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| 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]].
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| ==Theorem==
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| Let <math>T:\Omega\rightarrow\Omega'</math> be a function of a set <math>\Omega</math> in a [[measure space]] <math>(\Omega',\mathcal{A}')</math> and let <math>f:\Omega\rightarrow\overline{\mathbb{R}}</math> be a scalar function on <math>\Omega</math>. Then <math>f</math> is measurable with respect to the [[σ-algebra]] <math>\sigma(T)=T^{-1}(\mathcal{A}')</math> generated by <math>T</math> in <math>\Omega</math> if and only if there exists a measurable function <math>g:(\Omega',\mathcal{A}')\rightarrow(\overline{\mathbb{R}},\mathcal{B}(\overline{\mathbb{R}}))</math> such that <math>f=g\circ T</math>, where <math>\mathcal{B}(\overline{\mathbb{R}})</math> denotes the [[Borel set]] of the real numbers. If <math>f</math> only takes finite values, then <math>g</math> also only takes finite values.
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| ==Proof==
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| First, if <math>f=g\circ T</math>, then ''f'' is <math>\sigma(T)-\mathcal{B}(\overline{\mathbb{R}})</math> measurable because it is the composition of a <math>\sigma(T)-\mathcal{A}'</math> and of a <math>\mathcal{A}'-\mathcal{B}(\overline{\mathbb{R}})</math> 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.
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| ===''f'' is a step function===
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| Suppose <math>f=\sum_{i=1}^n\alpha_i 1_{A_i}</math> is a step function, i.e. <math>n\in\mathbb{N}^*, \forall i\in[\![1,n]\!], A_i\in\sigma(T)</math> and <math>\alpha_i\in\mathbb{R}^+</math>. As ''T'' is a measurable function, for all ''i'', there exists <math>A_i'\in\mathcal{A}'</math> such that <math>A_i=T^{-1}(A_i')</math>. <math>g=\sum_{i=1}^n\alpha_i 1_{A_i'}</math> fulfils the requirements.
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| ===''f'' takes only positive values===
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| If ''f'' takes only positive values, it is the limit of a sequence <math>(u_n)_{n\in\mathbb{N}}</math> of step functions. For each of these, by (1), there exists <math>g_n</math> such that <math>u_n=g_n\circ T</math>. The function <math>\lim_{n\rightarrow+\infty}g_n</math> fulfils the requirements.
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| ===General case===
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| We can decompose ''f'' in a positive part <math>f^+</math> and a negative part <math>f^-</math>. We can then find <math>g_0^+</math> and <math>g_0^-</math> such that <math>f^+=g_0^+\circ T</math> and <math>f^-=g_0^-\circ T</math>. The problem is that the difference <math>g:=g^+-g^-</math> is not defined on the set <math>U=\{x:g_0^+(x)=+\infty\}\cap\{x:g_0^-(x)=+\infty\}</math>. Fortunately, <math>T(\Omega)\cap U=\varnothing</math> because <math>g_0^+(T(\omega))=f^+(\omega)=+\infty</math> always implies <math>g_0^-(T(\omega))=f^-(\omega)=0</math>
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| We define <math>g^+=1_{\Omega'\backslash U}g_0^+</math> and <math>g^-=1_{\Omega'\backslash U}g_0^-</math>. <math>g=g^+-g^-</math> fulfils the requirements.
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| ===''f'' takes finite values only===
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| If ''f'' takes finite values only, we will show that ''g'' also only takes finite values. Let <math>U'=\{\omega:|g(\omega)|=+\infty\}</math>. Then <math>g_0=1_{\Omega'\backslash U'}g</math> fulfils the requirements because <math>U'\cap T(\Omega)=\varnothing</math>.
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| ===Importance of the measure space===
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| If the function <math> f </math> is not scalar, but takes values in a different measurable space, such as <math>\mathbb{R} </math> with its trivial σ-algebra (the empty set, and the whole real line) instead of <math>\mathcal{B}(\mathbb{R}) </math>, then the lemma becomes false (as the restrictions on <math> f </math> are much weaker).
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| ==References==
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| * Heinz Bauer, Ed. (1992) ''Maß- und Integrationstheorie''. Walter de Gruyter edition. 11.7 Faktorisierungslemma p. 71-72.
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| [[Category:Measure theory]]
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| [[Category:Lemmas]]
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She is known by the name of Myrtle Shryock. Since she was 18 she's been operating as a meter reader but she's usually needed her own business. One of the things he enjoys most is ice skating but he is having difficulties to find time for it. Puerto Rico is where he and his spouse live.
Visit my homepage - www.shipzula.com