# 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.
* Proof.
* Let $f$ be $\sigma (T)/{\mathcal {B}}[0,1]$-measurable.
* First, note that, by the above descriptive definition of $\sigma (T)$ as the set of preimages of ${\mathcal {A}}'$-measurable sets under $T$, we know that if $A\in \sigma (T)$, then there exists some $A'\in {\mathcal {A}}'$ such that $A=T^{-1}(A')$.
* Now, assume that $f=\mathbf {1} _{A}$ is an indicator of some set $A\in \sigma (T)$. If we identify $A'\in {\mathcal {A}}'$ such that $A=T^{-1}(A')$, then the function $g=\mathbf {1} _{A'}$ suits the requirement, and since $A\in \sigma (T)$, such a set $A'\in {\mathcal {A}}'$ always exists. 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 $f_{n}\geq 0$ of simple functions. The previous step guarantees that $f_{n}=g_{n}\circ T,$ for some measurable $g_{n}.$ The supremum $\textstyle g(x)=\sup _{n\geq 1}g_{n}(x)$ 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 $g_{n}(x)$ is non-decreasing, so $\textstyle g|_{\operatorname {Im} T}(x)=\lim _{n\to \infty }g_{n}|_{\operatorname {Im} T}(x)$ which shows that $f=g\circ T.$
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 $\sigma (f)\subseteq \sigma (T)$.