Factorization lemma
Encyclopedia
In measure theory, the factorization lemma allows us to express a function f with another function T if f is measurable with respect to T. An application of this is regression analysis
.
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.
, 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.
is a step function, i.e. 
and 
. As T is a measurable function, for all i, there exists 
such that 
. 
fulfills the requirements.
of step functions. For each of these, by (1), there exists 
such that 
. The function 
fulfils the requirements.
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.
. Then 
fulfils the requirements because 
.
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).
Regression analysis
In statistics, regression analysis includes many techniques for modeling and analyzing several variables, when the focus is on the relationship between a dependent variable and one or more independent variables...
.
Theorem
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 setBorel set
In mathematics, a Borel set is any set in a topological space that can be formed from open sets through the operations of countable union, countable intersection, and relative complement...
of the real numbers. If
only takes finite values, then also only takes finite values.Proof
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 functionStep function
In mathematics, a function on the real numbers is called a step function if it can be written as a finite linear combination of indicator functions of intervals...
, (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 . fulfills 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.General case
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).