# Math Help - integrable

1. ## integrable

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

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

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

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

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

Because $f(0)=0$