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Math Help - integrable

  1. #1
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    integrable

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  2. #2
    Super Member PaulRS's Avatar
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    (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 <br />
f\left( {\tfrac{p}<br />
{q}} \right) = \tfrac{1}<br />
{q}{\text{ if }}\left( {p,q} \right) = 1<br />

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

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

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

    Because f(0)=0
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