Let $$f(x)= \begin{cases} 0 & \left( x\in \Bbb Q^{\mathrm C} \right)\\ \cfrac{1}{p} & \left( x=\cfrac{q}{p},\; p, q\in \Bbb N,\; \gcd(p, q)=1 \right)\\ \end{cases} $$ Show that $f$ is Riemann integrable on $[0,1]$.

Since there is always an irrational number in every interval of every partition $P=\{x_0 ,\; \cdots,\; x_n \}$, $$ L(P,f)=\sum_{i=1}^{n}0 \cdot \Delta x_i =0 $$ and $\sup L(P,f)=0$.

I assumed that since $f$ is Riemann integrable, then $\inf U(P,f)$ must be $0$ as well. Thus $$ U(P,f)=\sum_{i=1}^{n}\cfrac{1}{p_i} \cdot \Delta x_i =0 $$ where $p_i \in \Bbb N$, since $p_i \rightarrow \infty$ as $n \rightarrow \infty$.

But I can't seem to prove it. Am I missing something, or is my assumption just wrong?


Similar to Is Thomae's function Riemann integrable? except the definition of Riemann integrability I'm using is:

Let $U(P,f)=\sum_{i=1}^{n}M_i \Delta x_i$, $L(P,f)=\sum_{i=1}^{n}m_i \Delta x_i$ where $M_i =\sup f(x)$, $m_i =\inf f(x)$ for $x \in \left[ x_{i-1} , x_i \right]$. $f$ is Riemann integrable if $\inf U(P,f) = \sup L(P,f)$.


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