A Basic Question about Integrals












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$begingroup$


I have a basic question about integrals. Suppose that $f: [0,1] mapsto mathbb{R}$ is a bounded function. Do not assume that $f$ is continuous. For each $n geq 1$, define $f_n: [0,1]mapsto mathbb{R}$ by:
$$f_n(x) = sum_{j=1}^n nint_{frac{j-1}{n}}^{frac{j}{n}} f(y) dy ~mathbf{1}left(frac{j-1}{n}< x leq frac{j}{n}right)~.$$
$f_n$ may be thought of as an $n-$step piecewise approximation of $f$. My question is:



Is it true that $int_0^1|f_n(x)-f(x)| dx rightarrow 0$ as $nrightarrow infty$? Also, does this answer depend on whether the integrals in question are Riemann or they are Lebesgue (in the sense of integration w.r.t. the Lebesgue measure)?



Any help will be appreciated! Thanks.






lebesgue-integral riemann-integration







share|cite|improve this question









$endgroup$












  • $begingroup$
    The equation for $f_n(x)$ is very confusing. j is a dummy index, so the term with $x$ on the right is hard to understand.
    $endgroup$
    – herb steinberg
    Nov 24 '18 at 18:53










  • $begingroup$
    @herb steinberg, when $x in ((j-1)/n,j/n]$ $(1leq jleq n)$, $f_n(x) = n int_{(j-1)/n}^{j/n} f$. Hence, $f_n$ is a step function.
    $endgroup$
    – abcd
    Nov 24 '18 at 19:27












  • $begingroup$
    And in my original post, everything (including the term in the left with the $x$ is inside the summation.
    $endgroup$
    – abcd
    Nov 24 '18 at 19:28










  • $begingroup$
    If it is Riemann integrable, using Darboux approach will give you what you need, since the upper and lower Darboux sums will bracket $f_n(x)$.
    $endgroup$
    – herb steinberg
    Nov 24 '18 at 22:25
















3












$begingroup$


I have a basic question about integrals. Suppose that $f: [0,1] mapsto mathbb{R}$ is a bounded function. Do not assume that $f$ is continuous. For each $n geq 1$, define $f_n: [0,1]mapsto mathbb{R}$ by:
$$f_n(x) = sum_{j=1}^n nint_{frac{j-1}{n}}^{frac{j}{n}} f(y) dy ~mathbf{1}left(frac{j-1}{n}< x leq frac{j}{n}right)~.$$
$f_n$ may be thought of as an $n-$step piecewise approximation of $f$. My question is:



Is it true that $int_0^1|f_n(x)-f(x)| dx rightarrow 0$ as $nrightarrow infty$? Also, does this answer depend on whether the integrals in question are Riemann or they are Lebesgue (in the sense of integration w.r.t. the Lebesgue measure)?



Any help will be appreciated! Thanks.






lebesgue-integral riemann-integration







share|cite|improve this question









$endgroup$












  • $begingroup$
    The equation for $f_n(x)$ is very confusing. j is a dummy index, so the term with $x$ on the right is hard to understand.
    $endgroup$
    – herb steinberg
    Nov 24 '18 at 18:53










  • $begingroup$
    @herb steinberg, when $x in ((j-1)/n,j/n]$ $(1leq jleq n)$, $f_n(x) = n int_{(j-1)/n}^{j/n} f$. Hence, $f_n$ is a step function.
    $endgroup$
    – abcd
    Nov 24 '18 at 19:27












  • $begingroup$
    And in my original post, everything (including the term in the left with the $x$ is inside the summation.
    $endgroup$
    – abcd
    Nov 24 '18 at 19:28










  • $begingroup$
    If it is Riemann integrable, using Darboux approach will give you what you need, since the upper and lower Darboux sums will bracket $f_n(x)$.
    $endgroup$
    – herb steinberg
    Nov 24 '18 at 22:25














3












3








3


1



$begingroup$


I have a basic question about integrals. Suppose that $f: [0,1] mapsto mathbb{R}$ is a bounded function. Do not assume that $f$ is continuous. For each $n geq 1$, define $f_n: [0,1]mapsto mathbb{R}$ by:
$$f_n(x) = sum_{j=1}^n nint_{frac{j-1}{n}}^{frac{j}{n}} f(y) dy ~mathbf{1}left(frac{j-1}{n}< x leq frac{j}{n}right)~.$$
$f_n$ may be thought of as an $n-$step piecewise approximation of $f$. My question is:



Is it true that $int_0^1|f_n(x)-f(x)| dx rightarrow 0$ as $nrightarrow infty$? Also, does this answer depend on whether the integrals in question are Riemann or they are Lebesgue (in the sense of integration w.r.t. the Lebesgue measure)?



Any help will be appreciated! Thanks.






lebesgue-integral riemann-integration







share|cite|improve this question









$endgroup$




I have a basic question about integrals. Suppose that $f: [0,1] mapsto mathbb{R}$ is a bounded function. Do not assume that $f$ is continuous. For each $n geq 1$, define $f_n: [0,1]mapsto mathbb{R}$ by:
$$f_n(x) = sum_{j=1}^n nint_{frac{j-1}{n}}^{frac{j}{n}} f(y) dy ~mathbf{1}left(frac{j-1}{n}< x leq frac{j}{n}right)~.$$
$f_n$ may be thought of as an $n-$step piecewise approximation of $f$. My question is:



Is it true that $int_0^1|f_n(x)-f(x)| dx rightarrow 0$ as $nrightarrow infty$? Also, does this answer depend on whether the integrals in question are Riemann or they are Lebesgue (in the sense of integration w.r.t. the Lebesgue measure)?



Any help will be appreciated! Thanks.






lebesgue-integral riemann-integration




lebesgue-integral riemann-integration






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 24 '18 at 17:37









abcdabcd

817




817












  • $begingroup$
    The equation for $f_n(x)$ is very confusing. j is a dummy index, so the term with $x$ on the right is hard to understand.
    $endgroup$
    – herb steinberg
    Nov 24 '18 at 18:53










  • $begingroup$
    @herb steinberg, when $x in ((j-1)/n,j/n]$ $(1leq jleq n)$, $f_n(x) = n int_{(j-1)/n}^{j/n} f$. Hence, $f_n$ is a step function.
    $endgroup$
    – abcd
    Nov 24 '18 at 19:27












  • $begingroup$
    And in my original post, everything (including the term in the left with the $x$ is inside the summation.
    $endgroup$
    – abcd
    Nov 24 '18 at 19:28










  • $begingroup$
    If it is Riemann integrable, using Darboux approach will give you what you need, since the upper and lower Darboux sums will bracket $f_n(x)$.
    $endgroup$
    – herb steinberg
    Nov 24 '18 at 22:25


















  • $begingroup$
    The equation for $f_n(x)$ is very confusing. j is a dummy index, so the term with $x$ on the right is hard to understand.
    $endgroup$
    – herb steinberg
    Nov 24 '18 at 18:53










  • $begingroup$
    @herb steinberg, when $x in ((j-1)/n,j/n]$ $(1leq jleq n)$, $f_n(x) = n int_{(j-1)/n}^{j/n} f$. Hence, $f_n$ is a step function.
    $endgroup$
    – abcd
    Nov 24 '18 at 19:27












  • $begingroup$
    And in my original post, everything (including the term in the left with the $x$ is inside the summation.
    $endgroup$
    – abcd
    Nov 24 '18 at 19:28










  • $begingroup$
    If it is Riemann integrable, using Darboux approach will give you what you need, since the upper and lower Darboux sums will bracket $f_n(x)$.
    $endgroup$
    – herb steinberg
    Nov 24 '18 at 22:25
















$begingroup$
The equation for $f_n(x)$ is very confusing. j is a dummy index, so the term with $x$ on the right is hard to understand.
$endgroup$
– herb steinberg
Nov 24 '18 at 18:53




$begingroup$
The equation for $f_n(x)$ is very confusing. j is a dummy index, so the term with $x$ on the right is hard to understand.
$endgroup$
– herb steinberg
Nov 24 '18 at 18:53












$begingroup$
@herb steinberg, when $x in ((j-1)/n,j/n]$ $(1leq jleq n)$, $f_n(x) = n int_{(j-1)/n}^{j/n} f$. Hence, $f_n$ is a step function.
$endgroup$
– abcd
Nov 24 '18 at 19:27






$begingroup$
@herb steinberg, when $x in ((j-1)/n,j/n]$ $(1leq jleq n)$, $f_n(x) = n int_{(j-1)/n}^{j/n} f$. Hence, $f_n$ is a step function.
$endgroup$
– abcd
Nov 24 '18 at 19:27














$begingroup$
And in my original post, everything (including the term in the left with the $x$ is inside the summation.
$endgroup$
– abcd
Nov 24 '18 at 19:28




$begingroup$
And in my original post, everything (including the term in the left with the $x$ is inside the summation.
$endgroup$
– abcd
Nov 24 '18 at 19:28












$begingroup$
If it is Riemann integrable, using Darboux approach will give you what you need, since the upper and lower Darboux sums will bracket $f_n(x)$.
$endgroup$
– herb steinberg
Nov 24 '18 at 22:25




$begingroup$
If it is Riemann integrable, using Darboux approach will give you what you need, since the upper and lower Darboux sums will bracket $f_n(x)$.
$endgroup$
– herb steinberg
Nov 24 '18 at 22:25










1 Answer
1






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1












$begingroup$

This reduces for a bounded Riemann integrable function to



$$begin{align}int_0^1 |f(x) - f_n(x)| , dx &= sum_{j=1}^n int_{frac{j-1}{n}}^{frac{j}{n}}left|f(x) - nint_{frac{j-1}{n}}^{frac{j}{n}}f(y) , dy right| , dx \ &= nsum_{j=1}^n int_{frac{j-1}{n}}^{frac{j}{n}}left|frac{f(x)}{n} - int_{frac{j-1}{n}}^{frac{j}{n}}f(y) , dy right|, dx \ &= nsum_{j=1}^n int_{frac{j-1}{n}}^{frac{j}{n}}left|int_{frac{j-1}{n}}^{frac{j}{n}}[f(x) - f(y)] , dy right|,dx \ &leqslant nsum_{j=1}^n int_{frac{j-1}{n}}^{frac{j}{n}}int_{frac{j-1}{n}}^{frac{j}{n}}|f(x) - f(y)| , dy , dx \ &leqslant nsum_{j=1}^n frac{sup_{x,y in left[frac{j-1}{n},frac{j}{n}right]} |f(x) - f(y)|}{n^2} \ &= frac{1}{n}sum_{j=1}^n sup_{x in left[frac{j-1}{n},frac{j}{n}right]} f(x) - frac{1}{n}sum_{j=1}^n inf_{x in left[frac{j-1}{n},frac{j}{n}right]} f(x) end{align}$$



The RHS is a difference of upper and lower sums which converges to $0$ as $n to infty$.



The steps leading to the final difference of sums are valid if $f$ is bounded and Lebesgue integrable. I'll leave it to you to determine if the limit is always $0$ in case $f$ is Lebesgue but not Riemann integrable. Can you think of a counterexample?






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    $begingroup$

    This reduces for a bounded Riemann integrable function to



    $$begin{align}int_0^1 |f(x) - f_n(x)| , dx &= sum_{j=1}^n int_{frac{j-1}{n}}^{frac{j}{n}}left|f(x) - nint_{frac{j-1}{n}}^{frac{j}{n}}f(y) , dy right| , dx \ &= nsum_{j=1}^n int_{frac{j-1}{n}}^{frac{j}{n}}left|frac{f(x)}{n} - int_{frac{j-1}{n}}^{frac{j}{n}}f(y) , dy right|, dx \ &= nsum_{j=1}^n int_{frac{j-1}{n}}^{frac{j}{n}}left|int_{frac{j-1}{n}}^{frac{j}{n}}[f(x) - f(y)] , dy right|,dx \ &leqslant nsum_{j=1}^n int_{frac{j-1}{n}}^{frac{j}{n}}int_{frac{j-1}{n}}^{frac{j}{n}}|f(x) - f(y)| , dy , dx \ &leqslant nsum_{j=1}^n frac{sup_{x,y in left[frac{j-1}{n},frac{j}{n}right]} |f(x) - f(y)|}{n^2} \ &= frac{1}{n}sum_{j=1}^n sup_{x in left[frac{j-1}{n},frac{j}{n}right]} f(x) - frac{1}{n}sum_{j=1}^n inf_{x in left[frac{j-1}{n},frac{j}{n}right]} f(x) end{align}$$



    The RHS is a difference of upper and lower sums which converges to $0$ as $n to infty$.



    The steps leading to the final difference of sums are valid if $f$ is bounded and Lebesgue integrable. I'll leave it to you to determine if the limit is always $0$ in case $f$ is Lebesgue but not Riemann integrable. Can you think of a counterexample?






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      This reduces for a bounded Riemann integrable function to



      $$begin{align}int_0^1 |f(x) - f_n(x)| , dx &= sum_{j=1}^n int_{frac{j-1}{n}}^{frac{j}{n}}left|f(x) - nint_{frac{j-1}{n}}^{frac{j}{n}}f(y) , dy right| , dx \ &= nsum_{j=1}^n int_{frac{j-1}{n}}^{frac{j}{n}}left|frac{f(x)}{n} - int_{frac{j-1}{n}}^{frac{j}{n}}f(y) , dy right|, dx \ &= nsum_{j=1}^n int_{frac{j-1}{n}}^{frac{j}{n}}left|int_{frac{j-1}{n}}^{frac{j}{n}}[f(x) - f(y)] , dy right|,dx \ &leqslant nsum_{j=1}^n int_{frac{j-1}{n}}^{frac{j}{n}}int_{frac{j-1}{n}}^{frac{j}{n}}|f(x) - f(y)| , dy , dx \ &leqslant nsum_{j=1}^n frac{sup_{x,y in left[frac{j-1}{n},frac{j}{n}right]} |f(x) - f(y)|}{n^2} \ &= frac{1}{n}sum_{j=1}^n sup_{x in left[frac{j-1}{n},frac{j}{n}right]} f(x) - frac{1}{n}sum_{j=1}^n inf_{x in left[frac{j-1}{n},frac{j}{n}right]} f(x) end{align}$$



      The RHS is a difference of upper and lower sums which converges to $0$ as $n to infty$.



      The steps leading to the final difference of sums are valid if $f$ is bounded and Lebesgue integrable. I'll leave it to you to determine if the limit is always $0$ in case $f$ is Lebesgue but not Riemann integrable. Can you think of a counterexample?






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        This reduces for a bounded Riemann integrable function to



        $$begin{align}int_0^1 |f(x) - f_n(x)| , dx &= sum_{j=1}^n int_{frac{j-1}{n}}^{frac{j}{n}}left|f(x) - nint_{frac{j-1}{n}}^{frac{j}{n}}f(y) , dy right| , dx \ &= nsum_{j=1}^n int_{frac{j-1}{n}}^{frac{j}{n}}left|frac{f(x)}{n} - int_{frac{j-1}{n}}^{frac{j}{n}}f(y) , dy right|, dx \ &= nsum_{j=1}^n int_{frac{j-1}{n}}^{frac{j}{n}}left|int_{frac{j-1}{n}}^{frac{j}{n}}[f(x) - f(y)] , dy right|,dx \ &leqslant nsum_{j=1}^n int_{frac{j-1}{n}}^{frac{j}{n}}int_{frac{j-1}{n}}^{frac{j}{n}}|f(x) - f(y)| , dy , dx \ &leqslant nsum_{j=1}^n frac{sup_{x,y in left[frac{j-1}{n},frac{j}{n}right]} |f(x) - f(y)|}{n^2} \ &= frac{1}{n}sum_{j=1}^n sup_{x in left[frac{j-1}{n},frac{j}{n}right]} f(x) - frac{1}{n}sum_{j=1}^n inf_{x in left[frac{j-1}{n},frac{j}{n}right]} f(x) end{align}$$



        The RHS is a difference of upper and lower sums which converges to $0$ as $n to infty$.



        The steps leading to the final difference of sums are valid if $f$ is bounded and Lebesgue integrable. I'll leave it to you to determine if the limit is always $0$ in case $f$ is Lebesgue but not Riemann integrable. Can you think of a counterexample?






        share|cite|improve this answer











        $endgroup$



        This reduces for a bounded Riemann integrable function to



        $$begin{align}int_0^1 |f(x) - f_n(x)| , dx &= sum_{j=1}^n int_{frac{j-1}{n}}^{frac{j}{n}}left|f(x) - nint_{frac{j-1}{n}}^{frac{j}{n}}f(y) , dy right| , dx \ &= nsum_{j=1}^n int_{frac{j-1}{n}}^{frac{j}{n}}left|frac{f(x)}{n} - int_{frac{j-1}{n}}^{frac{j}{n}}f(y) , dy right|, dx \ &= nsum_{j=1}^n int_{frac{j-1}{n}}^{frac{j}{n}}left|int_{frac{j-1}{n}}^{frac{j}{n}}[f(x) - f(y)] , dy right|,dx \ &leqslant nsum_{j=1}^n int_{frac{j-1}{n}}^{frac{j}{n}}int_{frac{j-1}{n}}^{frac{j}{n}}|f(x) - f(y)| , dy , dx \ &leqslant nsum_{j=1}^n frac{sup_{x,y in left[frac{j-1}{n},frac{j}{n}right]} |f(x) - f(y)|}{n^2} \ &= frac{1}{n}sum_{j=1}^n sup_{x in left[frac{j-1}{n},frac{j}{n}right]} f(x) - frac{1}{n}sum_{j=1}^n inf_{x in left[frac{j-1}{n},frac{j}{n}right]} f(x) end{align}$$



        The RHS is a difference of upper and lower sums which converges to $0$ as $n to infty$.



        The steps leading to the final difference of sums are valid if $f$ is bounded and Lebesgue integrable. I'll leave it to you to determine if the limit is always $0$ in case $f$ is Lebesgue but not Riemann integrable. Can you think of a counterexample?







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 25 '18 at 7:34

























        answered Nov 25 '18 at 7:10









        RRLRRL

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