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Criteriu pentru absolut continuitatea masurilor

 
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Liviu Paunescu
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Joined: 26 Sep 2007
Posts: 85

PostPosted: Sat Mar 15, 2008 1:12 am    Post subject: Criteriu pentru absolut continuitatea masurilor Reply with quote

Fie X un spatiu topologic local compact, \lambda,\mu doua masuri boreliene pozitive pe X. Aratati ca \lambda<<\mu daca si numai daca oricare ar fi f\in C_\infty(X) pozitiva si oricare \varepsilon>0 exista \delta>0 astfel incat pentru orice h\in C_\infty(X)\ 0\leq h\leq f avem implicatia \int hd\mu\leq\delta\Rightarrow\int hd\lambda\leq\varepsilon.
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Cristi
Euclid


Joined: 07 Nov 2007
Posts: 19

PostPosted: Sun May 02, 2010 11:36 am    Post subject: Reply with quote

Pentru implicatia directa, cred ca trebuie modificat enuntul. Altfel, avem urmatorul contraexemplu:

Contraexemplu: Fie  X=(0,1),\ \mu este masura Lebesgue si  \mathrm{d} \lambda= \frac 1 {t^2} \mathrm{d}t . Luam  f sa fie functia in forma de triunghi data de  \frac 1 2 - \left| x- \frac 1 2 \right| . Luam un  \varepsilon si alegem  h=\frac f c pentru c bine ales in functie de  \delta . Atunci  \int_X h \mathrm{d}t < \delta si  \int_X h(t) \frac 1 {t^2} \mathrm{d}t =\infty .

Propozita devine adevarata daca cerem condita doar pentru  f\in L^1(\lambda) \cap C_{\infty}(X)_+.

Demonstratie: Presupunem ca exista f a.i.  \int_X f \mathrm{d} \lambda  < \infty , exista un c>0 si  0 \leq h_n \leq f cu  \int_X h_n \mathrm{d} \mu < 2^{-n} si  \int_X h_n \mathrm{d} \lambda >c .
Fie  H=\limsup h_n \leq f . Avem  \int_{X} H \mathrm{d} \mu =0 deci  \int_{X} X \mathrm{d} \lambda =0 din absolut continuitate. Dar, din lema lui Fatou avem:
 \int_{X} \liminf [f-h_n] \mathrm{d} \lambda \leq \liminf \int_{X} f-h_n  \mathrm{d}\lambda deci  \limsup \int_{X} h_n \mathrm{d} \lambda =0 , contradictie.

Pentru reciproca, as vrea sa folosesc si conditii de regularitate pentru masuri. Putem sa le presupunem ? Sau macar masurile sa fie finite pe compacte. Pe X putem sa il consideram Hausdorff?
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