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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.

MarioM21
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    Arbitrary partitions will only get you so far. Instead, take an arbitrary partition, then add a few more points to it in strategic places. Also, to have a formal proof, you really should begin with a $\varepsilon>0$. Both because that's what the definition requires, but also because the exact meaning of "strategic places" relies on having $\varepsilon$ available. As for the supremum you have doubts about, consider what $\sup\limits_{x\in (1/3,2/3)}h(x)$ should be. – Arthur Dec 13 '20 at 21:53
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    There are not many points where $m/1$ (with $m\in \mathbb{N}$) is in $[0,1]$. Exactly two ($0/1$, $1/1$). Now try to think how many elements get mapped to $1/n$ under $h$. – Severin Schraven Dec 13 '20 at 21:57
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    See Thomae's function, and answers here, here, and here. – RRL Dec 13 '20 at 22:29

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