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I want to show that if C is a Jordan measurable and a measure 0 set then $\int_C 1 =0$ (this is, its Jordan Measure is zero)

My approach is to show that $L(X_C,P)=0$ for all partition $P$, But I am not sure in how to proceed in this. Thanks a lot for your help.

Note:

My definition of measure 0 is that there exist rectangles $R_1,R_2,\ldots$ such that: $C \subset \bigcup R_i$ and the the sum of the volumes $\sum v(R_i) < \epsilon$.

Jordan measurable for me is a set tha is bounded and its boundary has measure zero.

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  • $\begingroup$ Is the collection of rectangles finite? $\endgroup$ – copper.hat Mar 17 '15 at 19:18
  • $\begingroup$ No, it is infinite :) $\endgroup$ – user162343 Mar 17 '15 at 19:18
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If there is some non trivial rectangle $R$ such that $\inf_{x \in R} 1_C(x) \neq 0$ then you must have $R \subset C$, in which case $C$ cannot have measure zero. Hence $\inf_{x \in R} 1_C(x) = 0$ for all rectangles $R$, and so $L(1_C, P) = 0$.

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  • $\begingroup$ Thanks a lot, and this proves that the integral is zero right? $\endgroup$ – user162343 Mar 17 '15 at 18:58
  • $\begingroup$ But why is this: then you must have R⊂C, in which case C cannot have measure zero. $\endgroup$ – user162343 Mar 17 '15 at 18:59
  • $\begingroup$ This shows that if the integral exists it is zero. $\endgroup$ – copper.hat Mar 17 '15 at 18:59
  • $\begingroup$ Ok, and what about my above comment? $\endgroup$ – user162343 Mar 17 '15 at 19:00
  • $\begingroup$ Well, one way is to note that if $R \subset C$ then $C$ cannot have measure zero. Otherwise if $R \subset C \subset \cup_i R_i$ and so $\sum_i v(R_i) \ge v(R)>0$. Then $v(C) \ge v(R)$. $\endgroup$ – copper.hat Mar 17 '15 at 19:03

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