Factorization Lemma

THEOREM

Let

T:\Omega
ightarrow\Omega'

be a function of a set

\Omega

in a 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

, orall i\in {Link without Title} \!], 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 $A_i=T^{-1}(A_i')$ . $g=\sum_{i=1}^n\alpha_i 1_{A_i'}$ fullfills 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
ightarrow+\infty}g_n

fullfills 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=arnothing

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^-

fullfills the requirements.

''f'' takes finite values only

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	CATEGORIES ABOUT FACTORIZATION LEMMA

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Factorization Lemma