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21-356 Lecture 29

# Taking the supremum over all partitions p of r we get

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Unformatted text preview: 5 P } . Taking the supremum over all partitions P of R, we get A /E (x) dx % sup {meas P : P pluri-rectangle, E 5 P } . R This completes the proof of (29). To prove (30), note that if P is a pluri-rectangles with P 5 E , then /E % /P and so by Proposition 111, A A /E (x) dx % /P (x) dx = meas P R R and taking the inﬁmum over all such P , we get A /E (x) dx % sup {meas P : P pluri-rectangle, E 5 P } . R To prove the opposite inequality, consider again a partition P = {R1 , . . . , Rn } of R. For every i = 1, . . . , n, we have that supx'Ri /E (x) = 0 if and only if Ri + E = ., while in all the other cases supx'Ri /E (x) = 1. Hence, if we set E P2 := Ri , Ri -E .=/ we have that P2 is a pluri-rectangle containing E such that inf {meas P : P pluri-rectangle, E ) P } % meas P2 = U (/E ,...
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