Rudin real anlysis: prove that $f$ is Riemann integrable












2












$begingroup$


Let $f : [0,1] →Bbb R$ satisfy the property that for every $ε>0$ the set ${x∈[0,1]: |f(x)|ge ε}$ is finite. Prove that for every continuous increasing function $α :[0,1]→Bbb R$, that $f$ is Riemann integral and integral $int_0^1f,dα=0$.



I am confused about what is the meaning of the the set ${x∈[0,1]: |f(x)|ge ε}$ is finite.
And I want to apply theorem 6.10(rudin):suppose f is bounded on[a,b], f has only finitely many points of discontinuity on[a,b], and α is continuous at every point at which f is discontinuous. Then f is Riemann integral.






real-analysis integration riemann-integration







share|cite|improve this question











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  • 2




    $begingroup$
    in this site the questions that doesn't show some effort to solve it are not well-received. So I recommend to you that edit your question showing what had you tried to solve the posted exercise.
    $endgroup$
    – Masacroso
    Dec 9 '18 at 4:20










  • $begingroup$
    The set is finite, i.e. the set contains only finitely many elements.
    $endgroup$
    – xbh
    Dec 9 '18 at 5:37










  • $begingroup$
    Loosely, it means that if you pick an $epsilon>0$ then the number of points at which $f(x)$ lies outside the band $[-epsilon,epsilon]$ is finite, and it is straightforward to create a partition that essentially 'ignores' these finite points. In particular, the lower and upper sums for the integral will be bounded below by $-epsilon(alpha(1)-alpha(0))$ and above by $epsilon (alpha(1)-alpha(0))$. Hence the integral will be zero. Continuity of $alpha$ is used to 'ignore' the finite points.
    $endgroup$
    – copper.hat
    Dec 9 '18 at 5:42
















2












$begingroup$


Let $f : [0,1] →Bbb R$ satisfy the property that for every $ε>0$ the set ${x∈[0,1]: |f(x)|ge ε}$ is finite. Prove that for every continuous increasing function $α :[0,1]→Bbb R$, that $f$ is Riemann integral and integral $int_0^1f,dα=0$.



I am confused about what is the meaning of the the set ${x∈[0,1]: |f(x)|ge ε}$ is finite.
And I want to apply theorem 6.10(rudin):suppose f is bounded on[a,b], f has only finitely many points of discontinuity on[a,b], and α is continuous at every point at which f is discontinuous. Then f is Riemann integral.






real-analysis integration riemann-integration







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    in this site the questions that doesn't show some effort to solve it are not well-received. So I recommend to you that edit your question showing what had you tried to solve the posted exercise.
    $endgroup$
    – Masacroso
    Dec 9 '18 at 4:20










  • $begingroup$
    The set is finite, i.e. the set contains only finitely many elements.
    $endgroup$
    – xbh
    Dec 9 '18 at 5:37










  • $begingroup$
    Loosely, it means that if you pick an $epsilon>0$ then the number of points at which $f(x)$ lies outside the band $[-epsilon,epsilon]$ is finite, and it is straightforward to create a partition that essentially 'ignores' these finite points. In particular, the lower and upper sums for the integral will be bounded below by $-epsilon(alpha(1)-alpha(0))$ and above by $epsilon (alpha(1)-alpha(0))$. Hence the integral will be zero. Continuity of $alpha$ is used to 'ignore' the finite points.
    $endgroup$
    – copper.hat
    Dec 9 '18 at 5:42














2












2








2





$begingroup$


Let $f : [0,1] →Bbb R$ satisfy the property that for every $ε>0$ the set ${x∈[0,1]: |f(x)|ge ε}$ is finite. Prove that for every continuous increasing function $α :[0,1]→Bbb R$, that $f$ is Riemann integral and integral $int_0^1f,dα=0$.



I am confused about what is the meaning of the the set ${x∈[0,1]: |f(x)|ge ε}$ is finite.
And I want to apply theorem 6.10(rudin):suppose f is bounded on[a,b], f has only finitely many points of discontinuity on[a,b], and α is continuous at every point at which f is discontinuous. Then f is Riemann integral.






real-analysis integration riemann-integration







share|cite|improve this question











$endgroup$




Let $f : [0,1] →Bbb R$ satisfy the property that for every $ε>0$ the set ${x∈[0,1]: |f(x)|ge ε}$ is finite. Prove that for every continuous increasing function $α :[0,1]→Bbb R$, that $f$ is Riemann integral and integral $int_0^1f,dα=0$.



I am confused about what is the meaning of the the set ${x∈[0,1]: |f(x)|ge ε}$ is finite.
And I want to apply theorem 6.10(rudin):suppose f is bounded on[a,b], f has only finitely many points of discontinuity on[a,b], and α is continuous at every point at which f is discontinuous. Then f is Riemann integral.






real-analysis integration riemann-integration




real-analysis integration riemann-integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 4:28







Bearry

















asked Dec 9 '18 at 4:14









BearryBearry

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112








  • 2




    $begingroup$
    in this site the questions that doesn't show some effort to solve it are not well-received. So I recommend to you that edit your question showing what had you tried to solve the posted exercise.
    $endgroup$
    – Masacroso
    Dec 9 '18 at 4:20










  • $begingroup$
    The set is finite, i.e. the set contains only finitely many elements.
    $endgroup$
    – xbh
    Dec 9 '18 at 5:37










  • $begingroup$
    Loosely, it means that if you pick an $epsilon>0$ then the number of points at which $f(x)$ lies outside the band $[-epsilon,epsilon]$ is finite, and it is straightforward to create a partition that essentially 'ignores' these finite points. In particular, the lower and upper sums for the integral will be bounded below by $-epsilon(alpha(1)-alpha(0))$ and above by $epsilon (alpha(1)-alpha(0))$. Hence the integral will be zero. Continuity of $alpha$ is used to 'ignore' the finite points.
    $endgroup$
    – copper.hat
    Dec 9 '18 at 5:42














  • 2




    $begingroup$
    in this site the questions that doesn't show some effort to solve it are not well-received. So I recommend to you that edit your question showing what had you tried to solve the posted exercise.
    $endgroup$
    – Masacroso
    Dec 9 '18 at 4:20










  • $begingroup$
    The set is finite, i.e. the set contains only finitely many elements.
    $endgroup$
    – xbh
    Dec 9 '18 at 5:37










  • $begingroup$
    Loosely, it means that if you pick an $epsilon>0$ then the number of points at which $f(x)$ lies outside the band $[-epsilon,epsilon]$ is finite, and it is straightforward to create a partition that essentially 'ignores' these finite points. In particular, the lower and upper sums for the integral will be bounded below by $-epsilon(alpha(1)-alpha(0))$ and above by $epsilon (alpha(1)-alpha(0))$. Hence the integral will be zero. Continuity of $alpha$ is used to 'ignore' the finite points.
    $endgroup$
    – copper.hat
    Dec 9 '18 at 5:42








2




2




$begingroup$
in this site the questions that doesn't show some effort to solve it are not well-received. So I recommend to you that edit your question showing what had you tried to solve the posted exercise.
$endgroup$
– Masacroso
Dec 9 '18 at 4:20




$begingroup$
in this site the questions that doesn't show some effort to solve it are not well-received. So I recommend to you that edit your question showing what had you tried to solve the posted exercise.
$endgroup$
– Masacroso
Dec 9 '18 at 4:20












$begingroup$
The set is finite, i.e. the set contains only finitely many elements.
$endgroup$
– xbh
Dec 9 '18 at 5:37




$begingroup$
The set is finite, i.e. the set contains only finitely many elements.
$endgroup$
– xbh
Dec 9 '18 at 5:37












$begingroup$
Loosely, it means that if you pick an $epsilon>0$ then the number of points at which $f(x)$ lies outside the band $[-epsilon,epsilon]$ is finite, and it is straightforward to create a partition that essentially 'ignores' these finite points. In particular, the lower and upper sums for the integral will be bounded below by $-epsilon(alpha(1)-alpha(0))$ and above by $epsilon (alpha(1)-alpha(0))$. Hence the integral will be zero. Continuity of $alpha$ is used to 'ignore' the finite points.
$endgroup$
– copper.hat
Dec 9 '18 at 5:42




$begingroup$
Loosely, it means that if you pick an $epsilon>0$ then the number of points at which $f(x)$ lies outside the band $[-epsilon,epsilon]$ is finite, and it is straightforward to create a partition that essentially 'ignores' these finite points. In particular, the lower and upper sums for the integral will be bounded below by $-epsilon(alpha(1)-alpha(0))$ and above by $epsilon (alpha(1)-alpha(0))$. Hence the integral will be zero. Continuity of $alpha$ is used to 'ignore' the finite points.
$endgroup$
– copper.hat
Dec 9 '18 at 5:42










1 Answer
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$begingroup$

First of all, it's not true that $f$ has only finitely many points of discontinuity. For example, $g : [0,1] to mathbf R$ defined by



$$ g(x) = begin{cases} 0 & x notin mathbf Q \ frac{1}{q} & x = frac pq text{ where } gcd(p,q) = 1 end{cases} $$



Notice that there are only finitely many rational numbers with denominator $le N$ for any given integer $N$.



On the other hand, let



$$ A = {x : f(x) ne 0} = bigcup_{n ge 1} left{x in [0,1] : |f(x)| ge frac1n right} $$



You can show that $f$ is continuous on $A^c$. (Hint: it is a similar proof to $g$ being continuous at irrational numbers.)



For $alpha(x) = x$, the theorem from the Lebesgue theory portion of Rudin says that $f$ is Riemann integrable if and only if it is continuous almost everywhere. But we don't need to use this theorem, we can prove it by hand.



Here's the sketch when $alpha(x) = x$ and I will leave it to you to fill in the details and extend to a general continuous and increasing function $alpha$.




There are finitely many points, say $x_1,dots,x_N$ in $[0,1]$ where $|f(x)| > delta$. Surround these points $x_1,dots,x_N$ by small intervals, $I_1,dots,I_N$. Let $J = [0,1] setminus (I_1 cup cdots cup I_N)$. Then a Riemann sum over this partition is bounded by
$$ left(sup_J |f(x)|right) operatorname{length}(J) + left(sup_{I_1} |f(x)|right) operatorname{length}(I_1) + dots + left(sup_{I_N} |f(x)|right) operatorname{length}(I_N) $$
If you play with the numbers right you can get this $< varepsilon$.







share|cite|improve this answer









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    1 Answer
    1






    active

    oldest

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    active

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    active

    oldest

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    0












    $begingroup$

    First of all, it's not true that $f$ has only finitely many points of discontinuity. For example, $g : [0,1] to mathbf R$ defined by



    $$ g(x) = begin{cases} 0 & x notin mathbf Q \ frac{1}{q} & x = frac pq text{ where } gcd(p,q) = 1 end{cases} $$



    Notice that there are only finitely many rational numbers with denominator $le N$ for any given integer $N$.



    On the other hand, let



    $$ A = {x : f(x) ne 0} = bigcup_{n ge 1} left{x in [0,1] : |f(x)| ge frac1n right} $$



    You can show that $f$ is continuous on $A^c$. (Hint: it is a similar proof to $g$ being continuous at irrational numbers.)



    For $alpha(x) = x$, the theorem from the Lebesgue theory portion of Rudin says that $f$ is Riemann integrable if and only if it is continuous almost everywhere. But we don't need to use this theorem, we can prove it by hand.



    Here's the sketch when $alpha(x) = x$ and I will leave it to you to fill in the details and extend to a general continuous and increasing function $alpha$.




    There are finitely many points, say $x_1,dots,x_N$ in $[0,1]$ where $|f(x)| > delta$. Surround these points $x_1,dots,x_N$ by small intervals, $I_1,dots,I_N$. Let $J = [0,1] setminus (I_1 cup cdots cup I_N)$. Then a Riemann sum over this partition is bounded by
    $$ left(sup_J |f(x)|right) operatorname{length}(J) + left(sup_{I_1} |f(x)|right) operatorname{length}(I_1) + dots + left(sup_{I_N} |f(x)|right) operatorname{length}(I_N) $$
    If you play with the numbers right you can get this $< varepsilon$.







    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      First of all, it's not true that $f$ has only finitely many points of discontinuity. For example, $g : [0,1] to mathbf R$ defined by



      $$ g(x) = begin{cases} 0 & x notin mathbf Q \ frac{1}{q} & x = frac pq text{ where } gcd(p,q) = 1 end{cases} $$



      Notice that there are only finitely many rational numbers with denominator $le N$ for any given integer $N$.



      On the other hand, let



      $$ A = {x : f(x) ne 0} = bigcup_{n ge 1} left{x in [0,1] : |f(x)| ge frac1n right} $$



      You can show that $f$ is continuous on $A^c$. (Hint: it is a similar proof to $g$ being continuous at irrational numbers.)



      For $alpha(x) = x$, the theorem from the Lebesgue theory portion of Rudin says that $f$ is Riemann integrable if and only if it is continuous almost everywhere. But we don't need to use this theorem, we can prove it by hand.



      Here's the sketch when $alpha(x) = x$ and I will leave it to you to fill in the details and extend to a general continuous and increasing function $alpha$.




      There are finitely many points, say $x_1,dots,x_N$ in $[0,1]$ where $|f(x)| > delta$. Surround these points $x_1,dots,x_N$ by small intervals, $I_1,dots,I_N$. Let $J = [0,1] setminus (I_1 cup cdots cup I_N)$. Then a Riemann sum over this partition is bounded by
      $$ left(sup_J |f(x)|right) operatorname{length}(J) + left(sup_{I_1} |f(x)|right) operatorname{length}(I_1) + dots + left(sup_{I_N} |f(x)|right) operatorname{length}(I_N) $$
      If you play with the numbers right you can get this $< varepsilon$.







      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        First of all, it's not true that $f$ has only finitely many points of discontinuity. For example, $g : [0,1] to mathbf R$ defined by



        $$ g(x) = begin{cases} 0 & x notin mathbf Q \ frac{1}{q} & x = frac pq text{ where } gcd(p,q) = 1 end{cases} $$



        Notice that there are only finitely many rational numbers with denominator $le N$ for any given integer $N$.



        On the other hand, let



        $$ A = {x : f(x) ne 0} = bigcup_{n ge 1} left{x in [0,1] : |f(x)| ge frac1n right} $$



        You can show that $f$ is continuous on $A^c$. (Hint: it is a similar proof to $g$ being continuous at irrational numbers.)



        For $alpha(x) = x$, the theorem from the Lebesgue theory portion of Rudin says that $f$ is Riemann integrable if and only if it is continuous almost everywhere. But we don't need to use this theorem, we can prove it by hand.



        Here's the sketch when $alpha(x) = x$ and I will leave it to you to fill in the details and extend to a general continuous and increasing function $alpha$.




        There are finitely many points, say $x_1,dots,x_N$ in $[0,1]$ where $|f(x)| > delta$. Surround these points $x_1,dots,x_N$ by small intervals, $I_1,dots,I_N$. Let $J = [0,1] setminus (I_1 cup cdots cup I_N)$. Then a Riemann sum over this partition is bounded by
        $$ left(sup_J |f(x)|right) operatorname{length}(J) + left(sup_{I_1} |f(x)|right) operatorname{length}(I_1) + dots + left(sup_{I_N} |f(x)|right) operatorname{length}(I_N) $$
        If you play with the numbers right you can get this $< varepsilon$.







        share|cite|improve this answer









        $endgroup$



        First of all, it's not true that $f$ has only finitely many points of discontinuity. For example, $g : [0,1] to mathbf R$ defined by



        $$ g(x) = begin{cases} 0 & x notin mathbf Q \ frac{1}{q} & x = frac pq text{ where } gcd(p,q) = 1 end{cases} $$



        Notice that there are only finitely many rational numbers with denominator $le N$ for any given integer $N$.



        On the other hand, let



        $$ A = {x : f(x) ne 0} = bigcup_{n ge 1} left{x in [0,1] : |f(x)| ge frac1n right} $$



        You can show that $f$ is continuous on $A^c$. (Hint: it is a similar proof to $g$ being continuous at irrational numbers.)



        For $alpha(x) = x$, the theorem from the Lebesgue theory portion of Rudin says that $f$ is Riemann integrable if and only if it is continuous almost everywhere. But we don't need to use this theorem, we can prove it by hand.



        Here's the sketch when $alpha(x) = x$ and I will leave it to you to fill in the details and extend to a general continuous and increasing function $alpha$.




        There are finitely many points, say $x_1,dots,x_N$ in $[0,1]$ where $|f(x)| > delta$. Surround these points $x_1,dots,x_N$ by small intervals, $I_1,dots,I_N$. Let $J = [0,1] setminus (I_1 cup cdots cup I_N)$. Then a Riemann sum over this partition is bounded by
        $$ left(sup_J |f(x)|right) operatorname{length}(J) + left(sup_{I_1} |f(x)|right) operatorname{length}(I_1) + dots + left(sup_{I_N} |f(x)|right) operatorname{length}(I_N) $$
        If you play with the numbers right you can get this $< varepsilon$.








        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 9 '18 at 6:41









        Trevor GunnTrevor Gunn

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