Signed Measures - Mathematics

Chapter 4

Signed Measures

Up until now our measures have always assumed values that were greater than or equal to 0. In this chapter we will extend our definition to allow for both positive and negative values.

4.1 Basic Properties of Signed Measures

Definition 4.1.1. Let (X, A) be a measurable space. A signed measure on (X, A) is a function ? : A R such that

1) ? takes on at most one of the values - or .

2) ?() = 0

3) If {En} n=1 A is a sequence of pairwise disjoint sets, then

?( En) = ?(En).

n=1

n=1

Example 4.1.2. 1) Assume that (X, A) is a measurable space and that ?1, ?2 are measures on (X, A). If at least one of the ?i's is finite, then ? = ?1 - ?2

is a signed measure on (X, A).

2) Let (X, A, ?) be a measure space and let f be an integrable function. Then if f = f + - f -, we have that

(E) = f d? = f + d? - f - d?

E

E

E

is a signed measure on (X, A).

4.2 Jordan and Hahn Decompositions

Problem 4.2.1. We have just seen that if (X, A) is a measurable space and ?1, ?2 are measures on (X, A) with at least one of the ?i's being finite, then

? = ?1 - ?2

is a signed measure on (X, A). We can now ask: do all signed measures arise in this way?

61

The clue to answering the problem above comes from the second item in our previous example where we decomposed our integral function f into its positive and negative parts. This decomposition, often called the Jordan decomposition of f , can in fact be extended to signed measures, something which we shall see shortly.

We begin with the following natural definitions:

Definition 4.2.2. Let ? be a signed measure on (X, A). Let P, N, M A. Then we say that: 1) P is positive if ?(E P ) 0 for all E A. 2) N is negative if ?(E N ) 0 for all E A. 3) M is null if ?(E M ) = 0 for all E A. The following useful results follow almost immediately from the definitions above. As such the proofs

will be left as exercises.

Proposition 4.2.3. Let ? be a signed measure on (X, A).

1) If P A is positive (negative) [null] and if E A is such that E P , then E is positive (negative) [null].

2) Let {En} n=1 A, with each En being positive (negative) [null], then E = En is also positive

n=1

(negative) [null].

3) Let P be positive and N be negative, then P N is null.

Lemma 4.2.4. Let ? be a signed measure on (X, A). Let E A be such that 0 < ?(E) < . Then there exists a positive set A A with A E and ?(A) > 0.

Proof. If E is positive we are done. So we may assume without loss of generality that E is not positive. Let n1 N be the smallest natural number such that there exists an E1 E with

1 ?(E1) < - n1 .

If E \ E1 is positive we are done. If not then we choose the smallest natural number n2 N such that there

exists an E2 E \ E1 with

1 ?(E2) < - n2 .

Suppose that we have chosen {E1, E2, . . . , Ek-1} and natural numbers {n1, n2, . . . , nk} as above and that

k-1

E \ Ej is not positive, then we choose the smallest natural number nk N such that there exists an

j=1

k-1

Ek E \ Ej with

j=1

1 ?(Ek) < - nk .

Now either this process terminates, in which case we are done, or we are able to inductively construct a seqeunce {Ek} as above. In the latter case, let

A = E \ Ek.

k=1

62

Since {Ek} is pairwise disjoint, and each Ek is disjoint from A, we have that

?(E) = ?(A) + ?(Ek) ()

k=1

From () we may conclude the following:

1) Since ?(E) is finite, the series ?(Ek) converges.

k=1

2) Since the series

?(Ek) converges, so does the series

1 nk

,

and

hence

nk

.

k=1

k=1

3) Since each ?(Ek) < 0, we have

0 < ?(E) < ?(A).

Assume that there exists an E0 A with ?(E0) < 0. Then we can find a k large enough so that

1 ?(E0) < - nk - 1

But we have that

k-1

E0 A E \ Ej

j=1

so this would contradict how we went about choosing nk. It follows that A is positive.

Theorem 4.2.5. [Hahn Decomposition Theorem] Let ? be a signed measure on (X, A). The there is a positive set A A and a negative set B A so that

X = A B and A B = .

Proof. Since ? cannot take on both ? we may assume without loss of generality that ? never takes on the value .

Let = sup{?(E)|E A, E is positive}.

Since is positive, we have that 0. We can also find a sequence of positve sets {Ai} so that

= lim ?(Ai).

i

Next let

A = Ai

i=1

Then A is also positive and since Ai A for each i we have

?(A) = .

Let B = X \ A. Assume that B is not negative. Then there is an E B with ?(E) > 0. From the previous lemma we can find a positive set A0 B with ?(A0) > 0. But then if

A = A A0 we have that A is also positive and since A A0 = ,

?(A) = ?(A) + ?(A0) >

which is impossible.

63

Definition 4.2.6. Let ? be a signed measure on (X, A). A pair {P, N } of elements in A for which P is positive N is negative, P N = X and P N = is called a Hahn decomposition of X with respect to ?.

Remark 4.2.7. It is generally the case that the Hahn decomposition is not unique. In fact, let X = [0, 1]

and

let

A

=

P (X ).

If

?1 2

is

the

point

mass

at

1 2

,

then

if

P

=

{

1 2

}

and

N

=

[0,

1]

\

{

1 2

},

then

{P, N }

is

a

Hahn

decomposition

of

[0, 1]

with

respect

to

?.

However,

P1

=

[0,

1 2

]

and

N1

=

(

1 2

,

1]

is

also

a

Hahn

decomposition.

In fact, if {P, N } is a Hahn decomposition of X with respect to ? and if M A is null, then {P M, N \M }

is a Hahn decomposition of X with respect to ?.

Furthermore, if {P1, N2} and {P2, N2} are Hahn decompositions of X with respect to ?, then

M = P1P2 = (P1 N2) (N1 P2) = N1N2

is a null set. Furthermore, since E P1 \ P2 P1P2 and E P2 \ P1 P1P2, it follows that

?(E P1) = ?(E P1 P2) = ?(E P2)

for each E A. Similarly, ?(E N1) = ?(E N1 N2) = ?(E N2)

for each E A. What is true however, as we shall see, is that every Hahn decomposition induced a decomposition of ? into the difference of two positive measures and that any two Hahn decompositions induce the same decomposition of ?.

Definition 4.2.8. Let {P, N } be a Hahn decomposition of X with respect to ?. Define ?+(E) = ?(E P )

and ?-(E) = -?(E N ).

Remark 4.2.9. It is easy to see that ?+ and ?- are both positive measures and that ? = ?+ - ?-.

The pair {?+, ?-} is called a Jordan decomposition of ?.

Definition 4.2.10. Two measures ? and on (X, A) are said to be mutually singular if there are disjoint sets A, B A with X = A B and ?(A) = 0 while (B) = 0. In this case, we write

? .

Theorem 4.2.11. [Jordan Decomposition Theorem] Let ? be a signed measure on (X, A). Then there exist two mutually singular positive measures ?+ and

?- such that ? = ?+ - ?-.

Furthermore, if and are any two positive measures with

? = - ,

then for each E A we have and Finally, if , then = ?+ and = ?-.

(E) ?+(E) (E) ?-(E).

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Proof. Let {?+, ?-} be a Jordan decomposition of ? arising from the Hahn decomposition {P, N } . Then

clearly

? = ?+ - ?-.

Moreover, since

?+(N ) = ?(P N ) = ?-(P ) = 0

we have that

?+ ?-.

Let and be any two positive measures with

?=-

and let E A. Then

?+(E) = ?(P E) = (P E) - (P E) (P E) (E)

A similar argument shows that (E) ?-(E). Finally, assume that . Let {A, B} be a partition of X so that (B) = 0 and (A) = 0. For each E A we have

?(E A) = (E A) - (E A) = (E A) 0.

That is A is positive. Similarly B is negative, so {A, B} is a Hahn decomposition. It follows that for an

E A,

?+(E) = ?(E P ) = ?(E A) = (E A) = (E)

and ?-(E) = -?(E N ) = -?(E B) = (E B) = (E).

Definition 4.2.12. Let ? be a signed measure on (X, A). Then if {?+, ?-} is the Jordan decomposition of ?, then the measure

|?| d=ef ?+ + ?-

is called the total variation of ?.

Example 4.2.13. Let (X, A, ?) be a measure space and let f L(X, A, ?) be integrable. If

(E) = f d?

E

then is a signed measure with Jordan decomposition

and In particular,

+(E) = f + d?

E

-(E) = f - d?.

E

||(E) = f + d? + f - d? = |f | d?.

E

E

E

65

Definition 4.2.14. If ? and are signed measures on (X, A), then we say that is absolutely continuous with respect to ? and write

? if || |?|. We say that ? and are mutually singular and write

?

if |?| ||.

Remark 4.2.15. 1) If is a signed measure and ? is a positive measure on (X, A), then we claim that ? if and only if whenever E A with ?(E) = 0, we have (E) = 0.

Clearly if ? and ?(E) = 0, we have (E) = 0. To prove the converse assume that {P, N } is a Hahn-decomposition for . Let E A be such that m(E) = 0. Then clearly

m(E P ) = 0 = m(E N ).

It follows from our assumption that

+(E) = (E P ) = 0

and -(E) = (E N ) = 0.

Therefore, we have that ||(E) = 0 and || ? as desired.

2) If is a signed measure and ? is a positive measure on (X, A), then we claim that ? if and only if there are disjoint sets A, B A with X = AB and ?(A) = 0 while (E) = 0 for any E B, E A.

If ?, then there are disjoint sets A, B A with X = A B and ?(A) = 0 while ||(B) = 0. But then if E B, E A, we have ||(E) = 0 and hence (E) = 0.

To see that the converse holds let A, B A be disjoint sets with X = A B and ?(A) = 0 while (E) = 0 for any E B, E A. Let {P, N } be a Hahn decomposition for . Let E1 = B P and E2 = B n. Then

?+(B) = ?(E1) = 0

and ?-(B) = ?(E2) = 0.

It follows that ||(B) = 0 and that ?.

The following proposition gives us an alternative characterization of absolute continuity for finite positive measures which will prove useful later.

Proposition 4.2.16. Let and ? be finite positive measures on (X, A). Then the following are equivalent:

1) ?

2) For every > 0 there exist a > 0 such that if E A and ?(E) < , then (E) < .

Proof. 1) 2). Suppose that 2 fails. Then we can find an 0 > 0 and a sequence of sets {Ek} A such

that

?(Ek )

<

1 2k

but (Ek)

0.

Now let

Fn = Ek.

k=n

66

It

follows

that

?(Fn) <

1 2n-1

while

(Fn)

0. From here, since {Fn} is decreasing, we get that

?( Fn) = lim ?(Fn) = 0,

n n=1

while

(

Fn)

=

lim

n

(Fn)

0.

n=1

This shows that is not absolutely continuous with respect to ?. 2) 1). Suppose 2) holds and ?(E) = 0. Then for any > 0 we have that ?(E) < where > 0 is

chosen as in 2). It follows that (E) < . But since > 0 was arbitrary, we get that (E) = 0

4.3 Radon-Nikodym Theorem

Remark 4.3.1. We had previously asked about when, given a measure space (X, A, ?), and any measure on A, does there exists an f M+(X, A) with the property that for every E A,

(E) = f d?

E

for all E A. We have also seen that if such an f exists then it must be the case that ?. At this point we are in a position to use what we have learned about decompositions of signed measures to establish the converse of this result in the case of -finite measures.

Theorem 4.3.2. [Radon-Nikodym Theorem] Let and ? be -finite measures on (X, A). Suppose that is absloutely continuous with respect to ?.

Then there exists f M+(X, A) such that

(E) = f d?

E

for every E A. Moreover f is uniquely determined ?-almost everywhere.

Proof. Case 1: We assume that , ? are finite. For each c > 0 let {P (c), N (c)} be an Hahn decomposition for the signed measure - c?. Let

A1 = N (c)

and for each k N let

k

Ak+1 = N ((k + 1)c) \ Ai.

i=1

It follows that {Ai} i=1 is pairwise disjoint and

k

k

N (ic) = Ai.

i=1

i=1

Consequently, we have

k-1

Ak = N (kc) P (ic).

i=1

If E A and E Ak, then E N (kc) and E P ((k - 1)c). As such we have

(k - 1)c?(E) (E) kc?(E). ()

67

Next, let

B = X \ Ai = X \ N (ic) = P (ic).

i=1

i=1

i=1

Since B P (kc) for all k N, we get that

0 kc?(B) (B) (X) <

for each k N. Therefore, ?(B) = 0 and since ? we have that (B) = 0, as well. We now define for each c > 0,

fc(x) =

(k - 1)c 0

if x Ak, if x B.

For each E A, we have

E = (E B) ( (E Ak)).

i=1

Applying () to each of the component pieces above, we have that

fc d? (E) (fc + c) d? fc d? + c?(X).

E

E

E

Now for each n N let

gn

=

f1 2n

.

We get

?(X )

gn d? (E)

E

gn d? +

E

2n

If we let m n then () tells us that

().

?(X )

gn d? (E)

E

gm d? +

E

2m

and

?(X )

gm d? (E)

E

gn d? +

E

2n

.

Combining these two give us that

?(X )

| gn d? - gm d?|

E

E

2n

for each E A. In particular this holds for E1 = {x X|gn - gm| 0} and E2 = {x X|gn - gm| < 0}.

This allows us to deduce that

2?(X) ?(X)

|gn - gm| d?

X

2n

= 2n-1

and hence that {gn} n=1 is Cauchy in L1(X, A, ?). Assume that gn f in L1(X, A, ?). Since gn M+(X, A) we can also assume that f M+(X, A).

Moreover, for any E A we have

| gn d? - f d?| |gn - f | d? gn - f 1 0.

E

E

E

It then follows from () that

(E) = lim

n

gn d? =

E

f d?.

E

Suppose now that f, h M+(X, A) are such that

f d? = (E) = h d?

E

E

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