Factorization lemma

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

1 Theorem
2 Proof
3 Reference

[edit] Theorem

Let $T:\Omega\rightarrow\Omega'$ be a function of a set $Ω$ in a measure space $(\Omega,\mathcal{A}')$ and let $f:\Omega\rightarrow\overline{\mathbb{R}}$ be a scalar function on $Ω$ . Then $f$ is mesurable with respect to the σ-algebra $\sigma(T)=T^{-1}(\mathcal{A}')$ generated by $T$ in $Ω$ 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.

[edit] 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.

[edit] 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 $A i = T - 1 (A i')$ . $g=\sum_{i=1}^n\alpha_i 1_{A_i'}$ fullfills the requirements.

[edit] 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 $\lim_{n\rightarrow+\infty}g_n$ fullfills the requirements.

[edit] 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 $g^-=1_{\Omega'\backslash U}g_0^-$ . $g = g + - g -$ fullfills the requirements.

[edit] 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 $g_0=1_{\Omega'\backslash U'}g$ fullfills the requirements because $U'\cap T(\Omega)=\varnothing$ .