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Assume that f is integrable on [a,b]. How can we show that f is integrable on every interval $[c,d]\subseteq [a,b]$?

Here is what I've done so far:

If f is integrable on [a,b] then we know for any $\epsilon > 0 \text{ } \exists \text{ a partition on [a,b] such that } U(f,p) - L(f,p) < \epsilon$

If we consider P = [c,d] as a $\textbf{sub-partition of}$ Q = [a,b], then we can say:

$$ L(f,P) \le L(F, Q) \le U(F,Q) \le U(F,P) $$

But this actually is useless and does not help us.

I appreciate your guidance.

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From your question it seems that you mean Riemann integrable. One way, which is certainly an overkill, is the use the Riemann-Lebesgue theorem: $f:I\to \mathbb R$, where $I$ is an interval in $\mathbb R$, is Riemann integrable if, and only if, $f$ is bounded and its set of discontinuities has Lebesgue measure $0$. If $J \subseteq I$ is a subinterval of $I$, then the set of discontinuities of $f$ in $J$ is clearly a subset of the set of discontinuities of $f$ in $I$, and since it is it trivial that a subset of a set of Lebesgue measure $0$ has itself measure $0$, the result follows immediately.

A more elementary and direct proof can be given though. Using Cauchy's criterion for integrability, yields a direct proof.

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  • $\begingroup$ +1. Thanks for your answer. Also thank you for the hint on Cauchy's criterion, I was able to prove it with Cauchy criterion. $\endgroup$ – Sam Mar 7 '13 at 23:41
  • $\begingroup$ you are most welcome :) $\endgroup$ – Ittay Weiss Mar 7 '13 at 23:42

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