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This question already has an answer here:

which of the following function are reimann integrable on the interval $[0,1].$?

$1)$ $f(x) =\begin{cases} 1, &\text{if x is rational }\\ 0, &\text{if x is irrational } \end{cases}$

$2)$ $f(x) =\begin{cases} 1, &\text{if x } \in \{\alpha_1,\alpha_2,.......,\alpha_n\}\\ 0, &\text{otherwise } \end{cases}$

i know that option $1) $ will not reimann integrable because it is not bounded.

im confused about option $2)$

Any hints/solution

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marked as duplicate by Jyrki Lahtonen, Adrian Keister, Arnaud D., Paul Frost, user99914 Sep 3 '18 at 17:04

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    $\begingroup$ Strange..!! (1) is not bounded? $\endgroup$ – Empty Sep 2 '18 at 12:56
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    $\begingroup$ (2) There are finite number of discontinuities. So Riemann integrable $\endgroup$ – Empty Sep 2 '18 at 12:57
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    $\begingroup$ your question is already has an answer here $\endgroup$ – Chinnapparaj R Sep 2 '18 at 13:01
  • $\begingroup$ Thanks U @ChinnapparajR $\endgroup$ – Messi fifa Sep 2 '18 at 13:04
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    $\begingroup$ Reimann? $\endgroup$ – Qmechanic Sep 2 '18 at 13:30
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Clearly, for (1), f is bounded, since $|f| \leq 1$, so your reasoning is incorrect.

Hint: Take an arbitrary partition of [0,1] and show that $U(f,P) - L(f,P)$ can not be made smaller than $1$.

Alternatively, you can notice that the set of discontinuities does not have measure 0 ($f$ is discontinuous everywhere)

For (2), the function has finitely many discontinuities, so is integrable.

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