词条 | Factorization lemma |
释义 |
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. TheoremLet 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. ProofFirst, 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 functionSuppose 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 valuesIf f takes only positive values, it is the limit, for pointwise convergence, of a increasing sequence of step functions. For each of these, by (1), there exists such that . The function , which exists on the image of T for pointwise convergence because is monotonic, fulfils the requirements. General caseWe 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 onlyIf 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 spaceIf 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). See also
References
2 : Measure theory|Lemmas |
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