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Math Help - Analysis, integral

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Analysis, integral

    My book gave this as an excercise and I did it with a method similar to one given as an example, but I am curious of this alternate method will work. Forgive me if the proof is incoherent or just is wrong, I haven't slept in two days (long story).

    Question: Let f:x\longmapsto\left\{ \begin{array}{rcl} \frac{1}{2^n} & \mbox{if} & x=\frac{a}{2^n}~~a,n\in\mathbb{N}~2\not| a\\ 0 & \mbox{if} & \text{otherwise} \end{array}\right.. Is f\in\mathfrak{R}(\alpha) on [0,1] and if so what is the value of \int_0^1 f~d\alpha


    Notation: For my sake let S=\left\{x:\frac{a}{2^n}~~a,n\in\mathbb{N}~2\not| a\right\}


    Answer: By definition we must show that \forall \varepsilon>0 there exists a partition P of [0,1] such that U\left(P,f,\alpha\right)-L\left(P,f,\alpha\right)<\varepsilon~~{\color{red}  (*)}. Now it is clear that since on any interval [x_{i-1},x_i] there exists values of x such that x\not\in S, so consequently m_i=\inf_{[x_{i-1},x_i]}~f=0 which in turn implies that for any partiton L\left(P,f,\alpha\right)=\sum_{i=1}^{n}m_i\cdot \Delta \alpha_i=0. This realization enables to prove \color{red}(*) by proving that U\left(P,f,\alpha\right)<\varepsilon.

    To do this we must merely construct our partition carefully. Before we do this let me establish some "new" notation.

    Define \mathcal{O}_{\varphi}\left(x_i\right) as being a neighborhood of radius \varphi around x_i

    Now construct a partition of [0,1] as follows: Let a partition P_n=\left\{x_i\right\}~0\leqslant x\leqslant n of [0,1] posses the following charcteristics x_0=a, x_n=b, and x_i\not\in\mathbb{Q}~~1\leqslant i\leqslant n-1. Now it is clear that we must have 2\sum_{i} \mathcal{O}_{\varphi}\left(x_i\right)=1.


    So now it is obvious that we can make the 2^n in \frac{a}{2^n} as big as we want by making \varphi as small as we want, so we can make M_i=\sup_{[x_{i-1},x_i]}~f similarly as small as we want. Finally we can make \sup_{i}\left(M_i\right) as small as we want.

    Now let \delta be a number such that \varphi<\delta\implies \sup_{i}\left(M_i\right)\left(\alpha(b)-\alpha(a)\right)<\varepsilon. Choose a partition of [0,1] such that \max\left\{\varphi\right\}<\delta. Let \sup_{i}\left(M_i\right)=M, then

    \begin{aligned}U\left(P,f,\alpha\right)&=\sum_{i=1  }^{n}M_i\cdot\Delta \alpha_i\\<br />
&\leqslant M\sum_{i=1}^{n}\Delta\alpha_i\\<br />
&=M\left(\alpha(b)-\alpha(a)\right)\\<br />
&<\varepsilon\end{aligned}.

    This proves integrability of f. Now to find the value just note that since it is integrable that \int_0^1 f~d\alpha=\sup_{P}L\left(P,f,\alpha\right)=0


    Sorry if that is really far off.
    Last edited by Mathstud28; January 1st 2009 at 01:37 PM. Reason: Typo
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  2. #2
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    Quote Originally Posted by Mathstud28 View Post
    My book gave this as an excercise and I did it with a method similar to one given as an example, but I am curious of this alternate method will work. Forgive me if the proof is incoherent or just is wrong, I haven't slept in two days (long story).

    Question: Let f:x\longmapsto\left\{ \begin{array}{rcl} \frac{1}{2^n} & \mbox{if} & x=\frac{a}{2^n}~~a,n\in\mathbb{N}~2\not| a\\ 0 & \mbox{if} & \text{otherwise} \end{array}\right.. Is f\in\mathfrak{R}(\alpha) on [0,1] and if so what is the value of \int_0^1 f~d\alpha


    Notation: For my sake let S=\left\{x:\frac{a}{2^n}~~a,n\in\mathbb{N}~2\not| a\right\}


    Answer: By definition we must show that \forall \varepsilon>0 there exists a partition P of [0,1] such that U\left(P,f,\alpha\right)-L\left(P,f,\alpha\right)<\varepsilon~~{\color{red}  (*)}. Now it is clear that since on any interval [x_{i-1},x_i] there exists values of x such that x\not\in S, so consequently m_i=\inf_{[x_{i-1},x_i]}~f=0 which in turn implies that for any partiton L\left(P,f,\alpha\right)=\sum_{i=1}^{n}m_i\cdot \Delta \alpha_i=0. This realization enables to prove \color{red}(*) by proving that U\left(P,f,\alpha\right)<\varepsilon.

    To do this we must merely construct our partition carefully. Before we do this let me establish some "new" notation.

    Define \mathcal{O}_{\varphi}\left(x_i\right) as being a neighborhood of radius \varphi around x_i

    Now construct a partition of [0,1] as follows: Let a partition P_n=\left\{x_i\right\}~0\leqslant x\leqslant n of [0,1] posses the following charcteristics x_0=a, x_n=b, and x_i\not\in\mathbb{Q}~~1\leqslant i\leqslant n-1. Now it is clear that we must have 2\sum_{i} \mathcal{O}_{\varphi}\left(x_i\right)=1.


    So now it is obvious that we can make the 2^n in \frac{a}{2^n} as big as we want by making \varphi as small as we want, so we can make M_i=\sup_{[x_{i-1},x_i]}~f similarly as small as we want. Finally we can make \sup_{i}\left(M_i\right) as small as we want.

    Now let \delta be a number such that \varphi<\delta\implies \sup_{i}\left(M_i\right)\left(\alpha(b)-\alpha(a)\right)<\varepsilon. Choose a partition of [0,1] such that \max\left\{\varphi\right\}<\delta. Let \sup_{i}\left(M_i\right)=M, then

    \begin{aligned}U\left(P,f,\alpha\right)&=\sum_{i=1  }^{n}M_i\cdot\Delta \alpha_i\\<br />
&\leqslant M\sum_{i=1}^{n}\Delta\alpha_i\\<br />
&=M\left(\alpha(b)-\alpha(a)\right)\\<br />
&<\varepsilon\end{aligned}.

    This proves integrability of f. Now to find the value just note that since it is integrable that \int_0^1 f~d\alpha=\sup_{P}L\left(P,f,\alpha\right)=0


    Sorry if that is really far off.
    this proof is not correct, because for example since \frac{1}{2} \in [x_i, x_{i+1}], for some i, we'll have M_i=\frac{1}{2}. so we can't make all M_j as small as we want. i think (i'm not sure right now) that we also need \alpha

    to be continuous at least at elements of S in order to make \Delta \alpha_i as small as we want. you choose n large enough so that \frac{1}{2^n} < \epsilon. now let x_i=\frac{i}{2^n}, \ 1 \leq i \leq 2^{n-1}. choose disjoint neighbourhoods

    U_i=[a_i,b_i] of x_i such that \sum_i \Delta \alpha_i < \epsilon. then \sum_i (\sup f)\Delta \alpha_i < \frac{\epsilon}{2}. let V_i=[b_{i-1}, a_i], \ 1 \leq i \leq 2^{n-1} + 1, with b_0=0, \ a_{2^{n-1} + 1}=1. finally over \bigcup_i V_i we have:

    \sum (\sup f) \Delta \alpha_i \leq 2^{-n} ( \alpha(b) - \alpha(a)) < \epsilon( \alpha(b)- \alpha(a)).
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    this proof is not correct, because for example since \frac{1}{2} \in [x_i, x_{i+1}], for some i, we'll have M_i=\frac{1}{2}. so we can't make all M_j as small as we want. i think (i'm not sure right now) that we also need \alpha

    to be continuous at least at elements of S in order to make \Delta \alpha_i as small as we want. you choose n large enough so that \frac{1}{2^n} < \epsilon. now let x_i=\frac{i}{2^n}, \ 1 \leq i \leq 2^{n-1}. choose disjoint neighbourhoods

    U_i=[a_i,b_i] of x_i such that \sum_i \Delta \alpha_i < \epsilon. then \sum_i (\sup f)\Delta \alpha_i < \frac{\epsilon}{2}. let V_i=[b_{i-1}, a_i], \ 1 \leq i \leq 2^{n-1} + 1, with b_0=0, \ a_{2^{n-1} + 1}=1. finally over \bigcup_i V_i we have:

    \sum (\sup f) \Delta \alpha_i \leq 2^{-n} ( \alpha(b) - \alpha(a)) < \epsilon( \alpha(b)- \alpha(a)).
    Thanks NonCommAlg...in my state of coffee induced semi-catatonia I was thinking of \sup as being the biggest 2^n


    Can you also use this proof assuming that \alpha is continuous?

    Consider a partition P_n of [0,1] containing n elements. Since \alpha is continuous define \Delta \alpha_i=\frac{\alpha(b)-\alpha(a)}{n^2}. Now as you stated M_i\leqslant\frac{1}{2}. So

    \begin{aligned}U\left(P_n,f,\alpha\right)&=\sum_{i  =1}^{n}M_i\cdot\Delta\alpha_i\\<br />
&\leqslant \frac{1}{2}\sum_{i=1}^{n}\Delta\alpha_i\\<br />
&=\frac{\alpha(b)-\alpha(a)}{2n^2}\sum_{i}^{n}1\\<br />
&=\frac{\alpha(b)-\alpha(a)}{2n}<\varepsilon\end{aligned}

    for sufficiently large n. And since U\left(P_n,f,\alpha\right)-L\left(P_n,f,\alpha\right)=U\left(P_n,f,\alpha\rig  ht)<\varepsilon we have that f\in\mathfrak{R}(\alpha). My answer still stands that L\left(P_n,f,\alpha\right)=0\implies\int_0^1 f~d\alpha=\sup_{P_n}L\left(P_n,f,\alpha\right)=0

    Am I assuming something obviously not true in my above proof?

    EDIT: Yes, I see what is wrong...it would be \sum_{i=1}^{n^2}...I understand your proof so Ill just let this be lesson learned unless I think of something better.

    If it makes up for it, I think I realized that we can strengthen one of the theorems in my book...I think we can show that if \sum f is bounded and \alpha is continuous then f\in\mathfrak{R}(\alpha)
    Last edited by Mathstud28; January 2nd 2009 at 09:13 PM.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Hello everyone and thank you for reading this. On my way home from school I think I came up with another way to prove this. If any of the more knowledgable members here could take a look I would greatly appreciate it.


    Question: Let f:x\longmapsto\left\{ \begin{array}{rcl} \frac{1}{2^n} & \mbox{if} & x=\frac{a}{2^n}~~a,n\in\mathbb{N}~2\not| a\\ 0 & \mbox{if} & \text{otherwise} \end{array}\right.. Is f\in\mathfrak{R}(\alpha) on [0,1] and if so what is the value of \int_0^1 f~d\alpha


    Notation: For my sake let S=\left\{x:\frac{a}{2^n}~~a,n\in\mathbb{N}~2\not| a\right\}

    Addendum: Let \alpha be continuous on [0,1]
    Just as before we want to show that for every \varepsilon>0 there exists a partition P of [0,1] such that U\left(P,f,\alpha\right)-L\left(P,f,\alpha\right)<\varepsilon\quad{\color{r  ed}(*)}.

    Now note that since given any interval of [0,1] there must exist x\not\in S in that interval, which implies that for any interval m_i=\inf_{[x_{i-1},x_i]}f=0 which in turn implies that L\left(P,f,\alpha\right)=\sum_{i=1}^{n}m_i\cdot\De  lta\alpha_i=0. Due to this we can see that \color{red}(*) is equivalent to showing that U\left(P,f,\alpha\right)<\varepsilon

    We prove this as follows: For any partition P_{n+1} of [0,1] containing n+1] elements define \Delta \alpha_i=\frac{\alpha(b)-\alpha(a)}{n}, this can be done by the continuity of \alpha. Now we can see that the partition P_{n+1} divides [0,1] into n subintervals. Now it can be seen that

    \begin{aligned}U\left(P_{n+1},f,\alpha\right)&=\su  m_{i=1}^{n}M_i\cdot\Delta \alpha_i\\<br />
&=\frac{\alpha(b)-\alpha(a)}{n}\sum_{i=1}^{n}M_i\\<br />
&\leqslant\frac{\alpha(b)-\alpha(a)}{n}\sum_{i=1}^{n}\frac{1}{2^n}\\<br />
&\leqslant\frac{2\left(\alpha(b)-\alpha(a)\right)}{n}<\varepsilon\end{aligned}

    For sufficiently large enough n




    Is this correct? Or am I misinterpreting the function?
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