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 plurirectangles 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 plurirectangle, 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 plurirectangle contained in E and
L (/E , P ) = meas P1 % sup {meas P : P plurirectangle, E...
View
Full Document
 Spring '14
 Angles, Ri, measure, Lebesgue measure, Meas

Click to edit the document details