subset of N. So E and P are disjoint and E ∪ P is positive. Hence, p ≥ ν(E ∪ P ) = ν(E) + ν(p) = ν(E) + p. this shows ν(E) = 0. Therefore, N contains no positive subsets of positive measure. Consequenetly, by the contrapositive of the lemma, N contains no subsets of positive measure. The Hahn decomposition of a measurable space associated with a signed measure exists by the above theorem. However, it is not unique. For example. Let m be the Lebesgue measure on R. Define ν1(E) = m(E ∩[−1, 0]) and ν2(E) = m(E ∩[0, 1]). Let ν = ν1 −ν2. First of all, (−∞, 0] and (0, ∞) is a Hahn decomposition because for any measurable E ⊂ (−∞, 0], ν(E) = ν1(E∩[−1, 0])−ν2(E∩[0, 1]) = ν1(E ∩ [−1, 0]) ≥ 0 and any F ⊂ (0, ∞), ν(F ) = ν1(∅) − ν2(F ∩ [0, 1]) ≤ 0. On the other hand, it can be similarly checked that [−1, 0] and (−∞, −1) ∪ (0, ∞) is another Hahn decomposition. Nonetheless, Hahn decomposition is unique except for null sets. In the above example (−∞, −1) is a null set and the difference between the two composition is exactly due to the freedom of (−∞, −1)

If ν is a signed measure and P and N is a Hahn decomposition. Then we may define ν+ by ν+(E) = ν(E ∩ P ) and ν− by ν−(E) = −ν(E ∩ N).So ν = ν+ − ν−. Notice that ν+ and ν− are measures and they are mutually singular in the sense that there is binary partition {A, B} of X (in this case A = N and B = P ) such that ν+(A) = ν−(B) = 0. A decomposition of the measure ν as a difference of two mutually singlular measures ν+ and ν− exampled previously is called a Jordan decomposition. However, it is worthy to note that although we defined ν+ and ν− in terms of a particular Hahn decomposition of X, the Jordan decomposition is actually independent of the Hahn decomposition and there is only one pair of such decomposition. This may seem a little confusing. To see it clearly (the Jordan decomposition is unique), let ν = ν+ − ν− be a Jordan decomposition. Then by defintion, there is a partition A and B of X such that ν+(A) = ν−(B) = 0. We claim {B,A} is a Hahn decomposition. [Let E ⊂ B, then ν−(E) ≤ ν−(B) = 0, i.e., ν−(E) = 0. Thus, ν(E) = ν+(E) − ν−(E) ≥ 0. This shows B is positive. Similarly, A is negative.] Then since Hahn decomposition is up to null sets, the Jordan decomposition is unique.