Factorization lemma
In measure theory, the factorization lemma allows for expressing a function f with another function T if f is measurable with respect to T. An application of this is regression analysis.
Theorem
Let [math]\displaystyle{ T:\Omega\rightarrow\Omega' }[/math] be a function of a set [math]\displaystyle{ \Omega }[/math] in a measure space [math]\displaystyle{ (\Omega',\mathcal{A}') }[/math] and let [math]\displaystyle{ f:\Omega\rightarrow\overline{\mathbb{R}} }[/math] be a scalar function on [math]\displaystyle{ \Omega }[/math]. Then [math]\displaystyle{ f }[/math] is measurable with respect to the σ-algebra [math]\displaystyle{ \sigma(T)=T^{-1}(\mathcal{A}') }[/math] generated by [math]\displaystyle{ T }[/math] in [math]\displaystyle{ \Omega }[/math] if and only if there exists a measurable function [math]\displaystyle{ g:(\Omega',\mathcal{A}')\rightarrow(\overline{\mathbb{R}},\mathcal{B}(\overline{\mathbb{R}})) }[/math] such that [math]\displaystyle{ f=g\circ T }[/math], where [math]\displaystyle{ \mathcal{B}(\overline{\mathbb{R}}) }[/math] denotes the Borel set of the real numbers. If [math]\displaystyle{ f }[/math] only takes finite values, then [math]\displaystyle{ g }[/math] also only takes finite values.
Proof
First, if [math]\displaystyle{ f=g\circ T }[/math], then f is [math]\displaystyle{ \sigma(T)-\mathcal{B}(\overline{\mathbb{R}}) }[/math] measurable because it is the composition of a [math]\displaystyle{ \sigma(T)-\mathcal{A}' }[/math] and of a [math]\displaystyle{ \mathcal{A}'-\mathcal{B}(\overline{\mathbb{R}}) }[/math] measurable function. The proof of the converse falls into four parts: (1)f is a step function, (2)f is a positive function, (3) f is any scalar function, (4) f only takes finite values.
f is a step function
Suppose [math]\displaystyle{ f=\sum_{i=1}^n\alpha_i 1_{A_i} }[/math] is a step function, i.e. [math]\displaystyle{ n\in\mathbb{N}^*, \forall i\in[\![1,n]\!], A_i\in\sigma(T) }[/math] and [math]\displaystyle{ \alpha_i\in\mathbb{R}^+ }[/math]. As T is a measurable function, for all i, there exists [math]\displaystyle{ A_i'\in\mathcal{A}' }[/math] such that [math]\displaystyle{ A_i=T^{-1}(A_i') }[/math]. [math]\displaystyle{ g=\sum_{i=1}^n\alpha_i 1_{A_i'} }[/math] fulfils the requirements.
f takes only positive values
If f takes only positive values, it is the limit, for pointwise convergence, of an increasing sequence [math]\displaystyle{ (u_n)_{n\in\mathbb{N}} }[/math] of step functions. For each of these, by (1), there exists [math]\displaystyle{ g_n }[/math] such that [math]\displaystyle{ u_n=g_n\circ T }[/math]. The function [math]\displaystyle{ \lim_{n\rightarrow+\infty}g_n }[/math], which exists on the image of T for pointwise convergence because [math]\displaystyle{ (u_n)_{n\in\mathbb{N}} }[/math] is monotonic, fulfils the requirements.
General case
We can decompose f in a positive part [math]\displaystyle{ f^+ }[/math] and a negative part [math]\displaystyle{ f^- }[/math]. We can then find [math]\displaystyle{ g_0^+ }[/math] and [math]\displaystyle{ g_0^- }[/math] such that [math]\displaystyle{ f^+=g_0^+\circ T }[/math] and [math]\displaystyle{ f^-=g_0^-\circ T }[/math]. The problem is that the difference [math]\displaystyle{ g:=g^+-g^- }[/math] is not defined on the set [math]\displaystyle{ U=\{x:g_0^+(x)=+\infty\}\cap\{x:g_0^-(x)=+\infty\} }[/math]. Fortunately, [math]\displaystyle{ T(\Omega)\cap U=\varnothing }[/math] because [math]\displaystyle{ g_0^+(T(\omega))=f^+(\omega)=+\infty }[/math] always implies [math]\displaystyle{ g_0^-(T(\omega))=f^-(\omega)=0 }[/math] We define [math]\displaystyle{ g^+=1_{\Omega'\backslash U}g_0^+ }[/math] and [math]\displaystyle{ g^-=1_{\Omega'\backslash U}g_0^- }[/math]. [math]\displaystyle{ g=g^+-g^- }[/math] fulfils the requirements.
f takes finite values only
If f takes finite values only, we will show that g also only takes finite values. Let [math]\displaystyle{ U'=\{\omega:|g(\omega)|=+\infty\} }[/math]. Then [math]\displaystyle{ g_0=1_{\Omega'\backslash U'}g }[/math] fulfils the requirements because [math]\displaystyle{ U'\cap T(\Omega)=\varnothing }[/math].
Importance of the measure space
If the function [math]\displaystyle{ f }[/math] is not scalar, but takes values in a different measurable space, such as [math]\displaystyle{ \mathbb{R} }[/math] with its trivial σ-algebra (the empty set, and the whole real line) instead of [math]\displaystyle{ \mathcal{B}(\mathbb{R}) }[/math], then the lemma becomes false (as the restrictions on [math]\displaystyle{ f }[/math] are much weaker).
See also
References
- Heinz Bauer, Ed. (1992) Maß- und Integrationstheorie. Walter de Gruyter edition. 11.7 Faktorisierungslemma p. 71-72.
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