Equivalent condition of a non-negative real-valued function on $[0,1]$ being measurable












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I could use some help for the problem:




Let $f$ be a non-negative real-valued function defined on $[0,1]$.



True or False? $f$ is measurable if and only if there is a sequence ${E_n}$ (finite or infinite) of measurable subsets of $[0,1]$ and a sequence of non-negative constants ${c_n}$ such that $f(x) = sum_n c_n chi_{E_n}(x)$ for every $xin[0,1]$, where $chi_{E_n}$ is the the characteristic function of $E_n$ for each $n$.




I think the statement is true. For the direction $(Longleftarrow$), since $f(x) = sum_n c_n chi_{E_n}(x)$ for every $xin[0,1]$, we see that ${sum_n^k c_n chi_{E_n}}_k$ is a sequence of simple functions on $[0,1]$ converging pointwise to $f$ and $|sum_n^k c_n chi_{E_n}|le |f|$ on $[0,1]$ for all $k$. By simple approximation theorem, $f$ is measurable.



But I got stuck on the forward ($Longrightarrow$) direction. Also, if the above proof is not right, please point it out. Thank you!






real-analysis measure-theory lebesgue-measure







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$endgroup$

















    1












    $begingroup$


    I could use some help for the problem:




    Let $f$ be a non-negative real-valued function defined on $[0,1]$.



    True or False? $f$ is measurable if and only if there is a sequence ${E_n}$ (finite or infinite) of measurable subsets of $[0,1]$ and a sequence of non-negative constants ${c_n}$ such that $f(x) = sum_n c_n chi_{E_n}(x)$ for every $xin[0,1]$, where $chi_{E_n}$ is the the characteristic function of $E_n$ for each $n$.




    I think the statement is true. For the direction $(Longleftarrow$), since $f(x) = sum_n c_n chi_{E_n}(x)$ for every $xin[0,1]$, we see that ${sum_n^k c_n chi_{E_n}}_k$ is a sequence of simple functions on $[0,1]$ converging pointwise to $f$ and $|sum_n^k c_n chi_{E_n}|le |f|$ on $[0,1]$ for all $k$. By simple approximation theorem, $f$ is measurable.



    But I got stuck on the forward ($Longrightarrow$) direction. Also, if the above proof is not right, please point it out. Thank you!






    real-analysis measure-theory lebesgue-measure







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I could use some help for the problem:




      Let $f$ be a non-negative real-valued function defined on $[0,1]$.



      True or False? $f$ is measurable if and only if there is a sequence ${E_n}$ (finite or infinite) of measurable subsets of $[0,1]$ and a sequence of non-negative constants ${c_n}$ such that $f(x) = sum_n c_n chi_{E_n}(x)$ for every $xin[0,1]$, where $chi_{E_n}$ is the the characteristic function of $E_n$ for each $n$.




      I think the statement is true. For the direction $(Longleftarrow$), since $f(x) = sum_n c_n chi_{E_n}(x)$ for every $xin[0,1]$, we see that ${sum_n^k c_n chi_{E_n}}_k$ is a sequence of simple functions on $[0,1]$ converging pointwise to $f$ and $|sum_n^k c_n chi_{E_n}|le |f|$ on $[0,1]$ for all $k$. By simple approximation theorem, $f$ is measurable.



      But I got stuck on the forward ($Longrightarrow$) direction. Also, if the above proof is not right, please point it out. Thank you!






      real-analysis measure-theory lebesgue-measure







      share|cite|improve this question











      $endgroup$




      I could use some help for the problem:




      Let $f$ be a non-negative real-valued function defined on $[0,1]$.



      True or False? $f$ is measurable if and only if there is a sequence ${E_n}$ (finite or infinite) of measurable subsets of $[0,1]$ and a sequence of non-negative constants ${c_n}$ such that $f(x) = sum_n c_n chi_{E_n}(x)$ for every $xin[0,1]$, where $chi_{E_n}$ is the the characteristic function of $E_n$ for each $n$.




      I think the statement is true. For the direction $(Longleftarrow$), since $f(x) = sum_n c_n chi_{E_n}(x)$ for every $xin[0,1]$, we see that ${sum_n^k c_n chi_{E_n}}_k$ is a sequence of simple functions on $[0,1]$ converging pointwise to $f$ and $|sum_n^k c_n chi_{E_n}|le |f|$ on $[0,1]$ for all $k$. By simple approximation theorem, $f$ is measurable.



      But I got stuck on the forward ($Longrightarrow$) direction. Also, if the above proof is not right, please point it out. Thank you!






      real-analysis measure-theory lebesgue-measure




      real-analysis measure-theory lebesgue-measure






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 7 at 3:54







      Alex

















      asked Jan 7 at 3:45









      AlexAlex

      567




      567






















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          The forward part is also true. Let $f_n$ be simple functions increasing to $f$. Then $f=f_1+(f_2-f_1)+(f_3-f_2)+cdots$. Writing $f_{n+1}-f_n=sum_{j=1}^{k_n} a_{j,n} I_{E_{jn}}$ (with $a_{j,n}$'s non-negative) we
          see that $f$ is a countable sum of terms which are constant times characteristic functions.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for your answer. Slick! I would never have come up with something like this.
            $endgroup$
            – Alex
            Jan 7 at 8:18













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          $begingroup$

          The forward part is also true. Let $f_n$ be simple functions increasing to $f$. Then $f=f_1+(f_2-f_1)+(f_3-f_2)+cdots$. Writing $f_{n+1}-f_n=sum_{j=1}^{k_n} a_{j,n} I_{E_{jn}}$ (with $a_{j,n}$'s non-negative) we
          see that $f$ is a countable sum of terms which are constant times characteristic functions.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for your answer. Slick! I would never have come up with something like this.
            $endgroup$
            – Alex
            Jan 7 at 8:18


















          1












          $begingroup$

          The forward part is also true. Let $f_n$ be simple functions increasing to $f$. Then $f=f_1+(f_2-f_1)+(f_3-f_2)+cdots$. Writing $f_{n+1}-f_n=sum_{j=1}^{k_n} a_{j,n} I_{E_{jn}}$ (with $a_{j,n}$'s non-negative) we
          see that $f$ is a countable sum of terms which are constant times characteristic functions.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for your answer. Slick! I would never have come up with something like this.
            $endgroup$
            – Alex
            Jan 7 at 8:18
















          1












          1








          1





          $begingroup$

          The forward part is also true. Let $f_n$ be simple functions increasing to $f$. Then $f=f_1+(f_2-f_1)+(f_3-f_2)+cdots$. Writing $f_{n+1}-f_n=sum_{j=1}^{k_n} a_{j,n} I_{E_{jn}}$ (with $a_{j,n}$'s non-negative) we
          see that $f$ is a countable sum of terms which are constant times characteristic functions.






          share|cite|improve this answer









          $endgroup$



          The forward part is also true. Let $f_n$ be simple functions increasing to $f$. Then $f=f_1+(f_2-f_1)+(f_3-f_2)+cdots$. Writing $f_{n+1}-f_n=sum_{j=1}^{k_n} a_{j,n} I_{E_{jn}}$ (with $a_{j,n}$'s non-negative) we
          see that $f$ is a countable sum of terms which are constant times characteristic functions.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 7 at 5:34









          Kavi Rama MurthyKavi Rama Murthy

          56k42158




          56k42158












          • $begingroup$
            Thanks for your answer. Slick! I would never have come up with something like this.
            $endgroup$
            – Alex
            Jan 7 at 8:18




















          • $begingroup$
            Thanks for your answer. Slick! I would never have come up with something like this.
            $endgroup$
            – Alex
            Jan 7 at 8:18


















          $begingroup$
          Thanks for your answer. Slick! I would never have come up with something like this.
          $endgroup$
          – Alex
          Jan 7 at 8:18






          $begingroup$
          Thanks for your answer. Slick! I would never have come up with something like this.
          $endgroup$
          – Alex
          Jan 7 at 8:18




















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