2Measurable functions and random variables

II Probability and Measure

2.1 Measurable functions

The definition of a measurable function is somewhat like the definition of a

continuous function, except that we replace “open” with “in the σ-algebra”.

Definition

(Measurable functions)

.

Let (

E, E

) and (

G, G

) be measure spaces.

A map f : E → G is measurable if for every A ∈ G, we have

f

−1

(A) = {x ∈ E : f(x) ∈ E} ∈ E.

If (G, G) = (R, B), then we will just say that f is measurable on E.

If (

G, G

) = ([0

, ∞

]

, B

), then we will just say that

f

is non-negative measurable.

If E is a topological space and E = B(E), then we call f a Borel function.

How do we actually check in practice that a function is measurable? It turns

out we are lucky. We can simply check that

f

−1

(

A

)

∈ E

for

A

in any generating

set Q of G.

Lemma.

Let (

E, E

) and (

G, G

) be measurable spaces, and

G

=

σ

(

Q

) for some

Q. If f

−1

(A) ∈ E for all A ∈ Q, then f is measurable.

Proof. We claim that

{A ⊆ G : f

−1

(A) ∈ E}

is a

σ

-algebra on

G

. Then the result follows immediately by definition of

σ

(

Q

).

Indeed, this follows from the fact that

f

−1

preserves everything. More

precisely, we have

f

−1

[

n

A

n

!

=

[

n

f

−1

(A

n

), f

−1

(A

C

) = (f

−1

(A))

C

, f

−1

(∅) = ∅.

So if, say, all A

n

∈ A, then so is

S

n

A

n

.

Example.

In the particular case where we have a function

f

:

E → R

, we know

that

B

=

B

(

R

) is generated by (

−∞, y

] for

y ∈ R

. So we just have to check that

{x ∈ E : f(x) ≤ y} = f

−1

((−∞, y])) ∈ E.

Example.

Let

E, F

be topological spaces, and

f

:

E → F

be continuous. We

will see that

f

is a measurable function (under the Borel

σ

-algebras). Indeed,

by definition, whenever

U ⊆ F

is open, we have

f

−1

(

U

) open as well. So

f

−1

(

U

)

∈ B

(

E

) for all

U ⊆ F

open. But since

B

(

F

) is the

σ

-algebra generated

by the open sets, this implies that f is measurable.

This is one very important example. We can do another very important

example.

Example.

Suppose that

A ⊆ E

. The indicator function of

A

is

1

A

(

x

) :

E →

{0, 1} given by

1

A

(x) =

(

1 x ∈ A

0 x 6∈ A

.

Suppose we give

{

0

,

1

}

the non-trivial measure. Then

1

A

is a measurable function

iff A ∈ E.

Example. The identity function is always measurable.

Example.

Composition of measurable functions are measurable. More precisely,

if (

E, E

), (

F, F

) and (

G, G

) are measurable spaces, and the functions

f

:

E → F

and

g

:

F → G

are measurable, then the composition

g◦f

:

E → G

is measurable.

Indeed, if

A ∈ G

, then

g

−1

(

A

)

∈ F

, so

f

−1

(

g

−1

(

A

))

∈ E

. But

f

−1

(

g

−1

(

A

)) =

(g ◦ f)

−1

(A). So done.

Definition

(

σ

-algebra generated by functions)

.

Now suppose we have a set

E

,

and a family of real-valued functions {f

i

: i ∈ I} on E. We then define

σ(f

i

: i ∈ I) = σ(f

−1

i

(A) : A ∈ B, i ∈ I).

This is the smallest

σ

-algebra on

E

which makes all the

f

i

’s measurable.

This is analogous to the notion of initial topologies for topological spaces.

If we want to construct more measurable functions, the following definition

will be rather useful:

Definition

(Product measurable space)

.

Let (

E, E

) and (

G, G

) be measure

spaces. We define the product measure space as

E × G

whose

σ

-algebra is

generated by the projections

E × G

E G

π

1

π

2

.

More explicitly, the σ-algebra is given by

E ⊗ G = σ({A × B : A ∈ E, B ∈ G}).

More generally, if (

E

i

, E

i

) is a collection of measure spaces, the product measure

space has underlying set

Q

i

E

i

, and the

σ

-algebra generated by the projection

maps π

i

:

Q

j

E

j

→ E

i

.

This satisfies the following property:

Proposition.

Let

f

i

:

E → F

i

be functions. Then

{f

i

}

are all measurable iff

(

f

i

) :

E →

Q

F

i

is measurable, where the function (

f

i

) is defined by setting the

ith component of (f

i

)(x) to be f

i

(x).

Proof.

If the map (

f

i

) is measurable, then by composition with the projections

π

i

, we know that each f

i

is measurable.

Conversely, if all

f

i

are measurable, then since the

σ

-algebra of

Q

F

i

is

generated by sets of the form

π

−1

j

(

A

) :

A ∈ F

j

, and the pullback of such sets

along (f

i

) is exactly f

−1

j

(A), we know the function (f

i

) is measurable.

Using this, we can prove that a whole lot more functions are measurable.

Proposition.

Let (

E, E

) be a measurable space. Let (

f

n

:

n ∈ N

) be a sequence

of non-negative measurable functions on

E

. Then the following are measurable:

f

1

+ f

2

, f

1

f

2

, max{f

1

, f

2

}, min{f

1

, f

2

},

inf

n

f

n

, sup

n

f

n

, lim inf

n

f

n

, lim sup

n

f

n

.

The same is true with “real” replaced with “non-negative”, provided the new

functions are real (i.e. not infinity).

Proof.

This is an (easy) exercise on the example sheet. For example, the sum

f

1

+ f

2

can be written as the following composition.

E [0, ∞]

2

[0, ∞].

(f

1

,f

2

)

+

We know the second map is continuous, hence measurable. The first function is

also measurable since the f

i

are. So the composition is also measurable.

The product follows similarly, but for the infimum and supremum, we need to

check explicitly that the corresponding maps [0

, ∞

]

N

→

[0

, ∞

] is measurable.

Notation. We will write

f ∧ g = min{f, g}, f ∨g = max{f, g}.

We are now going to prove the monotone class theorem, which is a “Dynkin’s

lemma” for measurable functions. As in the case of Dynkin’s lemma, it will

sound rather awkward but will prove itself to be very useful.

Theorem

(Monotone class theorem)

.

Let (

E, E

) be a measurable space, and

A ⊆ E

be a

π

-system with

σ

(

A

) =

E

. Let

V

be a vector space of functions such

that

(i) The constant function 1 = 1

E

is in V.

(ii) The indicator functions 1

A

∈ V for all A ∈ A

(iii) V is closed under bounded, monotone limits.

More explicitly, if (

f

n

) is a bounded non-negative sequence in

V

,

f

n

% f

(pointwise) and f is also bounded, then f ∈ V.

Then V contains all bounded measurable functions.

Note that the conditions for

V

is pretty like the conditions for a d-system,

where taking a bounded, monotone limit is something like taking increasing

unions.

Proof. We first deduce that 1

A

∈ V for all A ∈ E.

D = {A ∈ E : 1

A

∈ V}.

We want to show that

D

=

E

. To do this, we have to show that

D

is a

d

-system.

(i) Since 1

E

∈ V, we know E ∈ D.

(ii) If 1

A

∈ V , then 1 − 1

A

= 1

E\A

∈ V. So E \ A ∈ D.

(iii)

If (

A

n

) is an increasing sequence in

D

, then

1

A

n

→ 1

S

A

n

monotonically

increasingly. So 1

S

A

n

is in D.

So, by Dynkin’s lemma, we know

D

=

E

. So

V

contains indicators of all measur-

able sets. We will now try to obtain any measurable function by approximating.

Suppose that

f

is bounded and non-negative measurable. We want to show

that f ∈ V. To do this, we approximate it by letting

f

n

= 2

−n

b2

n

fc =

∞

X

k=0

k2

−n

1

{k2

−n

≤f<(k+1)2

−n

}

.

Note that since

f

is bounded, this is a finite sum. So it is a finite linear

combination of indicators of elements in

E

. So

f

n

∈ V

, and 0

≤ f

n

→ f

monotonically. So f ∈ V.

More generally, if f is bounded and measurable, then we can write

f = (f ∨ 0) + (f ∧0) ≡ f

+

− f

−

.

Then f

+

and f

−

are bounded and non-negative measurable. So f ∈ V.

Unfortunately, we will not have a chance to use this result until the next

chapter where we discuss integration. There we will use this a lot.