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

Question

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

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

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

im confused about option $2)$

Any hints/solution

(2) There are finite number of discontinuities. So Riemann integrable — Empty, Sep 02 '18 at 12:57

score 2 · Accepted Answer · 2018-09-02T13:17:59.837

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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