Doob–Dynkin lemma

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). ==See also==