The extension of measures

In this paper we have to show that if A is any algebra of subsets of a set X and if μ is a measure defined on A then there exist a σ- algebra A^* containing A and a measure μ^* defined on A^* such that μ^* (E)=μ(E) in A. In other words the measure μ can be extended to a measure on a σ-algebra A^* of subsets of X, which contains A.

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Introduction 1.1 Definition

Let X be any set and 𝒜 be any non-empty subset of 𝑃(𝑋) the power set of X, 𝒜 is said to be an algebra if 1. 𝜙 and 𝑋 ∈ 𝒜 2. 𝐴, 𝐵 ∈ 𝒜 ⇒ 𝐴 ∪ 𝐵 ∈ 𝒜 3. 𝐴 ∈ 𝒜 ⇒ 𝐴 𝑐 ∈ 𝒜 Here it is easy to prove that if 𝒜 is an algebra and 𝐴 𝑎𝑛𝑑 𝐵 are in 𝒜 then 𝐴 ∩ 𝐵 and 𝐴/ 𝐵 are also in 𝒜 and also 𝐴 Δ 𝐵 = (𝐴 -𝐵) ∪ (𝐵 -𝐴) belongs to 𝒜. [𝑎 𝑖 , 𝑏 𝑖 ) where 0 ≤ 𝑎 𝑖 < 𝑏 𝑖 ≤ 𝑎 𝑖+1 ≤ 1 is an algebra. 2. If X be a finite set then consider the set E = {𝐴 ∈ P(X)/ A is finite or 𝐴 𝑐 is finite} then E is an algebra. It can be easily verified that E is closed under finite unions and finite intersections.

3.

Obviously for any set X the power set P(X) and E= { 𝜙, 𝑋 } are 𝜎 -𝑎𝑙gebras in X and P(X) is called the largest and E = { 𝜙, 𝑋 } is the smallest 𝜎 -𝑎𝑙gebra. (5) If (𝐸 𝑖 ) be a sequence of subsets of X, then

Proof: Statements (1), ( 2) and (3) are immediate consequences of the definition 1.1. (4) Since {A, 𝜙, 𝜙, … . . } is a countable collection of sets in 𝒜 whose union contains A, then it follows that 𝜇 * (𝐴) ≤ 𝜇 (𝐴) + 0 + 0 + …….….= 𝜇 (𝐴).

…………….. (1) Conversely, If (𝐸 𝑖 ) be any sequence of subsets from 𝒜 with 𝐴 ⊆ ∪ 𝐸 𝑖 , then 𝐴 = ∪ (A ∩ 𝐸 𝑖 ). Since 𝜇 is a measure on 𝒜, then

Hence from ( 1) and ( 2) we have 𝜇 * (𝐴) = 𝜇 (𝐴).

(5) Let 𝜀 > 0 be arbitrary and for each n chose a sequence (𝐸 𝑛𝑘 ) of the sets in 𝒜 such that

+ 𝜀. Since 𝜀 is arbitrary, hence proved.

Remark:

The property ( 5) is referred by saying that 𝜇 * in countably sub-additive. Although 𝜇 * has the advantage that it is defined for arbitrary subsets of X, it has the defect that it is not necessarily countably additive. Hence we are willing to restrict 𝜇 * to a smaller 𝜎 -algebra provided we can find one containing 𝒜 and over which 𝜇 * has the property of countable additivity. There is a remarkable condition due to Caratheodory which provides the desired restriction of the domain of 𝜇 * .

Definition:

A subset E of X is said to be to be 𝜇 * -measurable if

for all sub sets A of X. The collection of all 𝜇 * -measurable sets is denoted by 𝒜 * . Condition (A) indicated an addtivity property on 𝜇 *

Caratheodory Extension theorem:

The collection 𝒜 * of all 𝜇 * -measurable sets is a 𝜎 -algebra containing 𝒜. Moreover if (𝐸 𝑛 ) is a disjoint sequence in 𝒜 * , then

. Which shows that E ∩ F belongs to 𝒜 * . Since 𝒜 * is closed under intersection and complementation, it follows that 𝒜 * is an algebra.

Suppose that E, F ∈ 𝒜 * and E ∩ F = ∅. If we take A to be A ∩ (E ∪ F) in (A) we get

For A= X, this relation implies that 𝜇 * is additive on 𝒜 * .

We shall now show that 𝒜 * is a 𝜎 -algebra and that 𝜇 * is countably additive on 𝒜 * . Let (𝐸 𝑖 ) be a disjoint sequence in 𝒜 * and let E = ∪ 𝐸 𝑘 if A be any subset of X then

-𝜇 * (A\𝐹 𝑛 ). Since 𝐹 𝑛 ⊆ E, then A\E, then A\E ⊆A\𝐹 𝑛 and letting n → ∞ the above relation yields + 𝜇 * (A\𝐸).

In particular this shows that E = ⋃ k=1 ∞ 𝐸 𝑘 is 𝜇 * -measurable. On taking A= E, we obtain 𝜇 * ( ⋃

. Now it remains to prove that A ⊆ 𝐴 * . It was proved in lemma 1.5(4) that if E ∈ 𝒜, then 𝜇 * (𝐸) = 𝜇 (𝐸), but we need to show that E is 𝜇 * -measurable. Let A be an arbitrary subset of X; it follows from 1.5 (5) that 𝜇 * (A) ≤ 𝜇 * (𝐴 ∩ 𝐸) + 𝜇 * (𝐴\𝐸). To establish the opposite inequality, let 𝜀 > 0 be arbitrary and let (𝐹 𝑛 ) be a sequence in 𝒜 such that A ⊆ ∪ 𝐹 𝑛 and ∑ 𝜇(𝐹 𝑛 )

, 𝜇 * (𝐴\𝐸) ≤ ∑ 𝜇(𝐹 𝑛 \E).

Since 𝜀 is arbitrary, the desired inequality is established and the set E belongs to 𝒜 * . The Caratheodory Extension Theorem shows that a measure 𝜇 on an algebra 𝒜 can always be extended to a measure 𝜇 * on a 𝜎 -algebra 𝒜 * containing 𝒜. The 𝜎-algebra 𝒜 * obtained in this way is automatically complete in the sense that if E ∈ 𝒜 * with 𝜇 * (E) = 0 and if B ⊆ E, then B ∈ 𝒜 * and 𝜇 * (B) = 0. To prove this let A be an arbitrary subset of X and employ Lemma 1.5(3) to observe that 𝜇 * (A) = 𝜇 * (E) + 𝜇 * (A) ≥ 𝜇 * (A ∩ B) +𝜇 * (A\B); and as before the inequality 𝜇 * (A) ≤ 𝜇 * (A ∩ B) + 𝜇 * (A\B) follows from Lemma 1.5(5). Hence B is 𝜇 * -measurable and 0 ≤ 𝜇 * (B) ≤ 𝜇 * (E) ≤ 0. We shall now show that in the case that 𝜇 is a 𝜎finite measure, it has a unique extension to a measure on 𝒜 * .

Hahn Extension Theorem:

Suppose that 𝜇 is a 𝜎finite measure on an algebra 𝒜. Then there exist a unique extension of 𝜇 to a measure on 𝒜 * .

Proof:

The fact that 𝜇 * gives a measure on 𝒜 * proved in theorem 1.7 even without the 𝜎finiteness assumption. To establish the uniqueness, let 𝑣 be a measure on 𝒜 * which agrees with 𝜇 on 𝒜. . Therefore 𝑣(𝐸) ≤ 𝜇 * (E) for any E ∈ 𝒜 * . Since 𝜇 * and 𝑣 are additive. 𝜇 * (E) + 𝜇 * (X\E) + 𝑣(𝐸) + 𝑣(X\E). Since the terms on the right hand side are finite and not greater than the corresponding terms on the left hand side, we infer that 𝜇 * (E) + 𝑣(𝐸) for all E ∈ 𝒜 * . This establishes the uniqueness when 𝜇 is a finite measure. Suppose that 𝜇 is 𝜎-finite and let (F 𝑛 ) be an increasing sequence of sets in 𝒜 with 𝜇(F 𝑛 ) < ∞ and X = ∪ F 𝑛 . Then from the proceeding 𝜇 * (E ∩ F 𝑛 ) = 𝑣(E ∩ F 𝑛 ) for each E ∈ 𝒜 * . Therefore 𝜇 * (E) = lim 𝜇 * (E ∩ F 𝑛 ) = lim 𝑣(E ∩ F 𝑛 ) = 𝑣(E), hence 𝜇 and 𝜇 * agrees on 𝒜 * .

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