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 $\sigma$-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, $$\mathcal{B}[0,1]$$ is the $$\sigma$$-algebra of Borel sets on $$ [0,1]. $$ If $$T\colon X\to Y,$$ and $$(Y,{\mathcal Y})$$ is a measurable space, then
 * $$\sigma(T)\ \stackrel{\text{def}}{=}\ \{T^{-1}(S)\mid S\in {\mathcal Y}\} $$

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

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

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