Doob–Dynkin lemma

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 ${\displaystyle \sigma }$-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 ${\displaystyle \sigma }$-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 ${\displaystyle \sigma }$-algebra that is generated by the random variable.

In the lemma below, ${\displaystyle {\mathcal {B}}[0,1]}$ is the ${\displaystyle \sigma }$-algebra of Borel sets on ${\displaystyle [0,1].}$ If ${\displaystyle T\colon X\to Y,}$ and ${\displaystyle (Y,{\mathcal {Y}})}$ is a measurable space, then

${\displaystyle \sigma (T)\ {\stackrel {\text{def}}{=}}\ \{T^{-1}(S)\mid S\in {\mathcal {Y}}\}}$

is the smallest ${\displaystyle \sigma }$-algebra on ${\displaystyle X}$ such that ${\displaystyle T}$ is ${\displaystyle \sigma (T)/{\mathcal {Y}}}$-measurable.

Let ${\displaystyle T\colon \Omega \rightarrow \Omega '}$ be a function, and ${\displaystyle (\Omega ',{\mathcal {A}}')}$ a measurable space. A function ${\displaystyle f\colon \Omega \rightarrow [0,1]}$ is ${\displaystyle \sigma (T)/{\mathcal {B}}[0,1]}$-measurable if and only if ${\displaystyle f=g\circ T,}$ for some ${\displaystyle {\mathcal {A}}'/{\mathcal {B}}[0,1]}$-measurable ${\displaystyle g\colon \Omega '\to [0,1].}$¹

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

Remark. The lemma remains valid if the space ${\displaystyle ([0,1],{\mathcal {B}}[0,1])}$ is replaced with ${\displaystyle (S,{\mathcal {B}}(S)),}$ where ${\displaystyle S\subseteq [-\infty ,\infty ],}$ ${\displaystyle S}$ is bijective with ${\displaystyle [0,1],}$ and the bijection is measurable in both directions.

By definition, the measurability of ${\displaystyle f}$ means that ${\displaystyle f^{-1}(S)\in \sigma (T)}$ for every Borel set ${\displaystyle S\subseteq [0,1].}$ Therefore ${\displaystyle \sigma (f)\subseteq \sigma (T),}$ and the lemma may be restated as follows.

Lemma. Let ${\displaystyle T\colon \Omega \rightarrow \Omega ',}$ ${\displaystyle f\colon \Omega \rightarrow [0,1],}$ and ${\displaystyle (\Omega ',{\mathcal {A}}')}$ is a measurable space. Then ${\displaystyle f=g\circ T,}$ for some ${\displaystyle {\mathcal {A}}'/{\mathcal {B}}[0,1]}$-measurable ${\displaystyle g\colon \Omega '\to [0,1],}$ if and only if ${\displaystyle \sigma (f)\subseteq \sigma (T)}$.