Factorization lemma

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' be a function of a set Omega in a measure space (Omega,mathcal{A}') 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}') generated by T in Omega if and only if there exists a measurable function g:(Omega',mathcal{A}') 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}' 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}^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'inmathcal{A}' such that A_i=T^{-1}(A_i'). g=sum_{i=1}^nalpha_i 1_{A_i'} 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)=0We define g^+=1_{Omega'ackslash U}g_0^+ and g^-=1_{Omega'ackslash 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'ackslash U'}g fulfils the requirements because U'cap T(Omega)=varnothing.

References

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


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