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Main Problem. Let $f:[a,b]\to\mathbb{R}$ be a Riemann integrable function. If $I\subseteq [a,b]$ then prove or disprove that $f$ is Riemann integrable on $I$.

I know that if $I$ be a closed an bounded interval then we can say that $f$ is Riemann integrable. Basing upon this I tried to prove the claim but for arbitrary subsets, I can't figure out how to take the partitions. On the other hand trying to construct a counterexample seemed to be more difficult to me.

However, if the answer to the above question is "No", then my question is,

Let $f:[a,b]\to\mathbb{R}$ be a Riemann integrable function. If $I\subseteq [a,b]$ then what condition(s) on $I$ ensure that $f$ is Riemann integrable on $I$?

I tried to approach the above problem too but what I actually tried to prove was the following,

Let $f:[a,b]\to\mathbb{R}$ be a Riemann integrable function. If $I\subseteq [a,b]$ then $f$ is Riemann integrable on $I$ iff $I$ is closed and bounded.

But again I couldn't succeed. Can anyone help?

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    Riemann integrability is defined on closed bounded intervals anyway, so this is just a case when one omits to state explicitely a natural hypothesis of the exercise. – Did Aug 09 '16 at 14:36
  • @Did: So, the question as stated is incomplete, right? –  Aug 09 '16 at 14:38
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    @Did: Are you sure? At a superficial view, it seems to me that the theory of Riemann integration can be done on any Jordan measurable subset - but I could be wrong. – Alex M. Aug 09 '16 at 14:40
  • @AlexM. Then I suggest to pass to a less superficial view of the theory, based on actual definitions and the like... :-) (Jordan measurable subsets included in [a,b], you say? Hmmm...) – Did Aug 09 '16 at 14:42
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    @Did: If $I \subseteq [a,b]$ is Jordan measurable, then $\chi_I$ is Riemann-integrable on $[a,b]$, so its product with the integrable $f$ will again be integrable. Finally, the Jordan measurable subsets $I \subseteq \Bbb R$ are those for which $\partial I$ is Lebesgue-negligible, and this provides us with quite a rich class of subsets $I$ for which the question can be answered affirmatively. You seem to disagree; why? – Alex M. Aug 09 '16 at 15:14
  • @AlexM. While possibly interesting, these considerations do not seem canonical to me. Are they? – Did Aug 09 '16 at 15:20

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