Measurable Spaces And Functions

Sigma Algebra

Suppose that

X

is a set. A collection

𝒮 \subseteq 𝒫 (X)

is a $σ$ -algebra on

X

$\emptyset \in 𝒮$ ,
if $E \in 𝒮$ , then $X ∖ E \in 𝒮$ ,
if $E_{1}, E_{2}, \dots \in 𝒮$ is a sequence, then $⋃_{k = 1}^{\infty} E_{k} \in 𝒮$ .

Sigma Algebras are Closed under Standard Set Operations

Suppose that

𝒮

is a

σ

-algebra on

X

. Then

$X \in 𝒮$ ,
if $D, E \in 𝒮$ , then $D \cup E \in 𝒮$ , $D \cap E \in 𝒮$ , and $D ∖ E \in 𝒮$ ,
if $E_{1}, E_{2}, \dots \in 𝒮$ , then $⋂_{k = 1}^{\infty} E_{k} \in 𝒮$ .

Since $\emptyset \in 𝒮$ and $𝒮$ is closed under complements, $X = X ∖ \emptyset \in 𝒮$ by the definition of a $σ$ -algebra.

If $D, E \in 𝒮$ , then $D \cup E \in 𝒮$ by closure under countable unions, using $\emptyset$ for all remaining terms. Also, $D \cap E = X ∖ ((X ∖ D) \cup (X ∖ E)),$ so $D \cap E \in 𝒮$ . Finally, $D ∖ E = D \cap (X ∖ E),$ so $D ∖ E \in 𝒮$ .

If $E_{1}, E_{2}, \dots \in 𝒮$ , then by De Morgan's law, $⋂_{k = 1}^{\infty} E_{k} = X ∖ ⋃_{k = 1}^{\infty} (X ∖ E_{k}) .$ The right side belongs to $𝒮$ , again by the defining closure properties of a $σ$ -algebra.

Measurable Space

A measurable space is an ordered pair

(X, 𝒮)

, where

X

is a set and

𝒮

is a

σ

-algebra on

X

Measurable Set

Suppose that

(X, 𝒮)

is a measurable space. A set

E \subseteq X

is $𝒮$ -measurable if

E \in 𝒮

Smallest Sigma Algebra Containing a Collection

Suppose that

X

is a set and

𝒜 \subseteq 𝒫 (X)

. The intersection of all

σ

-algebras on

X

that contain

𝒜

is a

σ

-algebra on

X

Let $𝒯$ be the intersection of all $σ$ -algebras on $X$ that contain $𝒜$ . This collection is nonempty because $𝒫 (X)$ itself is a $σ$ -algebra containing $𝒜$ .

Every $σ$ -algebra in the intersection contains $\emptyset$ , so $\emptyset \in 𝒯$ . If $E \in 𝒯$ , then $E$ belongs to every $σ$ -algebra in the intersection, so $X ∖ E$ belongs to every one of them, and hence $X ∖ E \in 𝒯$ . If $E_{1}, E_{2}, \dots \in 𝒯$ , then each $E_{k}$ belongs to every $σ$ -algebra in the intersection, so $⋃_{k = 1}^{\infty} E_{k}$ belongs to every one of them, and hence belongs to $𝒯$ . Thus $𝒯$ is a $σ$ -algebra on $X$ .

Borel

σ

-Algebra

The Borel $σ$ -algebra on

ℝ

is the smallest

σ

-algebra on

ℝ

that contains every open subset of

ℝ

Borel Set

A Borel set is an element of the Borel

σ

-algebra.

Singletons are Borel Sets

For every

x \in ℝ

, the singleton set

{x}

is a Borel set.

The set

ℝ ∖ {x} = (- \infty, x) \cup (x, \infty)

is open. Since the Borel

σ

-algebra contains every open subset of

ℝ

, we have

ℝ ∖ {x} \in ℬ

. Since a

σ

-algebra is closed under set difference from the ambient set,

{x} = ℝ ∖ (ℝ ∖ {x}) \in ℬ .

Therefore

{x}

is a Borel set.

Inverse Image

Suppose that

f : X \to Y

is a function and

A \subseteq Y

. The inverse image of

A

under

f

f^{- 1} (A) : = {x \in X : f (x) \in A} .

Algebra of Inverse Images

Suppose that

f : X \to Y

. Then

$f^{- 1} (Y ∖ A) = X ∖ f^{- 1} (A)$ for every $A \subseteq Y$ ,
$f^{- 1} (⋃_{A \in 𝒜} A) = ⋃_{A \in 𝒜} f^{- 1} (A)$ for every collection $𝒜 \subseteq 𝒫 (Y)$ ,
$f^{- 1} (⋂_{A \in 𝒜} A) = ⋂_{A \in 𝒜} f^{- 1} (A)$ for every collection $𝒜 \subseteq 𝒫 (Y)$ .

We prove each identity by element chasing from the definition of inverse image.

For the complement identity, $x \in f^{- 1} (Y ∖ A)$ if and only if $f (x) \in Y ∖ A$ , if and only if $f (x) \notin A$ , if and only if $x \in X ∖ f^{- 1} (A)$ .

For unions, $x \in f^{- 1} (⋃_{A \in 𝒜} A)$ if and only if $f (x) \in A$ for some $A \in 𝒜$ , if and only if $x \in f^{- 1} (A)$ for some $A \in 𝒜$ , if and only if $x \in ⋃_{A \in 𝒜} f^{- 1} (A) .$ The proof for intersections is the same with "for some" replaced by "for every".

Inverse Image of a Composition

Suppose that

f : X \to Y

g : Y \to W

, and

A \subseteq W

. Then

{(g \circ f)}^{- 1} (A) = f^{- 1} (g^{- 1} (A)) .

For $x \in X$ , we have $x \in {(g \circ f)}^{- 1} (A)$ if and only if $g (f (x)) \in A$ , if and only if $f (x) \in g^{- 1} (A)$ , if and only if $x \in f^{- 1} (g^{- 1} (A)) .$ Therefore the two sets are equal by set equality by elements.

Measurable Function

Suppose that

(X, 𝒮)

is a measurable space. A function

f : X \to ℝ

is $𝒮$ -measurable if

f^{- 1} (B) \in 𝒮

for every Borel set

B \subseteq ℝ

Characteristic Function

Suppose that

E \subseteq X

. The characteristic function of

E

is the function

χ_{E} : X \to ℝ

defined by

χ_{E} (x) = {\begin{matrix} 1 & x \in E, \\ 0 & x \notin E . \end{matrix}

Ray Test for Measurability

Suppose that

(X, 𝒮)

is a measurable space and

f : X \to ℝ

. If

f^{- 1} ((a, \infty)) \in 𝒮

for every

a \in ℝ

, then

f

𝒮

-measurable.

Let $𝒯 : = {B \subseteq ℝ : f^{- 1} (B) \in 𝒮} .$ By the algebra of inverse images and the definition of a $σ$ -algebra, the collection $𝒯$ is a $σ$ -algebra on $ℝ$ . By hypothesis, $(a, \infty) \in 𝒯$ for every $a \in ℝ$ . The $σ$ -algebra generated by these rays is the Borel $σ$ -algebra, so every Borel set belongs to $𝒯$ . Hence $f$ is $𝒮$ -measurable.

Borel Measurable Function

Suppose that

X \subseteq ℝ

. A function

f : X \to ℝ

is Borel measurable if

f^{- 1} (B)

is a Borel set for every Borel set

B \subseteq ℝ

Continuous Functions are Borel Measurable

Every continuous real-valued function whose domain is a Borel subset of

ℝ

is Borel measurable.

Let $X \subseteq ℝ$ be a Borel set, and let $f : X \to ℝ$ be continuous. By the ray test for measurability, it is enough to prove that $f^{- 1} ((a, \infty))$ is Borel for each $a \in ℝ$ .

If $x \in f^{- 1} ((a, \infty))$ , then by continuity, there is $δ_{x} \in ℝ^{+}$ such that $f (y) > a$ whenever $y \in X \cap (x - δ_{x}, x + δ_{x})$ . Therefore $f^{- 1} ((a, \infty)) = X \cap ⋃_{x \in f^{- 1} ((a, \infty))} (x - δ_{x}, x + δ_{x}) .$ The union is open in $ℝ$ , hence Borel, and its intersection with the Borel set $X$ is Borel by closure of $σ$ -algebras under intersections. Thus $f$ is Borel measurable.

Increasing Function

Suppose that

X \subseteq ℝ

and

f : X \to ℝ

. The function

f

is increasing if

f (x) \leq f (y)

whenever

x, y \in X

and

x < y

Strictly Increasing Function

Suppose that

X \subseteq ℝ

and

f : X \to ℝ

. The function

f

is strictly increasing if

f (x) < f (y)

whenever

x, y \in X

and

x < y

Increasing Functions are Borel Measurable

Every increasing real-valued function whose domain is a Borel subset of

ℝ

is Borel measurable.

Let $X \subseteq ℝ$ be Borel and let $f : X \to ℝ$ be increasing. By the ray test, it is enough to show that $f^{- 1} ((a, \infty))$ is Borel for each $a \in ℝ$ .

If this inverse image is empty, it is Borel. Otherwise let $b = \inf f^{- 1} ((a, \infty))$ . Since $f$ is increasing, the set $f^{- 1} ((a, \infty))$ is either $X \cap (b, \infty)$ or $X \cap [b, \infty)$ . Both are Borel because $X$ is Borel and the Borel $σ$ -algebra is closed under intersections. Therefore $f$ is Borel measurable.

Composition of Measurable Functions

Suppose that

(X, 𝒮)

is a measurable space and

f : X \to ℝ

𝒮

-measurable. If

g

is a Borel measurable real-valued function on a subset of

ℝ

that contains the range of

f

, then

g \circ f

𝒮

-measurable.

Let $B \subseteq ℝ$ be a Borel set. By the inverse image formula for a composition, ${(g \circ f)}^{- 1} (B) = f^{- 1} (g^{- 1} (B)) .$ Since $g$ is Borel measurable, $g^{- 1} (B)$ is Borel. Since $f$ is $𝒮$ -measurable, the right side belongs to $𝒮$ . Thus $g \circ f$ is $𝒮$ -measurable.

Algebraic Operations with Measurable Functions

Suppose that

(X, 𝒮)

is a measurable space and

f, g : X \to ℝ

are

𝒮

-measurable. Then

f + g

f - g

, and

f g

are

𝒮

-measurable. If

g (x) \neq 0

for every

x \in X

, then

f / g

𝒮

-measurable.

For addition, fix $a \in ℝ$ . We have ${(f + g)}^{- 1} ((a, \infty)) = ⋃_{r \in ℚ} (f^{- 1} ((r, \infty)) \cap g^{- 1} ((a - r, \infty))) .$ The equality follows because if $f (x) + g (x) > a$ , then some rational $r$ lies between $a - g (x)$ and $f (x)$ . The right side belongs to $𝒮$ because $f$ and $g$ are measurable, $ℚ$ is countable, and $𝒮$ is a $σ$ -algebra. Thus $f + g$ is measurable by the ray test.

The function $- g$ is measurable because it is the composition of $g$ with the continuous function $x \mapsto - x$ , using continuous functions are Borel measurable and composition of measurable functions. Hence $f - g = f + (- g)$ is measurable.

The square of a measurable function is measurable for the same composition reason, using $x \mapsto x^{2}$ . Since $f g = \frac{{(f + g)}^{2} - f^{2} - g^{2}}{2},$ the product $f g$ is measurable. If $g (x) \neq 0$ for all $x$ , then $1 / g$ is the composition of $g$ with the continuous function $x \mapsto 1 / x$ on $ℝ ∖ {0}$ , so $1 / g$ is measurable. Therefore $f / g = f \cdot (1 / g)$ is measurable.

Pointwise Limit of Measurable Functions is Measurable

Suppose that

(X, 𝒮)

is a measurable space and

f_{1}, f_{2}, \dots : X \to ℝ

are

𝒮

-measurable. If

f (x) : = \lim_{k \to \infty} f_{k} (x)

exists for every

x \in X

, then

f

𝒮

-measurable.

Let $a \in ℝ$ . We claim that $f^{- 1} ((a, \infty)) = ⋃_{j = 1}^{\infty} ⋃_{m = 1}^{\infty} ⋂_{k = m}^{\infty} f_{k}^{- 1} ((a + \frac{1}{j}, \infty)) .$ If $f (x) > a$ , choose $j$ with $f (x) > a + 1 / j$ . Since $f_{k} (x) \to f (x)$ , eventually $f_{k} (x) > a + 1 / j$ , putting $x$ in the right side. Conversely, if $x$ is in the right side, then for some $j, m$ , all $k \geq m$ satisfy $f_{k} (x) > a + 1 / j$ , so taking limits gives $f (x) \geq a + 1 / j > a$ .

The right side belongs to $𝒮$ because each $f_{k}$ is measurable and $𝒮$ is closed under countable unions and intersections. By the ray test, $f$ is measurable.

Borel Subsets of the Extended Real Line

A subset

C \subseteq [- \infty, \infty]

is Borel if

C \cap ℝ

is a Borel subset of

ℝ

Extended Real-Valued Measurable Function

Suppose that

(X, 𝒮)

is a measurable space. A function

f : X \to [- \infty, \infty]

𝒮

-measurable if

f^{- 1} (B) \in 𝒮

for every Borel set

B \subseteq [- \infty, \infty]

Extended Ray Test for Measurability

Suppose that

(X, 𝒮)

is a measurable space and

f : X \to [- \infty, \infty]

. If

f^{- 1} ((a, \infty]) \in 𝒮

for every

a \in ℝ

, then

f

𝒮

-measurable.

Let $𝒯 : = {B \subseteq [- \infty, \infty] : f^{- 1} (B) \in 𝒮} .$ The algebra of inverse images shows that $𝒯$ is a $σ$ -algebra on $[- \infty, \infty]$ . By hypothesis, it contains every ray $(a, \infty]$ . These rays generate the Borel subsets of the extended real line, so every Borel $B \subseteq [- \infty, \infty]$ belongs to $𝒯$ . Therefore $f$ is $𝒮$ -measurable.

Infimum and Supremum of Measurable Functions

Suppose that

(X, 𝒮)

is a measurable space and

f_{1}, f_{2}, \dots : X \to [- \infty, \infty]

are

𝒮

-measurable. Define

g (x) : = \inf {f_{k} (x) : k \in ℤ^{+}} and h (x) : = \sup {f_{k} (x) : k \in ℤ^{+}} .

Then

g

and

h

are

𝒮

-measurable.

For $h$ , note that for every $a \in ℝ$ , $h^{- 1} ((a, \infty]) = ⋃_{k = 1}^{\infty} f_{k}^{- 1} ((a, \infty]) .$ The right side belongs to $𝒮$ , so $h$ is measurable by the extended ray test.

For $g$ , observe that $g (x) = - \sup {- f_{k} (x) : k \in ℕ_{1}} .$ Each $- f_{k}$ is measurable as a composition with $x \mapsto - x$ , and the supremum result just proved applies. Therefore $g$ is measurable.