Prove Lebesgue integrability for an (almost) trigonometric function

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Let $X = [0,1]$, $mathfrak{M}$ - is a $sigma$ algebra of Lebesgue measurable subsets of $X$, $mu$ - Lebesgue measure on $mathfrak{M}$ Function $f:Xtomathbb{R} $ is defined as: $f(x) = sin{nx}$ if $x in (frac{1}{2^n},frac{1}{2^{n-1}}], n in mathbb{N}, f(0) = 0$
Prove that $f$ is Lebesgue integrable and calculate $intlimits_{X}fdmu$



As far as I understand, we could use the Lebesgue dominated convergence theorem in this case showing that $sin{nx}$ converges pointwise on each interval and is bounded by a constant. Is that correct?



As for integration, using the sigma-additive property of the integral, we can express the integral this way:



$intlimits_{X}fdmu = sumlimits_{n=1}^{infty}intlimits_{A_n}fdmu, n in mathbb{N}$



Unfortunately, I don't exactly understand how to calculate this integral.






measure-theory lebesgue-integral lebesgue-measure







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Let $X = [0,1]$, $mathfrak{M}$ - is a $sigma$ algebra of Lebesgue measurable subsets of $X$, $mu$ - Lebesgue measure on $mathfrak{M}$ Function $f:Xtomathbb{R} $ is defined as: $f(x) = sin{nx}$ if $x in (frac{1}{2^n},frac{1}{2^{n-1}}], n in mathbb{N}, f(0) = 0$
    Prove that $f$ is Lebesgue integrable and calculate $intlimits_{X}fdmu$



    As far as I understand, we could use the Lebesgue dominated convergence theorem in this case showing that $sin{nx}$ converges pointwise on each interval and is bounded by a constant. Is that correct?



    As for integration, using the sigma-additive property of the integral, we can express the integral this way:



    $intlimits_{X}fdmu = sumlimits_{n=1}^{infty}intlimits_{A_n}fdmu, n in mathbb{N}$



    Unfortunately, I don't exactly understand how to calculate this integral.






    measure-theory lebesgue-integral lebesgue-measure







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Let $X = [0,1]$, $mathfrak{M}$ - is a $sigma$ algebra of Lebesgue measurable subsets of $X$, $mu$ - Lebesgue measure on $mathfrak{M}$ Function $f:Xtomathbb{R} $ is defined as: $f(x) = sin{nx}$ if $x in (frac{1}{2^n},frac{1}{2^{n-1}}], n in mathbb{N}, f(0) = 0$
      Prove that $f$ is Lebesgue integrable and calculate $intlimits_{X}fdmu$



      As far as I understand, we could use the Lebesgue dominated convergence theorem in this case showing that $sin{nx}$ converges pointwise on each interval and is bounded by a constant. Is that correct?



      As for integration, using the sigma-additive property of the integral, we can express the integral this way:



      $intlimits_{X}fdmu = sumlimits_{n=1}^{infty}intlimits_{A_n}fdmu, n in mathbb{N}$



      Unfortunately, I don't exactly understand how to calculate this integral.






      measure-theory lebesgue-integral lebesgue-measure







      share|cite|improve this question











      $endgroup$




      Let $X = [0,1]$, $mathfrak{M}$ - is a $sigma$ algebra of Lebesgue measurable subsets of $X$, $mu$ - Lebesgue measure on $mathfrak{M}$ Function $f:Xtomathbb{R} $ is defined as: $f(x) = sin{nx}$ if $x in (frac{1}{2^n},frac{1}{2^{n-1}}], n in mathbb{N}, f(0) = 0$
      Prove that $f$ is Lebesgue integrable and calculate $intlimits_{X}fdmu$



      As far as I understand, we could use the Lebesgue dominated convergence theorem in this case showing that $sin{nx}$ converges pointwise on each interval and is bounded by a constant. Is that correct?



      As for integration, using the sigma-additive property of the integral, we can express the integral this way:



      $intlimits_{X}fdmu = sumlimits_{n=1}^{infty}intlimits_{A_n}fdmu, n in mathbb{N}$



      Unfortunately, I don't exactly understand how to calculate this integral.






      measure-theory lebesgue-integral lebesgue-measure




      measure-theory lebesgue-integral lebesgue-measure






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 18 at 2:47







      Don Draper

















      asked Jan 17 at 17:42









      Don DraperDon Draper

      539




      539






















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          $begingroup$

          The sequence



          $$f_n(x) = sum_{k=1}^n sin kx , mathbf{1}_{left(2^{-k},2^{-k+1} right]}(x)$$



          converges pointwise to $f(x)$ on $[0,1]$.



          Since $0 < sin kx < 1$ for $x in left(2^{-k},2^{-k+1} right]$ we have $|f_n(x)| leqslant 1$ for all $x in [0,1]$. By the LDCT, $f$ is integrable and



          $$int_{[0,1]} f , dmu= lim_{nto infty}int_{[0,1]} f_n , dmu = lim_{nto infty}sum_{k=1}^n int_{(2^{-k},2^{-k+1}]} sin kx , dmu \ = sum_{k=1}^infty int_{(2^{-k},2^{-k+1}]} sin kx , dmu $$



          The integrals on the right-hand side can be evaluated as Riemann integrals (since Riemann integrable functions are Lebesgue integrable) and, thus,



          $$int_{[0,1]} f , dmu = sum_{k=1}^infty int_{2^{-k}}^{2^{-k+1}} sin kx , dx \ = sum_{k=1}^infty frac{cos frac{k}{2^{k}} - cos frac{k}{2^{k-1}}}{k} \ underbrace{approx 0.605359}_{text{WolframAlpha}}$$






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            1 Answer
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            1 Answer
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            active

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            $begingroup$

            The sequence



            $$f_n(x) = sum_{k=1}^n sin kx , mathbf{1}_{left(2^{-k},2^{-k+1} right]}(x)$$



            converges pointwise to $f(x)$ on $[0,1]$.



            Since $0 < sin kx < 1$ for $x in left(2^{-k},2^{-k+1} right]$ we have $|f_n(x)| leqslant 1$ for all $x in [0,1]$. By the LDCT, $f$ is integrable and



            $$int_{[0,1]} f , dmu= lim_{nto infty}int_{[0,1]} f_n , dmu = lim_{nto infty}sum_{k=1}^n int_{(2^{-k},2^{-k+1}]} sin kx , dmu \ = sum_{k=1}^infty int_{(2^{-k},2^{-k+1}]} sin kx , dmu $$



            The integrals on the right-hand side can be evaluated as Riemann integrals (since Riemann integrable functions are Lebesgue integrable) and, thus,



            $$int_{[0,1]} f , dmu = sum_{k=1}^infty int_{2^{-k}}^{2^{-k+1}} sin kx , dx \ = sum_{k=1}^infty frac{cos frac{k}{2^{k}} - cos frac{k}{2^{k-1}}}{k} \ underbrace{approx 0.605359}_{text{WolframAlpha}}$$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              The sequence



              $$f_n(x) = sum_{k=1}^n sin kx , mathbf{1}_{left(2^{-k},2^{-k+1} right]}(x)$$



              converges pointwise to $f(x)$ on $[0,1]$.



              Since $0 < sin kx < 1$ for $x in left(2^{-k},2^{-k+1} right]$ we have $|f_n(x)| leqslant 1$ for all $x in [0,1]$. By the LDCT, $f$ is integrable and



              $$int_{[0,1]} f , dmu= lim_{nto infty}int_{[0,1]} f_n , dmu = lim_{nto infty}sum_{k=1}^n int_{(2^{-k},2^{-k+1}]} sin kx , dmu \ = sum_{k=1}^infty int_{(2^{-k},2^{-k+1}]} sin kx , dmu $$



              The integrals on the right-hand side can be evaluated as Riemann integrals (since Riemann integrable functions are Lebesgue integrable) and, thus,



              $$int_{[0,1]} f , dmu = sum_{k=1}^infty int_{2^{-k}}^{2^{-k+1}} sin kx , dx \ = sum_{k=1}^infty frac{cos frac{k}{2^{k}} - cos frac{k}{2^{k-1}}}{k} \ underbrace{approx 0.605359}_{text{WolframAlpha}}$$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                The sequence



                $$f_n(x) = sum_{k=1}^n sin kx , mathbf{1}_{left(2^{-k},2^{-k+1} right]}(x)$$



                converges pointwise to $f(x)$ on $[0,1]$.



                Since $0 < sin kx < 1$ for $x in left(2^{-k},2^{-k+1} right]$ we have $|f_n(x)| leqslant 1$ for all $x in [0,1]$. By the LDCT, $f$ is integrable and



                $$int_{[0,1]} f , dmu= lim_{nto infty}int_{[0,1]} f_n , dmu = lim_{nto infty}sum_{k=1}^n int_{(2^{-k},2^{-k+1}]} sin kx , dmu \ = sum_{k=1}^infty int_{(2^{-k},2^{-k+1}]} sin kx , dmu $$



                The integrals on the right-hand side can be evaluated as Riemann integrals (since Riemann integrable functions are Lebesgue integrable) and, thus,



                $$int_{[0,1]} f , dmu = sum_{k=1}^infty int_{2^{-k}}^{2^{-k+1}} sin kx , dx \ = sum_{k=1}^infty frac{cos frac{k}{2^{k}} - cos frac{k}{2^{k-1}}}{k} \ underbrace{approx 0.605359}_{text{WolframAlpha}}$$






                share|cite|improve this answer









                $endgroup$



                The sequence



                $$f_n(x) = sum_{k=1}^n sin kx , mathbf{1}_{left(2^{-k},2^{-k+1} right]}(x)$$



                converges pointwise to $f(x)$ on $[0,1]$.



                Since $0 < sin kx < 1$ for $x in left(2^{-k},2^{-k+1} right]$ we have $|f_n(x)| leqslant 1$ for all $x in [0,1]$. By the LDCT, $f$ is integrable and



                $$int_{[0,1]} f , dmu= lim_{nto infty}int_{[0,1]} f_n , dmu = lim_{nto infty}sum_{k=1}^n int_{(2^{-k},2^{-k+1}]} sin kx , dmu \ = sum_{k=1}^infty int_{(2^{-k},2^{-k+1}]} sin kx , dmu $$



                The integrals on the right-hand side can be evaluated as Riemann integrals (since Riemann integrable functions are Lebesgue integrable) and, thus,



                $$int_{[0,1]} f , dmu = sum_{k=1}^infty int_{2^{-k}}^{2^{-k+1}} sin kx , dx \ = sum_{k=1}^infty frac{cos frac{k}{2^{k}} - cos frac{k}{2^{k-1}}}{k} \ underbrace{approx 0.605359}_{text{WolframAlpha}}$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 18 at 5:50









                RRLRRL

                51k42573




                51k42573






























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