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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 plurirectangle, E 5 P } .
R This completes the proof of (29).
To prove (30), note that if P is a plurirectangles 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 plurirectangle, 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 plurirectangle containing E such that
inf {meas P : P plurirectangle, E ) P } % meas P2 = U (/E ,...
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This document was uploaded on 03/31/2014.
 Spring '14
 Angles

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