If a function is integrable, does it also have finite integral given any counting measure?












1














If the integral of a function $f$ is $L^1$,



begin{equation} int f dx < infty end{equation}



Does the same function have finite integral under any countable counting measure,



begin{equation} sum f(x_{i}) dmu(x_{i}) < infty end{equation}



for any sequance ${ x_{i} }_{i in mathbb{N}}$?



I wanna say yes, due to the latter being the same as simple functions on point sets. But also we might loose cancelation of areas when throwing away alot of stuff in this fashion.






real-analysis







share|cite|improve this question




















  • 2




    I don't think that $sum f(x_i)dmu (x_i) $ is a standard notation
    – Jakobian
    38 mins ago










  • @Jakobian it is any counting measure with weight 1 on each point.
    – user7534
    35 mins ago










  • @user7534 Then you should write $int fdmu$ or $sum_{i} f(x_i)mu({x_i})$. Not a mixup of both.
    – drhab
    26 mins ago


















1














If the integral of a function $f$ is $L^1$,



begin{equation} int f dx < infty end{equation}



Does the same function have finite integral under any countable counting measure,



begin{equation} sum f(x_{i}) dmu(x_{i}) < infty end{equation}



for any sequance ${ x_{i} }_{i in mathbb{N}}$?



I wanna say yes, due to the latter being the same as simple functions on point sets. But also we might loose cancelation of areas when throwing away alot of stuff in this fashion.






real-analysis







share|cite|improve this question




















  • 2




    I don't think that $sum f(x_i)dmu (x_i) $ is a standard notation
    – Jakobian
    38 mins ago










  • @Jakobian it is any counting measure with weight 1 on each point.
    – user7534
    35 mins ago










  • @user7534 Then you should write $int fdmu$ or $sum_{i} f(x_i)mu({x_i})$. Not a mixup of both.
    – drhab
    26 mins ago
















1












1








1







If the integral of a function $f$ is $L^1$,



begin{equation} int f dx < infty end{equation}



Does the same function have finite integral under any countable counting measure,



begin{equation} sum f(x_{i}) dmu(x_{i}) < infty end{equation}



for any sequance ${ x_{i} }_{i in mathbb{N}}$?



I wanna say yes, due to the latter being the same as simple functions on point sets. But also we might loose cancelation of areas when throwing away alot of stuff in this fashion.






real-analysis







share|cite|improve this question















If the integral of a function $f$ is $L^1$,



begin{equation} int f dx < infty end{equation}



Does the same function have finite integral under any countable counting measure,



begin{equation} sum f(x_{i}) dmu(x_{i}) < infty end{equation}



for any sequance ${ x_{i} }_{i in mathbb{N}}$?



I wanna say yes, due to the latter being the same as simple functions on point sets. But also we might loose cancelation of areas when throwing away alot of stuff in this fashion.






real-analysis




real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 28 mins ago

























asked 43 mins ago









user7534

595




595








  • 2




    I don't think that $sum f(x_i)dmu (x_i) $ is a standard notation
    – Jakobian
    38 mins ago










  • @Jakobian it is any counting measure with weight 1 on each point.
    – user7534
    35 mins ago










  • @user7534 Then you should write $int fdmu$ or $sum_{i} f(x_i)mu({x_i})$. Not a mixup of both.
    – drhab
    26 mins ago
















  • 2




    I don't think that $sum f(x_i)dmu (x_i) $ is a standard notation
    – Jakobian
    38 mins ago










  • @Jakobian it is any counting measure with weight 1 on each point.
    – user7534
    35 mins ago










  • @user7534 Then you should write $int fdmu$ or $sum_{i} f(x_i)mu({x_i})$. Not a mixup of both.
    – drhab
    26 mins ago










2




2




I don't think that $sum f(x_i)dmu (x_i) $ is a standard notation
– Jakobian
38 mins ago




I don't think that $sum f(x_i)dmu (x_i) $ is a standard notation
– Jakobian
38 mins ago












@Jakobian it is any counting measure with weight 1 on each point.
– user7534
35 mins ago




@Jakobian it is any counting measure with weight 1 on each point.
– user7534
35 mins ago












@user7534 Then you should write $int fdmu$ or $sum_{i} f(x_i)mu({x_i})$. Not a mixup of both.
– drhab
26 mins ago






@user7534 Then you should write $int fdmu$ or $sum_{i} f(x_i)mu({x_i})$. Not a mixup of both.
– drhab
26 mins ago












3 Answers
3






active

oldest

votes


















3














So, first of all, please make sure you get your notation correct, otherwise it's hard to be sure what you're asking. I think you mean to say that $fin L_1$, not $L_2$. It would also be nice to have the domain of $f$, which I assume from context is $(0,infty)$. Also, if you want to talk about an integral over the counting measure on $mathbb{N}$, the proper notation is
$$sum_{i=1}^infty f(x_i).$$
Of course, the answer is no. Consider the function $f=boldsymbol{1}_mathbb{N}$.



There are special conditions under which the answer is yes. By the Integral Test, if $f$ is nonnegative, continuous, and decreasing on $(0,infty)$ and $int f<infty$ then $sum_{n=1}^infty f(n)$ converges. In fact, so does $sum_{i=1}^infty f(x_i)$, provided $(x_i)_{i=1}^infty$ is increasing with $inf|x_{i+1}-x_i|>0$.






share|cite|improve this answer























  • Thanks, you mind clarify the last part with a lower bound on the increment?
    – user7534
    21 mins ago










  • @user7534 Sure. The gritty details are similar to the proof of the Integral Test itself (see wikipedia for an outline), but the underlying idea is quite intuitive. Suppose each $x_iin(i,i+1]$. Then $f(x_i)leq f(i)$ and so $sum f(x_i)leq sum f(i)$, the latter of which we already decided converges. Of course, it doesn't matter if we scale the $x_i$'s or cut out some of them.
    – Ben W
    16 mins ago





















1














Your question is confusing, but I think the answer is essentially, "no".



You can change the value of an integrable function on any countable set of points without changing integrability or the integral. So just define it to be large enough at each point at which your "counting measure" is supported to make the sum diverge.






share|cite|improve this answer





























    1














    If $f:mathbb Rtomathbb R$ is defined as $f=mathbf1_{mathbb Z}$ then $int f(x)^2dx=0$ but $sum_{ninmathbb Z}f(n)=infty$.






    share|cite|improve this answer





















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3














      So, first of all, please make sure you get your notation correct, otherwise it's hard to be sure what you're asking. I think you mean to say that $fin L_1$, not $L_2$. It would also be nice to have the domain of $f$, which I assume from context is $(0,infty)$. Also, if you want to talk about an integral over the counting measure on $mathbb{N}$, the proper notation is
      $$sum_{i=1}^infty f(x_i).$$
      Of course, the answer is no. Consider the function $f=boldsymbol{1}_mathbb{N}$.



      There are special conditions under which the answer is yes. By the Integral Test, if $f$ is nonnegative, continuous, and decreasing on $(0,infty)$ and $int f<infty$ then $sum_{n=1}^infty f(n)$ converges. In fact, so does $sum_{i=1}^infty f(x_i)$, provided $(x_i)_{i=1}^infty$ is increasing with $inf|x_{i+1}-x_i|>0$.






      share|cite|improve this answer























      • Thanks, you mind clarify the last part with a lower bound on the increment?
        – user7534
        21 mins ago










      • @user7534 Sure. The gritty details are similar to the proof of the Integral Test itself (see wikipedia for an outline), but the underlying idea is quite intuitive. Suppose each $x_iin(i,i+1]$. Then $f(x_i)leq f(i)$ and so $sum f(x_i)leq sum f(i)$, the latter of which we already decided converges. Of course, it doesn't matter if we scale the $x_i$'s or cut out some of them.
        – Ben W
        16 mins ago


















      3














      So, first of all, please make sure you get your notation correct, otherwise it's hard to be sure what you're asking. I think you mean to say that $fin L_1$, not $L_2$. It would also be nice to have the domain of $f$, which I assume from context is $(0,infty)$. Also, if you want to talk about an integral over the counting measure on $mathbb{N}$, the proper notation is
      $$sum_{i=1}^infty f(x_i).$$
      Of course, the answer is no. Consider the function $f=boldsymbol{1}_mathbb{N}$.



      There are special conditions under which the answer is yes. By the Integral Test, if $f$ is nonnegative, continuous, and decreasing on $(0,infty)$ and $int f<infty$ then $sum_{n=1}^infty f(n)$ converges. In fact, so does $sum_{i=1}^infty f(x_i)$, provided $(x_i)_{i=1}^infty$ is increasing with $inf|x_{i+1}-x_i|>0$.






      share|cite|improve this answer























      • Thanks, you mind clarify the last part with a lower bound on the increment?
        – user7534
        21 mins ago










      • @user7534 Sure. The gritty details are similar to the proof of the Integral Test itself (see wikipedia for an outline), but the underlying idea is quite intuitive. Suppose each $x_iin(i,i+1]$. Then $f(x_i)leq f(i)$ and so $sum f(x_i)leq sum f(i)$, the latter of which we already decided converges. Of course, it doesn't matter if we scale the $x_i$'s or cut out some of them.
        – Ben W
        16 mins ago
















      3












      3








      3






      So, first of all, please make sure you get your notation correct, otherwise it's hard to be sure what you're asking. I think you mean to say that $fin L_1$, not $L_2$. It would also be nice to have the domain of $f$, which I assume from context is $(0,infty)$. Also, if you want to talk about an integral over the counting measure on $mathbb{N}$, the proper notation is
      $$sum_{i=1}^infty f(x_i).$$
      Of course, the answer is no. Consider the function $f=boldsymbol{1}_mathbb{N}$.



      There are special conditions under which the answer is yes. By the Integral Test, if $f$ is nonnegative, continuous, and decreasing on $(0,infty)$ and $int f<infty$ then $sum_{n=1}^infty f(n)$ converges. In fact, so does $sum_{i=1}^infty f(x_i)$, provided $(x_i)_{i=1}^infty$ is increasing with $inf|x_{i+1}-x_i|>0$.






      share|cite|improve this answer














      So, first of all, please make sure you get your notation correct, otherwise it's hard to be sure what you're asking. I think you mean to say that $fin L_1$, not $L_2$. It would also be nice to have the domain of $f$, which I assume from context is $(0,infty)$. Also, if you want to talk about an integral over the counting measure on $mathbb{N}$, the proper notation is
      $$sum_{i=1}^infty f(x_i).$$
      Of course, the answer is no. Consider the function $f=boldsymbol{1}_mathbb{N}$.



      There are special conditions under which the answer is yes. By the Integral Test, if $f$ is nonnegative, continuous, and decreasing on $(0,infty)$ and $int f<infty$ then $sum_{n=1}^infty f(n)$ converges. In fact, so does $sum_{i=1}^infty f(x_i)$, provided $(x_i)_{i=1}^infty$ is increasing with $inf|x_{i+1}-x_i|>0$.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited 11 mins ago

























      answered 31 mins ago









      Ben W

      1,493513




      1,493513












      • Thanks, you mind clarify the last part with a lower bound on the increment?
        – user7534
        21 mins ago










      • @user7534 Sure. The gritty details are similar to the proof of the Integral Test itself (see wikipedia for an outline), but the underlying idea is quite intuitive. Suppose each $x_iin(i,i+1]$. Then $f(x_i)leq f(i)$ and so $sum f(x_i)leq sum f(i)$, the latter of which we already decided converges. Of course, it doesn't matter if we scale the $x_i$'s or cut out some of them.
        – Ben W
        16 mins ago




















      • Thanks, you mind clarify the last part with a lower bound on the increment?
        – user7534
        21 mins ago










      • @user7534 Sure. The gritty details are similar to the proof of the Integral Test itself (see wikipedia for an outline), but the underlying idea is quite intuitive. Suppose each $x_iin(i,i+1]$. Then $f(x_i)leq f(i)$ and so $sum f(x_i)leq sum f(i)$, the latter of which we already decided converges. Of course, it doesn't matter if we scale the $x_i$'s or cut out some of them.
        – Ben W
        16 mins ago


















      Thanks, you mind clarify the last part with a lower bound on the increment?
      – user7534
      21 mins ago




      Thanks, you mind clarify the last part with a lower bound on the increment?
      – user7534
      21 mins ago












      @user7534 Sure. The gritty details are similar to the proof of the Integral Test itself (see wikipedia for an outline), but the underlying idea is quite intuitive. Suppose each $x_iin(i,i+1]$. Then $f(x_i)leq f(i)$ and so $sum f(x_i)leq sum f(i)$, the latter of which we already decided converges. Of course, it doesn't matter if we scale the $x_i$'s or cut out some of them.
      – Ben W
      16 mins ago






      @user7534 Sure. The gritty details are similar to the proof of the Integral Test itself (see wikipedia for an outline), but the underlying idea is quite intuitive. Suppose each $x_iin(i,i+1]$. Then $f(x_i)leq f(i)$ and so $sum f(x_i)leq sum f(i)$, the latter of which we already decided converges. Of course, it doesn't matter if we scale the $x_i$'s or cut out some of them.
      – Ben W
      16 mins ago













      1














      Your question is confusing, but I think the answer is essentially, "no".



      You can change the value of an integrable function on any countable set of points without changing integrability or the integral. So just define it to be large enough at each point at which your "counting measure" is supported to make the sum diverge.






      share|cite|improve this answer


























        1














        Your question is confusing, but I think the answer is essentially, "no".



        You can change the value of an integrable function on any countable set of points without changing integrability or the integral. So just define it to be large enough at each point at which your "counting measure" is supported to make the sum diverge.






        share|cite|improve this answer
























          1












          1








          1






          Your question is confusing, but I think the answer is essentially, "no".



          You can change the value of an integrable function on any countable set of points without changing integrability or the integral. So just define it to be large enough at each point at which your "counting measure" is supported to make the sum diverge.






          share|cite|improve this answer












          Your question is confusing, but I think the answer is essentially, "no".



          You can change the value of an integrable function on any countable set of points without changing integrability or the integral. So just define it to be large enough at each point at which your "counting measure" is supported to make the sum diverge.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 36 mins ago









          Ethan Bolker

          41.3k547108




          41.3k547108























              1














              If $f:mathbb Rtomathbb R$ is defined as $f=mathbf1_{mathbb Z}$ then $int f(x)^2dx=0$ but $sum_{ninmathbb Z}f(n)=infty$.






              share|cite|improve this answer


























                1














                If $f:mathbb Rtomathbb R$ is defined as $f=mathbf1_{mathbb Z}$ then $int f(x)^2dx=0$ but $sum_{ninmathbb Z}f(n)=infty$.






                share|cite|improve this answer
























                  1












                  1








                  1






                  If $f:mathbb Rtomathbb R$ is defined as $f=mathbf1_{mathbb Z}$ then $int f(x)^2dx=0$ but $sum_{ninmathbb Z}f(n)=infty$.






                  share|cite|improve this answer












                  If $f:mathbb Rtomathbb R$ is defined as $f=mathbf1_{mathbb Z}$ then $int f(x)^2dx=0$ but $sum_{ninmathbb Z}f(n)=infty$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 33 mins ago









                  drhab

                  97.8k544129




                  97.8k544129






























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