Consider the function $h:[0,1] \to R$ defined as $h(x)= \begin {cases} 0 & x \notin Q \\ 1/n & x=m/n, m,n \in N, mcd(n,m)=1 \end {cases}$. We also define $h(0)=1$. Prove that h is integrable and that $\int_0^1 h=0$.
I tried this: Consider a partition $P=(x_0,x_1, x_2,...x_n)$ an arbitrary partition of $[0,1]$
Considering the density of the irrationals we have that for every subinterval $[x_{k-1},x_k]$ of P there is at least one irrational in it, therefore $m_k=inf\{f(x):x\in (x_{k-1},x_k\}=0$.
Also we have that $M_k=sup\{f(x):x\in (x_{k-1},x_k)\}=1$.
(I have doubts if this is the suprema).
Then $L(P;f)=\sum_{k=1^n} m_k(x_k-x_{k-1})=0$ and $U(P;f)=1(x_1-0)+1/n(x_k-x_{k-1})$
(I'm not sure this is the correct upper sum).
And I'm stuck there, any thoughts? Thanks for your help everybody.