Doob–Dynkin lemma

In probability theory, the Doob–Dynkin lemma, named after Joseph L. Doob and Eugene Dynkin, 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, ${\overline {\mathbb {R} }}=\mathbb {R} \cup \{-\infty \}\cup \{+\infty \}$ is the extended real number line, and ${\displaystyle {\mathcal {B}}({\overline {\mathbb {R} }})} is the $ \sigma$ $ -algebra of Borel sets on ${\overline {\mathbb {R} }}.$ The notation $g:(X,{\mathcal {X}})\rightarrow (Y,{\mathcal {Y}})$ indicates that g is a function from X to Y, and that g is measurable relative to the $\sigma$ -algebras ${\mathcal {X}}$ and ${\mathcal {Y}}.$

Furthermore, if $T:X\to Y,$ and $(Y,{\mathcal {Y}})$ is a measurable space, we define

$\sigma (T)=\{T^{-1}(S)\mid S\in {\mathcal {Y}}\}.$

One can easily check that $\sigma(T)$ is the minimal $\sigma$ -algebra on X under which T is measurable, i.e.

$T:(X,\sigma (T))\to (Y,{\mathcal {Y}}).$

Statement of the lemma

Let $T:\Omega \rightarrow \Omega '$ be a function from a set $\Omega$ to a measurable space $(\Omega ',{\mathcal {A}}'),$ and $\operatorname {Im} T$ is ${\mathcal {A}}'$ -measurable. Further, let $f:\Omega \rightarrow {\overline {\mathbb {R} }}$ be a scalar function on $\Omega$ . Then f {\displaystyle f} f is $\sigma(T)-$ measurable if and only if $f=g\circ T,$ for some measurable function $g:(\Omega ',{\mathcal {A}}')\rightarrow ({\overline {\mathbb {R} }},{\mathcal {B}}({\overline {\mathbb {R} }})).$

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

By definition, f being $\sigma(T)$ -measurable is the same as $f^{-1}(S)\in \sigma (T)$ for every Borel set S, which is the same as $\sigma (f)\subseteq \sigma (T)$ . So, the lemma can be rewritten in the following, equivalent form.

Lemma. Let f and T be as above. Then $f=g\circ T,$ for some Borel function g , if and only if $\sigma (f)\subseteq \sigma (T)$ .