1. ## integrable

2. (ii) We'll prove that it is continuous at $\displaystyle x=0^+$

Indeed, we needn't care about the irrationals for $\displaystyle f(x)=0$ if x is irrational, otherwise we have $\displaystyle f\left( {\tfrac{p} {q}} \right) = \tfrac{1} {q}{\text{ if }}\left( {p,q} \right) = 1$

Suppose $\displaystyle \left( {p,q} \right) = 1$ then if $\displaystyle \delta > \tfrac{p} {q} > 0$ we have $\displaystyle \delta > \tfrac{p} {q} > 0 \Rightarrow \delta \geqslant \tfrac{\delta } {p} > f\left( {\tfrac{p} {q}} \right) = \tfrac{1} {q} \geqslant 0$

And if $\displaystyle \left( {p,q} \right) > 1 \Rightarrow 0 \leqslant f\left( {\tfrac{p} {q}} \right) < \tfrac{1} {q}$ we have $\displaystyle \delta > \tfrac{p} {q} > 0 \Rightarrow \delta \geqslant \tfrac{\delta } {p} > \tfrac{1} {q} > f\left( {\tfrac{p} {q}} \right) \geqslant 0$

Set $\displaystyle \delta = \varepsilon$ then we have: $\displaystyle \forall \varepsilon > 0,\exists \delta > 0/{\text{if }}\delta > x \geqslant 0{\text{ we have }}\left| {f\left( x \right)} \right| < \varepsilon$

Because $\displaystyle f(0)=0$