Factorization lemma

Theorem

Let T : Ω → Ω ′ be a function of a set Ω in a measure space ( Ω ′ , A ′ ) and let f : Ω → R ¯ be a scalar function on Ω . Then f is measurable with respect to the σ-algebra σ ( T ) = T − 1 ( A ′ ) generated by T in Ω if and only if there exists a measurable function g : ( Ω ′ , A ′ ) → ( R ¯ , B ( R ¯ ) ) such that f = g ∘ T , where B ( 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 ∘ T , then f is σ ( T ) − B ( R ¯ ) measurable because it is the composition of a σ ( T ) − A ′ and of a A ′ − B ( 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 = ∑ i = 1 n α i 1 A i is a step function, i.e. n ∈ N ∗ , ∀ i ∈ [ [ 1 , n ] ] , A i ∈ σ ( T ) and α i ∈ R + . As T is a measurable function, for all i, there exists A i ′ ∈ A ′ such that A i = T − 1 ( A i ′ ) . g = ∑ i = 1 n α 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 ( u n ) n ∈ N of step functions. For each of these, by (1), there exists g n such that u n = g n ∘ T . The function lim n → + ∞ g n , which exists on the image of T for pointwise convergence because ( u n ) n ∈ 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 + ∘ T and f − = g 0 − ∘ T . The problem is that the difference g := g + − g − is not defined on the set U = { x : g 0 + ( x ) = + ∞ } ∩ { x : g 0 − ( x ) = + ∞ } . Fortunately, T ( Ω ) ∩ U = ∅ because g 0 + ( T ( ω ) ) = f + ( ω ) = + ∞ always implies g 0 − ( T ( ω ) ) = f − ( ω ) = 0 We define g + = 1 Ω ′ ∖ U g 0 + and g − = 1 Ω ′ ∖ 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 ′ = { ω : | g ( ω ) | = + ∞ } . Then g 0 = 1 Ω ′ ∖ U ′ g fulfils the requirements because U ′ ∩ T ( Ω ) = ∅ .

Importance of the measure space

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