Integration of set-valued map

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[tex]If \ \mu \ is\ nonatomic, \text{ }then \begin{equation*} \overline{\int_{X}Fd\mu }=\overline{co}\int_{X}F\text{ }d\mu =\int_{X}\overline{co}\text{ }F\text{ }d\mu , \end{equation*}[/tex][tex]Where, F :X\longrightarrow P(Y)\backslash \{\phi \}\ is \text{ } called\text{ }a\text{ } set-valued\text{ } map(or\text{ } multifunction.)[/tex] [tex]\text{ }And \text{ }a \text{ }measure\text{ } \mu \text{ }is \text{ }nonatomic \text{ }if \text{ }for \text{ }any \text{ }measurable \text{ }set \text{ }A \text{ }\text{ } with\text{ } \mu(A)>0 \text{ }\\there \text{ }exists \text{ }a \text{ }measurable \text{ }subset \text{ }A_{1} \text{ }of\text{ } A\text{ }such\text{ } that\text{ } \mu(A)>\mu (A_{1})>0.\\ Is \text{ }the \text{ }conversely \text{ }is \text{ }true \text{ }of \text{ }the \text{ }previous \text{ }property. \text{ }Please \text{ }help, \text{ }give\text{ }me \text{ }example.\\Thank \text{ }u.[/tex]
 
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  • #3
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sorry, F : X→P(Y)\{φ} be a measurable, integrably bounded set-valued maps with closed images.
 

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