Doob–Dynkin lemma explained

In probability theory, the Doob–Dynkin lemma, named after Joseph L. Doob and Eugene Dynkin (also known as the factorization lemma), characterizes the situation when one random variable is a function of another by the inclusion of the

\sigma

-algebras generated by the random variables. The usual statement of the lemma is formulated in terms of one random variable being measurable with respect to the

\sigma

-algebra generated by the other.

The lemma plays an important role in the conditional expectation in probability theory, where it allows replacement of the conditioning on a random variable by conditioning on the

\sigma

-algebra that is generated by the random variable.

Notations and introductory remarks

In the lemma below,

l{B}[0,1]

is the

\sigma

-algebra of Borel sets on

[0,1].

If

T\colonX\toY,

and

(Y,{lY})

is a measurable space, then

\sigma(T)\stackrel{def

}\ \ is the smallest

\sigma

-algebra on

X

such that

T

is

\sigma(T)/{lY}

-measurable.

Statement of the lemma

Let

T\colon\Omega\Omega'

be a function, and

(\Omega',l{A}')

a measurable space. A function

f\colon\Omega[0,1]

is

\sigma(T)/l{B}[0,1]

-measurable if and only if

f=g\circT,

for some

l{A}'/l{B}[0,1]

-measurable

g\colon\Omega'\to[0,1].

[1]

Remark. The "if" part simply states that the composition of two measurable functions is measurable. The "only if" part is proven below.

Proof.
Let

f

be

\sigma(T)/l{B}[0,1]

-measurable.

Assume that

f=1A

is an indicator of some set

A\in\sigma(T).

If

A=T-1(A'),

then the function

g={1}A'

suits the requirement. By linearity, the claim extends to any simple measurable function

f.

Let

f

be measurable but not necessarily simple. As explained in the article on simple functions,

f

is a pointwise limit of a monotonically non-decreasing sequence

fn\geq0

of simple functions. The previous step guarantees that

fn=gn\circT,

for some measurable

gn.

The supremum

styleg(x)=\supgn(x)

exists on the entire

\Omega'

and is measurable. (The article on measurable functions explains why supremum of a sequence of measurable functions is measurable). For every

x\in\operatorname{Im}T,

the sequence

gn(x)

is non-decreasing, so

styleg

_(x) = \lim_ g_n_(x) which shows that

f=g\circT.

Remark. The lemma remains valid if the space

([0,1],l{B}[0,1])

is replaced with

(S,l{B}(S)),

where

S\subseteq[-infty,infty],

S

is bijective with

[0,1],

and the bijection is measurable in both directions.

By definition, the measurability of

f

means that

f-1(S)\in\sigma(T)

for every Borel set

S\subseteq[0,1].

Therefore

\sigma(f)\subseteq\sigma(T),

and the lemma may be restated as follows.

Lemma. Let

T\colon\Omega\Omega',

f\colon\Omega[0,1],

and

(\Omega',l{A}')

is a measurable space. Then

f=g\circT,

for some

l{A}'/l{B}[0,1]

-measurable

g\colon\Omega'\to[0,1],

if and only if

\sigma(f)\subseteq\sigma(T)

.

See also

References

Notes and References

  1. Book: Kallenberg, Olav. Foundations of Modern Probability. 1997. 7. Springer. 0-387-94957-7 . Olav Kallenberg .