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30 r in particular measi e measo e and e is is

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Unformatted text preview: E = A /E (x) dx. (30) R In particular, measi E % measo E and E is is Peano—Jordan measurable if and only measi E = measo E , in which case meas E = measi E = measo E . Proof. Note that if P is a pluri-rectangles with P ) E , then /P % /E and so by Proposition 111, A A meas P = /P (x) dx % /E (x) dx R R and taking the supremum over all such P , we get sup {meas P : P pluri-rectangle, E 5 P } % A /E (x) dx. R To prove the converse inequality, let P = {R1 , . . . , Rn } be a partition of R. Observe that for every i = 1, . . . , n, inf x'Ri /E (x) = 1 if and only if Ri ) E , while in all the other cases inf x'Ri /E (x) = 0. Hence, if we set E P1 := Ri , Ri ,E 76 we have that P1 is a pluri-rectangle contained in E and L (/E , P ) = meas P1 % sup {meas P : P pluri-rectangle, E...
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