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, \( {\displaystyle {\overline {\mathbb {R} }}=\mathbb {R} \cup \{-\infty \}\cup \{+\infty \}} \) is the extended real number line, and \( {\displaystyle {\mathcal {B}}({\overline {\mathbb {R} }})} is the \( \sigma \) \) -algebra of Borel sets on \( {\displaystyle {\overline {\mathbb {R} }}.} \) The notation \( {\displaystyle g:(X,{\mathcal {X}})\rightarrow (Y,{\mathcal {Y}})} \) indicates that g is a function from X to Y, and that g is measurable relative to the \( \sigma \)-algebras \( {\mathcal {X}} \) and \( {\displaystyle {\mathcal {Y}}.} \)

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

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

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

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

Statement of the lemma

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

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

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