Show $int_{mathbb R^{d}}f(alpha x)dlambda^{d}(x)=frac{1}{|alpha|^{d}}int_{mathbb R^{d}}f(x)dlambda^{d}(x)$












0












$begingroup$


Let $alpha neq 0$



Use $|alpha|^{d}lambda^{d}(B)=lambda^{d}(alpha B)$ $(*)$ to:



Show that $int_{mathbb R^{d}}f(alpha x)dlambda^{d}(x)=frac{1}{|alpha|^{d}}int_{mathbb R^{d}}f(x)dlambda^{d}(x)$



Ideas:



Let $fgeq0$ and measurable then we can find $(g_{n})_{n}$ of step functions that approximate $f$, i.e. $sup g_{n}:=g=f$. Therefore



$int_{mathbb R^{d}}f(alpha x)dlambda^{d}(x)=int_{mathbb R^{d}}g(alpha x)dlambda^{d}(x)$ and by monotone convergence, we get:



$int_{mathbb R^{d}}g(alpha x)dlambda^{d}(x)=sup_{g_{n} in E, 0 leq g_{n} leq f}int_{mathbb R^{d}}sum_{i=1}^{N}alpha_{i}chi_{A_{i}}(alpha x)dlambda^{d}(x)=sup_{g_{n} in E, 0 leq g_{n} leq f}sum_{i=1}^{N}alpha_{i}lambda^{d}(A_{i})$



and using $(*)$ this means $sum_{i=1}^{N}alpha_{i}lambda^{d}(A_{i})=|alpha|^{d}sup_{g_{n} in E, 0 leq g_{n} leq f}sum_{i=1}^{N}alpha_{i}lambda^{d}(alpha^{-1}A_{i})$



But this leads nowhere...



Any ideas?






real-analysis measure-theory lebesgue-integral lebesgue-measure







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Let $alpha neq 0$



    Use $|alpha|^{d}lambda^{d}(B)=lambda^{d}(alpha B)$ $(*)$ to:



    Show that $int_{mathbb R^{d}}f(alpha x)dlambda^{d}(x)=frac{1}{|alpha|^{d}}int_{mathbb R^{d}}f(x)dlambda^{d}(x)$



    Ideas:



    Let $fgeq0$ and measurable then we can find $(g_{n})_{n}$ of step functions that approximate $f$, i.e. $sup g_{n}:=g=f$. Therefore



    $int_{mathbb R^{d}}f(alpha x)dlambda^{d}(x)=int_{mathbb R^{d}}g(alpha x)dlambda^{d}(x)$ and by monotone convergence, we get:



    $int_{mathbb R^{d}}g(alpha x)dlambda^{d}(x)=sup_{g_{n} in E, 0 leq g_{n} leq f}int_{mathbb R^{d}}sum_{i=1}^{N}alpha_{i}chi_{A_{i}}(alpha x)dlambda^{d}(x)=sup_{g_{n} in E, 0 leq g_{n} leq f}sum_{i=1}^{N}alpha_{i}lambda^{d}(A_{i})$



    and using $(*)$ this means $sum_{i=1}^{N}alpha_{i}lambda^{d}(A_{i})=|alpha|^{d}sup_{g_{n} in E, 0 leq g_{n} leq f}sum_{i=1}^{N}alpha_{i}lambda^{d}(alpha^{-1}A_{i})$



    But this leads nowhere...



    Any ideas?






    real-analysis measure-theory lebesgue-integral lebesgue-measure







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $alpha neq 0$



      Use $|alpha|^{d}lambda^{d}(B)=lambda^{d}(alpha B)$ $(*)$ to:



      Show that $int_{mathbb R^{d}}f(alpha x)dlambda^{d}(x)=frac{1}{|alpha|^{d}}int_{mathbb R^{d}}f(x)dlambda^{d}(x)$



      Ideas:



      Let $fgeq0$ and measurable then we can find $(g_{n})_{n}$ of step functions that approximate $f$, i.e. $sup g_{n}:=g=f$. Therefore



      $int_{mathbb R^{d}}f(alpha x)dlambda^{d}(x)=int_{mathbb R^{d}}g(alpha x)dlambda^{d}(x)$ and by monotone convergence, we get:



      $int_{mathbb R^{d}}g(alpha x)dlambda^{d}(x)=sup_{g_{n} in E, 0 leq g_{n} leq f}int_{mathbb R^{d}}sum_{i=1}^{N}alpha_{i}chi_{A_{i}}(alpha x)dlambda^{d}(x)=sup_{g_{n} in E, 0 leq g_{n} leq f}sum_{i=1}^{N}alpha_{i}lambda^{d}(A_{i})$



      and using $(*)$ this means $sum_{i=1}^{N}alpha_{i}lambda^{d}(A_{i})=|alpha|^{d}sup_{g_{n} in E, 0 leq g_{n} leq f}sum_{i=1}^{N}alpha_{i}lambda^{d}(alpha^{-1}A_{i})$



      But this leads nowhere...



      Any ideas?






      real-analysis measure-theory lebesgue-integral lebesgue-measure







      share|cite|improve this question









      $endgroup$




      Let $alpha neq 0$



      Use $|alpha|^{d}lambda^{d}(B)=lambda^{d}(alpha B)$ $(*)$ to:



      Show that $int_{mathbb R^{d}}f(alpha x)dlambda^{d}(x)=frac{1}{|alpha|^{d}}int_{mathbb R^{d}}f(x)dlambda^{d}(x)$



      Ideas:



      Let $fgeq0$ and measurable then we can find $(g_{n})_{n}$ of step functions that approximate $f$, i.e. $sup g_{n}:=g=f$. Therefore



      $int_{mathbb R^{d}}f(alpha x)dlambda^{d}(x)=int_{mathbb R^{d}}g(alpha x)dlambda^{d}(x)$ and by monotone convergence, we get:



      $int_{mathbb R^{d}}g(alpha x)dlambda^{d}(x)=sup_{g_{n} in E, 0 leq g_{n} leq f}int_{mathbb R^{d}}sum_{i=1}^{N}alpha_{i}chi_{A_{i}}(alpha x)dlambda^{d}(x)=sup_{g_{n} in E, 0 leq g_{n} leq f}sum_{i=1}^{N}alpha_{i}lambda^{d}(A_{i})$



      and using $(*)$ this means $sum_{i=1}^{N}alpha_{i}lambda^{d}(A_{i})=|alpha|^{d}sup_{g_{n} in E, 0 leq g_{n} leq f}sum_{i=1}^{N}alpha_{i}lambda^{d}(alpha^{-1}A_{i})$



      But this leads nowhere...



      Any ideas?






      real-analysis measure-theory lebesgue-integral lebesgue-measure




      real-analysis 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










      asked Jan 13 at 19:43









      MinaThumaMinaThuma

      798




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          1 Answer
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          $begingroup$

          You made a small mistake when integrating the indicator function.
          $$int chi_{A_i}(alpha x) , dlambda^d(x) = lambda^d(A_i/alpha) = |alpha|^{-d} lambda^d(A_i).$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for that. Is there an intuitive explanation why $int chi_{A_{i}}(alpha x)dlambda^{d} x=lambda^{d}(A_{i}/alpha)$
            $endgroup$
            – MinaThuma
            Jan 13 at 19:56








          • 2




            $begingroup$
            @MinaThuma Try checking that $chi_{A_i}(alpha x ) = chi_{A_i / alpha}(x)$ for all $x$.
            $endgroup$
            – angryavian
            Jan 13 at 19:59











          Your Answer





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          active

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          1












          $begingroup$

          You made a small mistake when integrating the indicator function.
          $$int chi_{A_i}(alpha x) , dlambda^d(x) = lambda^d(A_i/alpha) = |alpha|^{-d} lambda^d(A_i).$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for that. Is there an intuitive explanation why $int chi_{A_{i}}(alpha x)dlambda^{d} x=lambda^{d}(A_{i}/alpha)$
            $endgroup$
            – MinaThuma
            Jan 13 at 19:56








          • 2




            $begingroup$
            @MinaThuma Try checking that $chi_{A_i}(alpha x ) = chi_{A_i / alpha}(x)$ for all $x$.
            $endgroup$
            – angryavian
            Jan 13 at 19:59
















          1












          $begingroup$

          You made a small mistake when integrating the indicator function.
          $$int chi_{A_i}(alpha x) , dlambda^d(x) = lambda^d(A_i/alpha) = |alpha|^{-d} lambda^d(A_i).$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for that. Is there an intuitive explanation why $int chi_{A_{i}}(alpha x)dlambda^{d} x=lambda^{d}(A_{i}/alpha)$
            $endgroup$
            – MinaThuma
            Jan 13 at 19:56








          • 2




            $begingroup$
            @MinaThuma Try checking that $chi_{A_i}(alpha x ) = chi_{A_i / alpha}(x)$ for all $x$.
            $endgroup$
            – angryavian
            Jan 13 at 19:59














          1












          1








          1





          $begingroup$

          You made a small mistake when integrating the indicator function.
          $$int chi_{A_i}(alpha x) , dlambda^d(x) = lambda^d(A_i/alpha) = |alpha|^{-d} lambda^d(A_i).$$






          share|cite|improve this answer









          $endgroup$



          You made a small mistake when integrating the indicator function.
          $$int chi_{A_i}(alpha x) , dlambda^d(x) = lambda^d(A_i/alpha) = |alpha|^{-d} lambda^d(A_i).$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 13 at 19:49









          angryavianangryavian

          40.6k23380




          40.6k23380












          • $begingroup$
            Thanks for that. Is there an intuitive explanation why $int chi_{A_{i}}(alpha x)dlambda^{d} x=lambda^{d}(A_{i}/alpha)$
            $endgroup$
            – MinaThuma
            Jan 13 at 19:56








          • 2




            $begingroup$
            @MinaThuma Try checking that $chi_{A_i}(alpha x ) = chi_{A_i / alpha}(x)$ for all $x$.
            $endgroup$
            – angryavian
            Jan 13 at 19:59


















          • $begingroup$
            Thanks for that. Is there an intuitive explanation why $int chi_{A_{i}}(alpha x)dlambda^{d} x=lambda^{d}(A_{i}/alpha)$
            $endgroup$
            – MinaThuma
            Jan 13 at 19:56








          • 2




            $begingroup$
            @MinaThuma Try checking that $chi_{A_i}(alpha x ) = chi_{A_i / alpha}(x)$ for all $x$.
            $endgroup$
            – angryavian
            Jan 13 at 19:59
















          $begingroup$
          Thanks for that. Is there an intuitive explanation why $int chi_{A_{i}}(alpha x)dlambda^{d} x=lambda^{d}(A_{i}/alpha)$
          $endgroup$
          – MinaThuma
          Jan 13 at 19:56






          $begingroup$
          Thanks for that. Is there an intuitive explanation why $int chi_{A_{i}}(alpha x)dlambda^{d} x=lambda^{d}(A_{i}/alpha)$
          $endgroup$
          – MinaThuma
          Jan 13 at 19:56






          2




          2




          $begingroup$
          @MinaThuma Try checking that $chi_{A_i}(alpha x ) = chi_{A_i / alpha}(x)$ for all $x$.
          $endgroup$
          – angryavian
          Jan 13 at 19:59




          $begingroup$
          @MinaThuma Try checking that $chi_{A_i}(alpha x ) = chi_{A_i / alpha}(x)$ for all $x$.
          $endgroup$
          – angryavian
          Jan 13 at 19:59


















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