Defocus aberration - Revision history

en>Srleffler: Fix section target

2014-05-14T01:00:09Z

Fix section target

← Older revision		Revision as of 03:00, 14 May 2014
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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~~.<br><br>~~Visit~~ my ~~homepage -~~ [http://www.~~shipzula~~.com/~~blogs~~/~~post/16377 www.shipzula.com~~]		The name of the writer is Figures. One of the things she enjoys most is to read comics and she'll be beginning some thing else alongside with it. Hiring is his profession. California is our birth location.<br><br>Feel free to surf to my blog post; [http://www.youporntime.com/blog/12800 std testing at home]

en>Wavelength: inserting 1 hyphen: —> "three-dimensional"—wikt:three-dimensional

2014-02-26T02:09:50Z

inserting 1 hyphen: —> "three-dimensional"—wikt:three-dimensional

← Older revision		Revision as of 04:09, 26 February 2014
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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]].		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.<br><br>Visit my homepage - [http://www.shipzula.com/blogs/post/16377 www.shipzula.com]

	~~==Theorem==~~
	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~~.

	~~==Proof==~~
	~~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.~~

	~~===''f'' is a step function===~~
	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.~~

	~~===''f'' takes only positive values===~~
	~~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.

	~~===General case===~~
	~~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>~~
	~~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.~~

	~~===''f'' takes finite values only===~~
	~~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>.~~

	~~===Importance of the measure space===~~
	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).

	~~==References==~~
	* Heinz Bauer, Ed. (1992) ''Maß- und Integrationstheorie''. Walter de Gruyter edition. 11.~~7 Faktorisierungslemma p~~.~~ 71-72.~~

	~~[[Category:Measure theory]]~~
	~~[[Category:Lemmas]~~]

en>Epicgenius: Reverted 3 good faith edits by 2.246.114.74 using STiki (Mistake? Report it.)

2013-12-03T15:51:49Z

Reverted 3 good faith edits by 2.246.114.74 using STiki (Mistake? Report it.)

← Older revision		Revision as of 17:51, 3 December 2013
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	~~Emilia Shryock~~ is ~~my name but you can contact me something you like~~. [~~http~~://~~www~~.~~Nakedtruth~~.~~idaho~~.~~gov~~/do-I-~~have~~-~~an-std~~.~~aspx South Dakota~~] is ~~exactly where I~~'~~ve always at home std testing been living~~. ~~One of the extremely best things~~ in ~~the world at home std testing for~~ [~~http:~~//~~www~~.~~pornextras~~.~~info~~/~~user/G49Z std testing at home] him is to collect badges but he is having difficulties to discover time for~~ it~~. Hiring~~ is ~~her working day job now and she will not alter it anytime soon~~.<br><br>~~[http:~~//~~Www~~.~~channel4embarrassingillnesses~~.~~com~~/~~men~~-in-~~white~~-~~coats~~/~~genital~~-~~herpes~~/~~symptoms~~/ ~~Feel free] to visit at home std testing [http:~~//~~www~~.~~videoworld~~.~~com~~/~~user~~/~~SMaloney www~~.~~videoworld.com] my weblog~~ - ~~std testing at home~~ (~~[http:~~//~~moodle~~.~~kspu~~.~~karelia~~.~~ru/user/view~~.~~php?id=23302~~&~~course=1 Our Webpage~~])		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]].

			==Theorem==
			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.

			==Proof==
			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.

			===''f'' is a step function===
			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.

			===''f'' takes only positive values===
			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.

			===General case===
			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>
			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.

			===''f'' takes finite values only===
			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>.

			===Importance of the measure space===
			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).

			==References==
			* Heinz Bauer, Ed. (1992) ''Maß- und Integrationstheorie''. Walter de Gruyter edition. 11.7 Faktorisierungslemma p. 71-72.

			[[Category:Measure theory]]
			[[Category:Lemmas]]

en>Michael Hardy at 02:56, 19 February 2012

2012-02-19T02:56:54Z

New page

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