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 "T". An application of this is regression analysis.

Theorem

Let $T:Omega\; ightarrowOmega\text{'}$ be a function of a set $Omega$ in a measure space $(Omega,mathcal\{A\}\text{'})$ and let $f:Omega\; ightarrowoverline\{mathbb\{R$ be a scalar function on $Omega$. Then $f$ is measurable with respect to the σ-algebra $sigma(T)=T^\{-1\}(mathcal\{A\}\text{'})$ generated by $T$ in $Omega$ if and only if there exists a measurable function $g:(Omega\text{'},mathcal\{A\}\text{'})\; ightarrow(overline\{mathbb\{R,mathcal\{B\}(overline\{mathbb\{R))$ such that $f=gcirc\; T$, 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 $f=gcirc\; T$, then "f" is $sigma(T)-mathcal\{B\}(overline\{mathbb\{R)$ measurable because it is the composition of a $sigma(T)-mathcal\{A\}\text{'}$ and of a $mathcal\{A\}\text{'}-mathcal\{B\}(overline\{mathbb\{R)$ 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 $f=sum\_\{i=1\}^nalpha\_i\; 1\_\{A\_i\}$ is a step function, i.e. $ninmathbb\{N\}^*,\; forall\; iin\; [!\; [1,n]\; !]\; ,\; A\_iinsigma(T)$ and $alpha\_iinmathbb\{R\}^+$. As "T" is a measurable function, for all "i", there exists $A\_i\text{'}inmathcal\{A\}\text{'}$ such that $A\_i=T^\{-1\}(A\_i\text{'})$. $g=sum\_\{i=1\}^nalpha\_i\; 1\_\{A\_i\text{'}\}$ fulfills the requirements.

"f" takes only positive values

If "f" takes only positive values, it is the limit of a sequence $(u\_n)\_\{ninmathbb\{N$ of step functions. For each of these, by (1), there exists $g\_n$ such that $u\_n=g\_ncirc\; T$. The function $lim\_\{n\; ightarrow+infty\}g\_n$ 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 and . $g=g^+-g^-$ 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 $U\text{'}=\{omega:|g(omega)|=+infty\}$. Then fulfils the requirements because $U\text{'}cap\; T(Omega)=varnothing$.

References

* Heinz Bauer, Ed. (1992) "Maß- und Integrationstheorie". Walter de Gruyter edition. 11.7 Faktorisierungslemma p.71-72.