Unformatted text preview: = 0. It follows by the previous theorem that # E is Peano—Jordan measurable with
measure zero.
Step 2: Assume that # E ) RN is Peano—Jordan measurable with measure
zero. Since E is bounded, so is E and so there exists a rectangle R containing
# E . Since meas # E = 0 by the previous theorem there exists a a plurirectangle
P containing # E such that
meas P % (.
78 The set R \ P is a pluriinterval and thus we can write it as disjoint unions of
rectangles,
n
E
R\P =
Ri .
i=1 Let P1 be the pluriinterval given by the union of all the rectangles Ri that are
contained in E , so that P1 ) E . Let P2 := P  P1 . We claim that
E ) P2 . Fix x ! E . If x does not belong to P2 , then in particular it cannot belong to P
and so it belongs to R \ P . Hence, there exists Ri such that x ! Ri . But then
Ri must be contained in E . Indeed, if not, then there exists y ! Ri + (R \ E ).
It follows by Exercise 139 that the segment S joining x and y must contained
a point on the boundary of # E , which is a contradiction since the segment S is
contained in Ri and Ri does not intersect P...
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 Spring '14
 Trigraph, Ri, measure, Lebesgue integration, Meas

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