Factorization lemma

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

Theorem

Let $T:\mathrm{\Omega }\to {\mathrm{\Omega }}^{\prime }$ be a function of a set $\mathrm{\Omega }$ in a measure space $({\mathrm{\Omega }}^{\prime },{{𝒜}}^{\prime })$ and let $f:\mathrm{\Omega }\to \overline{\mathbb{ℝ}}$ be a scalar function on $\mathrm{\Omega }$. Then $f$ is measurable with respect to the σ-algebra $\sigma (T)=T^{-1}({{𝒜}}^{\prime })$ generated by $T$ in $\mathrm{\Omega }$ if and only if there exists a measurable function $g:({\mathrm{\Omega }}^{\prime },{{𝒜}}^{\prime })\to (\overline{\mathbb{ℝ}},{ℬ}(\overline{\mathbb{ℝ}}))$ such that $f=g\circ T$, where ${ℬ}(\overline{\mathbb{ℝ}})$ 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 $\sigma (T)-{ℬ}(\overline{\mathbb{ℝ}})$ measurable because it is the composition of a $\sigma (T)-{{𝒜}}^{\prime }$ and of a ${{𝒜}}^{\prime }-{ℬ}(\overline{\mathbb{ℝ}})$ 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. $n\in {\mathbb{\mathbb{N}}}^{*},\mathrm{\forall }i\in [[1,n]],A_i\in \sigma (T)$ and ${\alpha }_i\in {\mathbb{ℝ}}^{+}$. As T is a measurable function, for all i, there exists ${A_i}^{\prime }\in {{𝒜}}^{\prime }$ such that $A_i=T^{-1}({A_i}^{\prime })$. $g=\sum _{i=1}^n{\alpha }_i{1}_{{A_i}^{\prime }}$ fulfils the requirements.

f takes only positive values

If f takes only positive values, it is the limit, for pointwise convergence, of an increasing sequence $(u_n{)}_{n\in \mathbb{\mathbb{N}}}$ of step functions. For each of these, by (1), there exists $g_n$ such that $u_n=g_n\circ T$. The function $\underset{n\to +\mathrm{\infty }}{\lim }{g}_n$, which exists on the image of T for pointwise convergence because $(u_n{)}_{n\in \mathbb{\mathbb{N}}}$ is monotonic, 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 $U=\{x:g_0^{+}(x)=+\mathrm{\infty }\}\cap \{x:g_0^{-}(x)=+\mathrm{\infty }\}$. Fortunately, $T(\mathrm{\Omega })\cap U=\varnothing$ because $g_0^{+}(T(\omega ))=f^{+}(\omega )=+\mathrm{\infty }$ always implies $g_0^{-}(T(\omega ))=f^{-}(\omega )=0$ We define $g^{+}=1_{{\mathrm{\Omega }}^{\prime }\mathrm{\setminus }U}{g}_0^{+}$ and $g^{-}=1_{{\mathrm{\Omega }}^{\prime }\mathrm{\setminus }U}{g}_0^{-}$. $g=g^{+}-g^{-}$ 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 $U^{\prime }=\{\omega :|g(\omega )|=+\mathrm{\infty }\}$. Then $g_0=1_{{\mathrm{\Omega }}^{\prime }\mathrm{\setminus }{U}^{\prime }}g$ fulfils the requirements because $U^{\prime }\cap T(\mathrm{\Omega })=\varnothing$.

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 ${ℬ}(\mathbb{ℝ})$, then the lemma becomes false (as the restrictions on $f$ are much weaker).

See also

• Doob–Dynkin lemma
• Conditional expectation

References

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