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Is it true that if $E$ is a Jordan measurable set of measure zero, then $\int _E f=0 $ ? note that I'm talking about Riemann integral here. I managed to prove it when $E$ is compact: in that case I can take a finite amount of rectangles $R_i$ such that $E \subset \bigcup R_i $ and $\sum Vol(R_i)< \epsilon$, for every $\epsilon >0$ and then use Riemann sums on a rectangle that contains $E$. In the general case I might only have an infinite amount of rectangles that contains $E$, and in this case I'm stuck. How can I prove it?

Bernard
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user401516
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  • You should be able to manage if you really understand (1) what Jordan measurability means, (2) how a Riemann integral $\int_E f$ where $E$ is not a closed rectangle is defined, and (3) if you carried out the proof you described. If you are still stuck take a look at this. – RRL Dec 18 '17 at 17:31
  • Summing up: if $E$ has measure $0$, then $f$ can also be not integrable but, if it is, the integral is 0; if $E$ has even content $0$, then $f$ is integrable anyway. The crucial point is that $E$ cannot include any closed rectangle if it has measure $0$. – Tony Piccolo Dec 19 '17 at 12:46

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It is true.

And a set must be bounded to be Jordan measurable, so you don't need to worry about proving 'the general case'.

1likeMaths
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  • But what if this set is not closed? in this case maybe I don't have a finite number of rectangles that cover $E$... – user401516 Dec 15 '17 at 19:50
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    @user401516 - way late to the game, but for future posterity: Jordan outer measure is the infimum of the total areas of all finite collections of rectangles covering the set. If $E$ were not coverable by a finite number of rectangles, then $E$ would not have a Jordan measure at all. But in fact "bounded" and "can be covered by a finite set of rectangles" are equivalent. Being open or closed does not matter. – Paul Sinclair Jun 11 '20 at 01:09