Let be a function of a set in a measure space and let be a scalar function on . Then is measurable with respect to the σ-algebra generated by in if and only if there exists a measurable function such that , where denotes the Borel set of the real numbers. If only takes finite values, then also only takes finite values.
First, if , then f is measurable because it is the composition of a and of a 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 is a step function, i.e. and . As T is a measurable function, for all i, there exists such that . fulfils the requirements.
f takes only positive values
If f takes only positive values, it is the limit of a sequence of step functions. For each of these, by (1), there exists such that . The function fulfils the requirements.
We can decompose f in a positive part and a negative part . We can then find and such that and . The problem is that the difference is not defined on the set . Fortunately, because always implies We define and . 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 . Then fulfils the requirements because .
Importance of the measure space
If the function is not scalar, but takes values in a different measurable space, such as with its trivial σ-algebra (the empty set, and the whole real line) instead of , then the lemma becomes false (as the restrictions on are much weaker).
- Heinz Bauer, Ed. (1992) Maß- und Integrationstheorie. Walter de Gruyter edition. 11.7 Faktorisierungslemma p. 71-72.