Measures

Measure

Suppose that

X

is a set and

𝒮

is a

σ

-algebra on

X

. A measure on

(X, 𝒮)

is a function

μ : 𝒮 \to [0, \infty]

such that

μ (\emptyset) = 0

and

μ (⋃_{k = 1}^{\infty} E_{k}) = \sum_{k = 1}^{\infty} μ (E_{k})

for every pairwise disjoint sequence

E_{1}, E_{2}, \dots \in 𝒮

Measure Space

A measure space is an ordered triple

(X, 𝒮, μ)

, where

X

is a set,

𝒮

is a

σ

-algebra on

X

, and

μ

is a measure on

(X, 𝒮)

Measure Preserves Order and Set Difference

Suppose that

(X, 𝒮, μ)

is a measure space and

D, E \in 𝒮

with

D \subseteq E

. Then

$μ (D) \leq μ (E)$ ,
if $μ (D) < \infty$ , then $μ (E ∖ D) = μ (E) - μ (D)$ .

Since $D \subseteq E$ , we can write $E = D ⊔ (E ∖ D)$ , where the union is disjoint. By the definition of a measure, $μ (E) = μ (D) + μ (E ∖ D) .$ Since $μ (E ∖ D) \geq 0$ , we get $μ (D) \leq μ (E)$ . If $μ (D) < \infty$ , subtracting $μ (D)$ from both sides gives $μ (E ∖ D) = μ (E) - μ (D)$ .

Countable Subadditivity of Measures

Suppose that

(X, 𝒮, μ)

is a measure space and

E_{1}, E_{2}, \dots \in 𝒮

. Then

μ (⋃_{k = 1}^{\infty} E_{k}) \leq \sum_{k = 1}^{\infty} μ (E_{k}) .

Define disjoint sets $F_{1} : = E_{1}, F_{k} : = E_{k} ∖ ⋃_{j = 1}^{k - 1} E_{j} for k \geq 2.$ Each $F_{k} \in 𝒮$ because $σ$ -algebras are closed under standard set operations. Also $F_{k} \subseteq E_{k}$ , and $⋃_{k = 1}^{\infty} E_{k} = ⋃_{k = 1}^{\infty} F_{k} .$ By countable additivity and order preservation, $μ (⋃_{k = 1}^{\infty} E_{k}) = \sum_{k = 1}^{\infty} μ (F_{k}) \leq \sum_{k = 1}^{\infty} μ (E_{k}) .$

Measure of an Increasing Union

Suppose that

(X, 𝒮, μ)

is a measure space and

E_{1} \subseteq E_{2} \subseteq \dots

is an increasing sequence in

𝒮

. Then

μ (⋃_{k = 1}^{\infty} E_{k}) = \lim_{k \to \infty} μ (E_{k}) .

Set $F_{1} : = E_{1}$ and $F_{k} : = E_{k} ∖ E_{k - 1}$ for $k \geq 2$ . Then $F_{1}, F_{2}, \dots$ are pairwise disjoint, each belongs to $𝒮$ , and $⋃_{k = 1}^{\infty} E_{k} = ⋃_{k = 1}^{\infty} F_{k} .$ By the definition of a measure, $μ (⋃_{k = 1}^{\infty} E_{k}) = \sum_{k = 1}^{\infty} μ (F_{k}) = \lim_{n \to \infty} \sum_{k = 1}^{n} μ (F_{k}) .$ For each $n$ , the finite disjoint union $F_{1} ⊔ \dots ⊔ F_{n}$ equals $E_{n}$ , so $\sum_{k = 1}^{n} μ (F_{k}) = μ (E_{n})$ . Therefore $μ (⋃_{k = 1}^{\infty} E_{k}) = \lim_{n \to \infty} μ (E_{n}) .$

Measure of a Decreasing Intersection

Suppose that

(X, 𝒮, μ)

is a measure space and

E_{1} \supseteq E_{2} \supseteq \dots

is a decreasing sequence in

𝒮

. If

μ (E_{1}) < \infty

, then

μ (⋂_{k = 1}^{\infty} E_{k}) = \lim_{k \to \infty} μ (E_{k}) .

Let $F_{k} : = E_{1} ∖ E_{k} .$ Then $F_{1} \subseteq F_{2} \subseteq \dots$ , and $⋃_{k = 1}^{\infty} F_{k} = E_{1} ∖ ⋂_{k = 1}^{\infty} E_{k} .$ By measure of an increasing union, $μ (E_{1} ∖ ⋂_{k = 1}^{\infty} E_{k}) = \lim_{k \to \infty} μ (E_{1} ∖ E_{k}) .$ Since $μ (E_{1}) < \infty$ , set-difference for finite measure gives $μ (E_{1}) - μ (⋂_{k = 1}^{\infty} E_{k}) = \lim_{k \to \infty} (μ (E_{1}) - μ (E_{k})) .$ Cancelling the finite value $μ (E_{1})$ gives the desired equality.

Measure of a Union

Suppose that

(X, 𝒮, μ)

is a measure space and

D, E \in 𝒮

. If

μ (D \cap E) < \infty

, then

μ (D \cup E) = μ (D) + μ (E) - μ (D \cap E) .

Write the union as the disjoint union $D \cup E = D ⊔ (E ∖ (D \cap E)) .$ By finite additivity, $μ (D \cup E) = μ (D) + μ (E ∖ (D \cap E)) .$ Since $μ (D \cap E) < \infty$ , set-difference for finite measure gives $μ (E ∖ (D \cap E)) = μ (E) - μ (D \cap E) .$ Substituting this into the previous equation proves the formula.