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Suppose that $f$ is a nonnegative Riemann integrable function on $[a,b]$ satisfying $f(r) = 0$ for all $r\in\mathbb{Q}\cap [a,b]$. Prove that $\int_a^b f\,dx = 0$.

Since all rational function values give $f = 0$, does $f(x) = 0$? If yes, how can I show that formally?

If no, how would I approach this proof?

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    For the first question, this would happen if $f$ is continuous. Else, consdier ${\bf 1}_{\Bbb R\setminus \Bbb Q}$, the indicator function of the irrationals.2012-12-13
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    You cannot conclude that $f(x)= 0$ for all $x$. However, note that if $f(x_0)>0$ for some $x_0\in [a,b]$, then $x_0$ must be a point of discontinuity of $f$ (why?). What do you know about the set of points where a Riemann integrable function is discontinuous?2012-12-13
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    Hint: If $f$ is Riemann integrable, then the value of the integral can be approximated as closely as you wish by any Riemann Sum corresponding to a partition of $[a,b]$ with sufficiently small norm.2012-12-13

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