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Thread: 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 $\displaystyle x=0^+$

    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 $\displaystyle f(0)=0$
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