Doob–Dynkin lemma

Let ${\displaystyle f}$ be ${\displaystyle \sigma (T)}$-measurable.

Step 1: assume that ${\displaystyle f}$ is a simple function, i.e. ${\displaystyle \textstyle f=\sum _{i=1}^{n}f_{i}\,\cdot {\mathbf {1} }_{A_{i}},}$ for some non-empty pairwise disjoint sets ${\displaystyle \{A_{i}\}_{i=1}^{n}}$ from ${\displaystyle \sigma (T).}$ If ${\displaystyle A_{i}=T^{-1}(A_{i}'),}$ then the function ${\displaystyle g=\textstyle \sum _{i=1}^{n}f_{i}\,\cdot {\mathbf {1} }_{A_{i}'}}$ suits the requirement.

Step 2: if ${\displaystyle f\geq 0}$, then ${\displaystyle f}$ is a pointwise limit of a non-decreasing sequence ${\displaystyle f_{n}}$ of simple functions (see the article on simple functions for the proof). Step 1 guarantees that ${\displaystyle f_{n}=g_{n}\circ T.}$ This equality, in turn, implies that the sequence ${\displaystyle g_{n}(x)}$ is non-decreasing, as long as ${\displaystyle x\in \operatorname {Im} T,}$ so the function

${\displaystyle {\widetilde {g}}(x)=\lim _{n\to \infty }g_{n}(x)}$

is well defined (finite or infinite) for every ${\displaystyle x\in \operatorname {Im} T.}$ As a pointwise limit of measurable ${\displaystyle {\overline {\mathbb {R} }}}$-valued functions, ${\displaystyle {\widetilde {g}}}$ is itself measurable (see the article on measurable functions). Define

${\displaystyle g(x)={\begin{cases}{\widetilde {g}}(x)&\quad {\text{if }}x\in \operatorname {Im} T\\[1mm]0&\quad {\text{if }}x\notin \operatorname {Im} T.\end{cases}}}$

The measurability of ${\displaystyle g}$ is based on the assumption that ${\displaystyle \operatorname {Im} T\in {\mathcal {A}}'.}$ Thus, ${\displaystyle g}$ suits the requirement.

Step 3: every measurable function ${\displaystyle f}$ is the difference of its positive and negative parts, i.e. ${\displaystyle f=f^{+}-f^{-},}$ where both ${\displaystyle f^{+}}$ and ${\displaystyle f^{-}}$ are measurable and non-negative. Step 2 guarantees that ${\displaystyle f^{+}=g^{+}\circ T}$ and ${\displaystyle f^{-}=g^{-}\circ T.}$ Define

${\displaystyle g(x)={\begin{cases}g^{+}(x)-g^{-}(x)&\quad {\text{if }}x\in \operatorname {Im} T\\[1mm]0&\quad {\text{if }}x\notin \operatorname {Im} T.\end{cases}}}$

Since it is not possible that ${\displaystyle f^{+}(x)>0}$ and ${\displaystyle f^{-}(x)>0}$ for the same ${\displaystyle x,}$ the equality ${\displaystyle g^{+}(x)=g^{-}(x)=\infty }$ never holds, and hence ${\displaystyle g}$ is well defined.

Being the difference of two measurable functions, ${\displaystyle g^{+}-g^{-}}$ is also measurable. Since ${\displaystyle \operatorname {Im} T}$ is measurable, so is ${\displaystyle g.}$ Thus, ${\displaystyle g}$ suits the requirement.