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I am supposed to determine if function:

$$g(t)=\left\{\begin{matrix} \frac{1}{p+q}, t \in \left ( 0,1 \right )\cap \mathbb{Q}, p,q \in \mathbb{N}, t=\frac{p}{q}, GCD\left ( p,q \right )=1\\ 0, t \in\left ( 0,1 \right )\setminus \mathbb{Q} \end{matrix}\right.$$

  1. Is continous in $t\in \left ( 0,1 \right )\setminus \mathbb{Q}$
  2. Is integrable on $\left ( 0,1 \right )$ and if so, determine the value of integral

My solution:

  1. $\left ( \forall \varepsilon> 0 \right )\left ( \exists \sigma > 0 \right ) $ such as interval $\left ( x_{0}-\sigma , x_{0}+\sigma \right ) $ does not contain any number in the form of $\frac{1}{p+q} $, because numbers in those form are finite. So, for $t\in \left ( x_{0}-\sigma , x_{0}+\sigma \right ) $ is $g(x)=0$ and therefore function is continous in $t \in\left ( 0,1 \right )\setminus \mathbb{Q}$.
  2. $P(n)=\frac{1}{n}$, $L\left (P,g \right )=0$

Can anyone help me, how to continue, or if my solution is so far correct?

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  • $\begingroup$ I'm sorry, but your proof of (1) is completely false. The set of numbers of the form $\frac 1{p+q}$ is infinite. And they are the values of $g(t)$, not of $t$, but your argument claims they restrict $t$. $g(t) = 1/(p+q)$ for any rational value of $t$, and there are rational numbers in every open interval. Thus every neighborhood of a point in $(0,1) \setminus \Bbb Q$ will contain points where $g(t) \ne 0$. To prove this, you need to show that if the interval is small enough, the value of $p+q$ is large (just showing $q$ is large is sufficient). $\endgroup$ Nov 2, 2019 at 3:47
  • $\begingroup$ As for integrability, the lower sum for any partition is clearly $0$, so you will need to show that the upper sum goes to $0$ as the partition refines. $\endgroup$ Nov 2, 2019 at 3:59
  • $\begingroup$ @PaulSinclair can you elaborate, how I show that upper sum goes to 0? $\endgroup$
    – Peter F.
    Nov 2, 2019 at 8:25
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    $\begingroup$ if $\epsilon > 0$, then there is a $q$ such that $1/q < \epsilon$, and there are only finitely many rational numbers in $(0,1)$ whose reduced denominator is $< q$. Each of these can be contained in partition intervals as small as you want. $\endgroup$ Nov 2, 2019 at 15:55
  • $\begingroup$ @PaulSinclair thank you $\endgroup$
    – Peter F.
    Nov 2, 2019 at 19:52

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