$mu$ is $sigma-$finite $iff exists$ function $f in mathcal{L}^{1}(mu)$ with $f(x)>0$ for all $x in X$












5












$begingroup$


Let $(X,mathcal{A}, mu)$ be a measure space. Prove:



$mu$ is $sigma-$finite $iff exists$ function $f in mathcal{L}^{1}(mu)$ with $f(x)>0$ for all $x in X$



My ideas



$"Leftarrow"$



Let $f in mathcal{L}^{1}(mu)$ with $f(x)>0$, $forall x in X$. So, $f$ is measurable. This implies that for a $B_{n}$ defined as $B_{n}:={f>frac{1}{n}}$ which is measurable, so $in mathcal{A}$. By definition, ${f>0}=bigcup_{nin mathbb N}B_{n}in mathcal{A}$,
but since $f > 0, forall x in X$ then $Xsubseteqbigcup_{nin mathbb N}B_{n}$ and $mu(B_{n})<infty, forall n in mathbb N$, since $int_{X}fdmu < infty$$Rightarrow mu$ is $sigma-$finite.



$"Rightarrow"$
I have no idea how to define this function, particularly as $f>0$, $forall x in X$






real-analysis integration measure-theory







share|cite|improve this question











$endgroup$

















    5












    $begingroup$


    Let $(X,mathcal{A}, mu)$ be a measure space. Prove:



    $mu$ is $sigma-$finite $iff exists$ function $f in mathcal{L}^{1}(mu)$ with $f(x)>0$ for all $x in X$



    My ideas



    $"Leftarrow"$



    Let $f in mathcal{L}^{1}(mu)$ with $f(x)>0$, $forall x in X$. So, $f$ is measurable. This implies that for a $B_{n}$ defined as $B_{n}:={f>frac{1}{n}}$ which is measurable, so $in mathcal{A}$. By definition, ${f>0}=bigcup_{nin mathbb N}B_{n}in mathcal{A}$,
    but since $f > 0, forall x in X$ then $Xsubseteqbigcup_{nin mathbb N}B_{n}$ and $mu(B_{n})<infty, forall n in mathbb N$, since $int_{X}fdmu < infty$$Rightarrow mu$ is $sigma-$finite.



    $"Rightarrow"$
    I have no idea how to define this function, particularly as $f>0$, $forall x in X$






    real-analysis integration measure-theory







    share|cite|improve this question











    $endgroup$















      5












      5








      5


      1



      $begingroup$


      Let $(X,mathcal{A}, mu)$ be a measure space. Prove:



      $mu$ is $sigma-$finite $iff exists$ function $f in mathcal{L}^{1}(mu)$ with $f(x)>0$ for all $x in X$



      My ideas



      $"Leftarrow"$



      Let $f in mathcal{L}^{1}(mu)$ with $f(x)>0$, $forall x in X$. So, $f$ is measurable. This implies that for a $B_{n}$ defined as $B_{n}:={f>frac{1}{n}}$ which is measurable, so $in mathcal{A}$. By definition, ${f>0}=bigcup_{nin mathbb N}B_{n}in mathcal{A}$,
      but since $f > 0, forall x in X$ then $Xsubseteqbigcup_{nin mathbb N}B_{n}$ and $mu(B_{n})<infty, forall n in mathbb N$, since $int_{X}fdmu < infty$$Rightarrow mu$ is $sigma-$finite.



      $"Rightarrow"$
      I have no idea how to define this function, particularly as $f>0$, $forall x in X$






      real-analysis integration measure-theory







      share|cite|improve this question











      $endgroup$




      Let $(X,mathcal{A}, mu)$ be a measure space. Prove:



      $mu$ is $sigma-$finite $iff exists$ function $f in mathcal{L}^{1}(mu)$ with $f(x)>0$ for all $x in X$



      My ideas



      $"Leftarrow"$



      Let $f in mathcal{L}^{1}(mu)$ with $f(x)>0$, $forall x in X$. So, $f$ is measurable. This implies that for a $B_{n}$ defined as $B_{n}:={f>frac{1}{n}}$ which is measurable, so $in mathcal{A}$. By definition, ${f>0}=bigcup_{nin mathbb N}B_{n}in mathcal{A}$,
      but since $f > 0, forall x in X$ then $Xsubseteqbigcup_{nin mathbb N}B_{n}$ and $mu(B_{n})<infty, forall n in mathbb N$, since $int_{X}fdmu < infty$$Rightarrow mu$ is $sigma-$finite.



      $"Rightarrow"$
      I have no idea how to define this function, particularly as $f>0$, $forall x in X$






      real-analysis integration measure-theory




      real-analysis integration measure-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 4 '18 at 19:10







      SABOY

















      asked Dec 4 '18 at 19:01









      SABOYSABOY

      615311




      615311






















          1 Answer
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          9












          $begingroup$

          Let $(E_n)_{n in mathbb{N}}$ be measurable (disjoint) sets with $mu(E_n) < infty$ and $Omega = bigcup_{n=1}^infty E_n$. Now define
          $$f(x) := sum_{n=1}^infty frac{1}{2^n} frac{1}{1 + mu(E_n)} 1_{E_n}(x).$$
          By definition, we have $f(x) >0$ for all $x in Omega$. On the other hand, we have
          $$int f(x) d mu(x) le sum_{n=1}^infty 2^{-n} frac{mu(E_n)}{1+mu(E_n)} le 1. $$






          share|cite|improve this answer









          $endgroup$













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            1 Answer
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            active

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            oldest

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            active

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            9












            $begingroup$

            Let $(E_n)_{n in mathbb{N}}$ be measurable (disjoint) sets with $mu(E_n) < infty$ and $Omega = bigcup_{n=1}^infty E_n$. Now define
            $$f(x) := sum_{n=1}^infty frac{1}{2^n} frac{1}{1 + mu(E_n)} 1_{E_n}(x).$$
            By definition, we have $f(x) >0$ for all $x in Omega$. On the other hand, we have
            $$int f(x) d mu(x) le sum_{n=1}^infty 2^{-n} frac{mu(E_n)}{1+mu(E_n)} le 1. $$






            share|cite|improve this answer









            $endgroup$


















              9












              $begingroup$

              Let $(E_n)_{n in mathbb{N}}$ be measurable (disjoint) sets with $mu(E_n) < infty$ and $Omega = bigcup_{n=1}^infty E_n$. Now define
              $$f(x) := sum_{n=1}^infty frac{1}{2^n} frac{1}{1 + mu(E_n)} 1_{E_n}(x).$$
              By definition, we have $f(x) >0$ for all $x in Omega$. On the other hand, we have
              $$int f(x) d mu(x) le sum_{n=1}^infty 2^{-n} frac{mu(E_n)}{1+mu(E_n)} le 1. $$






              share|cite|improve this answer









              $endgroup$
















                9












                9








                9





                $begingroup$

                Let $(E_n)_{n in mathbb{N}}$ be measurable (disjoint) sets with $mu(E_n) < infty$ and $Omega = bigcup_{n=1}^infty E_n$. Now define
                $$f(x) := sum_{n=1}^infty frac{1}{2^n} frac{1}{1 + mu(E_n)} 1_{E_n}(x).$$
                By definition, we have $f(x) >0$ for all $x in Omega$. On the other hand, we have
                $$int f(x) d mu(x) le sum_{n=1}^infty 2^{-n} frac{mu(E_n)}{1+mu(E_n)} le 1. $$






                share|cite|improve this answer









                $endgroup$



                Let $(E_n)_{n in mathbb{N}}$ be measurable (disjoint) sets with $mu(E_n) < infty$ and $Omega = bigcup_{n=1}^infty E_n$. Now define
                $$f(x) := sum_{n=1}^infty frac{1}{2^n} frac{1}{1 + mu(E_n)} 1_{E_n}(x).$$
                By definition, we have $f(x) >0$ for all $x in Omega$. On the other hand, we have
                $$int f(x) d mu(x) le sum_{n=1}^infty 2^{-n} frac{mu(E_n)}{1+mu(E_n)} le 1. $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 4 '18 at 19:16









                p4schp4sch

                4,995217




                4,995217






























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