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

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

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

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

Statement of the lemma Let T : Ω → Ω ′ {\displaystyle T\colon \Omega \rightarrow \Omega '} be a function, and ( Ω ′ , A ′ ) {\displaystyle (\Omega ',{\mathcal {A}}')} a measurable space. A function f : Ω → [ 0 , 1 ] {\displaystyle f\colon \Omega \rightarrow [0,1]} is σ ( T ) / B [ 0 , 1 ] {\displaystyle \sigma (T)/{\mathcal {B}}[0,1]} -measurable if and only if f = g ∘ T , {\displaystyle f=g\circ T,} for some A ′ / B [ 0 , 1 ] {\displaystyle {\mathcal {A}}'/{\mathcal {B}}[0,1]} -measurable g : Ω ′ → [ 0 , 1 ] . {\displaystyle 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 ] , B [ 0 , 1 ] ) {\displaystyle ([0,1],{\mathcal {B}}[0,1])} is replaced with ( S , B ( S ) ) , {\displaystyle (S,{\mathcal {B}}(S)),} where S ⊆ [ − ∞ , ∞ ] , {\displaystyle S\subseteq [-\infty ,\infty ],} S {\displaystyle S} is bijective with [ 0 , 1 ] , {\displaystyle [0,1],} and the bijection is measurable in both directions. By definition, the measurability of f {\displaystyle f} means that f − 1 ( S ) ∈ σ ( T ) {\displaystyle f^{-1}(S)\in \sigma (T)} for every Borel set S ⊆ [ 0 , 1 ] . {\displaystyle S\subseteq [0,1].} Therefore σ ( f ) ⊆ σ ( T ) , {\displaystyle \sigma (f)\subseteq \sigma (T),} and the lemma may be restated as follows. Lemma. Let T : Ω → Ω ′ , {\displaystyle T\colon \Omega \rightarrow \Omega ',} f : Ω → [ 0 , 1 ] , {\displaystyle f\colon \Omega \rightarrow [0,1],} and ( Ω ′ , A ′ ) {\displaystyle (\Omega ',{\mathcal {A}}')} is a measurable space. Then f = g ∘ T , {\displaystyle f=g\circ T,} for some A ′ / B [ 0 , 1 ] {\displaystyle {\mathcal {A}}'/{\mathcal {B}}[0,1]} -measurable g : Ω ′ → [ 0 , 1 ] , {\displaystyle g\colon \Omega '\to [0,1],} if and only if σ ( f ) ⊆ σ ( T ) {\displaystyle \sigma (f)\subseteq \sigma (T)} .

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