Defocus aberration

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

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