2Measurable functions and random variables

II Probability and Measure

2.2 Constructing new measures

We are going to look at two ways to construct new measures on spaces based on

some measurable function we have.

Definition

(Image measure)

.

Let (

E, E

) and (

G, G

) be measure spaces. Suppose

µ

is a measure on

E

and

f

:

E → G

is a measurable function. We define the

image measure ν = µ ◦ f

−1

on G by

ν(A) = µ(f

−1

(A)).

It is a routine check that this is indeed a measure.

If we have a strictly increasing continuous function, then we know it is

invertible (if we restrict the codomain appropriately), and the inverse is also

strictly increasing. It is also clear that these conditions are necessary for an

inverse to exist. However, if we relax the conditions a bit, we can get some sort

of “pseudoinverse” (some categorists may call them “left adjoints” (and will tell

you that it is a trivial consequence of the adjoint functor theorem)).

Recall that a function

g

is right continuous if

x

n

& x

implies

g

(

x

n

)

→ g

(

x

),

and similarly f is left continuous if x

n

% x implies f(x

n

) → f(x).

Lemma. Let g : R → R be non-constant, non-decreasing and right continuous.

We set

g(±∞) = lim

x→±∞

g(x).

We set I = (g(−∞), g(∞)). Since g is non-constant, this is non-empty.

Then there is a non-decreasing, left continuous function

f

:

I → R

such that

for all x ∈ I and y ∈ R, we have

x ≤ g(y) ⇔ f(x) ≤ y.

Thus, taking the negation of this, we have

x > g(y) ⇔ f(x) > y.

Explicitly, for x ∈ I, we define

f(x) = inf{y ∈ R : x ≤ g(y)}.

Proof. We just have to verify that it works. For x ∈ I, consider

J

x

= {y ∈ R : x ≤ g(y)}.

Since

g

is non-decreasing, if

y ∈ J

x

and

y

0

≥ y

, then

y

0

∈ J

x

. Since

g

is

right-continuous, if y

n

∈ J

x

is such that y

n

& y, then y ∈ J

x

. So we have

J

x

= [f(x), ∞).

Thus, for f ∈ R, we have

x ≤ g(y) ⇔ f(x) ≤ y.

So we just have to prove the remaining properties of

f

. Now for

x ≤ x

0

, we have

J

x

⊆ J

x

0

. So f(x) ≤ f(x

0

). So f is non-decreasing.

Similarly, if

x

n

% x

, then we have

J

x

=

T

n

J

x

n

. So

f

(

x

n

)

→ f

(

x

). So this

is left continuous.

Example. If g is given by the function

then f is given by

This allows us to construct new measures on R with ease.

Theorem.

Let

g

:

R → R

be non-constant, non-decreasing and right continuous.

Then there exists a unique Radon measure dg on B such that

dg((a, b]) = g(b) − g(a).

Moreover, we obtain all non-zero Radon measures on R in this way.

We have already seen an instance of this when we

g

was the identity function.

Given the lemma, this is very easy.

Proof.

Take

I

and

f

as in the previous lemma, and let

µ

be the restriction of

the Lebesgue measure to Borel subsets of

I

. Now

f

is measurable since it is left

continuous. We define dg = µ ◦ f

−1

. Then we have

dg((a, b]) = µ({x ∈ I : a < f(x) ≤ b})

= µ({x ∈ I : g(a) < x ≤ g(b)})

= µ((g(a), g(b)]) = g(b) − g(a).

So dg is a Radon measure with the required property.

There are no other such measures by the argument used for uniqueness of

the Lebesgue measure.

To show we get all non-zero Radon measures this way, suppose we have a

Radon measure ν on R, we want to produce a g such that ν = dg. We set

g(y) =

(

−ν((y, 0]) y ≤ 0

ν((0, y]) y > 0

.

Then

ν

((

a, b

]) =

g

(

b

)

− g

(

a

). We see that

ν

is non-zero, so

g

is non-constant.

It is also easy to see it is non-decreasing and right continuous. So

ν

= d

g

by

continuity.