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

## Proof

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

### f takes only positive values

If f takes only positive values, it is the limit, for pointwise convergence, of a increasing sequence ${\displaystyle (u_{n})_{n\in \mathbb {N} }}$ of step functions. For each of these, by (1), there exists ${\displaystyle g_{n}}$ such that ${\displaystyle u_{n}=g_{n}\circ T}$. The function ${\displaystyle \lim _{n\rightarrow +\infty }g_{n}}$, which exists on the image of T for pointwise convergence because ${\displaystyle (u_{n})_{n\in \mathbb {N} }}$ is monotonic, fulfils the requirements.

### General case

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

### Importance of the measure space

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

## References

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