Factorization lemma

In measure theory, the factorization lemma allows us to express a function f with another function T if f is measurable with respect to T. An application of this is regression analysis.

Theorem

Let T:\Omega\rightarrow\Omega' be a function of a set \Omega in a measure space (\Omega',\mathcal{A}') and let f:\Omega\rightarrow\overline{\mathbb{R}} be a scalar function on \Omega. Then f is measurable with respect to the σ-algebra \sigma(T)=T^{-1}(\mathcal{A}') generated by T in \Omega if and only if there exists a measurable function g:(\Omega',\mathcal{A}')\rightarrow(\overline{\mathbb{R}},\mathcal{B}(\overline{\mathbb{R}})) such that f=g\circ T, where \mathcal{B}(\overline{\mathbb{R}}) denotes the Borel set of the real numbers. If f only takes finite values, then g also only takes finite values.

Proof

First, if f=g\circ T, then f is \sigma(T)-\mathcal{B}(\overline{\mathbb{R}}) measurable because it is the composition of a \sigma(T)-\mathcal{A}' and of a \mathcal{A}'-\mathcal{B}(\overline{\mathbb{R}}) 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 f=\sum_{i=1}^n\alpha_i 1_{A_i} is a step function, i.e. n\in\mathbb{N}^*, \forall i\in[\![1,n]\!], A_i\in\sigma(T) and \alpha_i\in\mathbb{R}^+. As T is a measurable function, for all i, there exists A_i'\in\mathcal{A}' such that A_i=T^{-1}(A_i'). g=\sum_{i=1}^n\alpha_i 1_{A_i'} fulfils the requirements.

f takes only positive values

If f takes only positive values, it is the limit of a sequence (u_n)_{n\in\mathbb{N}} of step functions. For each of these, by (1), there exists g_n such that u_n=g_n\circ T. The function \lim_{n\rightarrow+\infty}g_n fulfils the requirements.

General case

We can decompose f in a positive part f^+ and a negative part f^-. We can then find g_0^+ and g_0^- such that f^+=g_0^+\circ T and f^-=g_0^-\circ T. The problem is that the difference g:=g^+-g^- is not defined on the set U=\{x:g_0^+(x)=+\infty\}\cap\{x:g_0^-(x)=+\infty\}. Fortunately, T(\Omega)\cap U=\varnothing because g_0^+(T(\omega))=f^+(\omega)=+\infty always implies g_0^-(T(\omega))=f^-(\omega)=0 We define g^+=1_{\Omega'\backslash U}g_0^+ and g^-=1_{\Omega'\backslash U}g_0^-. g=g^+-g^- 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 U'=\{\omega:|g(\omega)|=+\infty\}. Then g_0=1_{\Omega'\backslash U'}g fulfils the requirements because U'\cap T(\Omega)=\varnothing.

Importance of the measure space

If the function f is not scalar, but takes values in a different measurable space, such as \mathbb{R} with its trivial σ-algebra (the empty set, and the whole real line) instead of \mathcal{B}(\mathbb{R}) , then the lemma becomes false (as the restrictions on f are much weaker).

References