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







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  • 4




    I don't think that $sum f(x_i)dmu (x_i) $ is a standard notation
    – Jakobian
    Jan 1 at 16:16










  • @Jakobian it is any counting measure with weight 1 on each point.
    – user7534
    Jan 1 at 16:20






  • 3




    @user7534 Then you should write $int fdmu$ or $sum_{i} f(x_i)mu({x_i})$. Not a mixup of both.
    – drhab
    Jan 1 at 16:29


















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




















  • 4




    I don't think that $sum f(x_i)dmu (x_i) $ is a standard notation
    – Jakobian
    Jan 1 at 16:16










  • @Jakobian it is any counting measure with weight 1 on each point.
    – user7534
    Jan 1 at 16:20






  • 3




    @user7534 Then you should write $int fdmu$ or $sum_{i} f(x_i)mu({x_i})$. Not a mixup of both.
    – drhab
    Jan 1 at 16:29
















1












1








1


0





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 Jan 1 at 16:26

























asked Jan 1 at 16:12









user7534

645




645








  • 4




    I don't think that $sum f(x_i)dmu (x_i) $ is a standard notation
    – Jakobian
    Jan 1 at 16:16










  • @Jakobian it is any counting measure with weight 1 on each point.
    – user7534
    Jan 1 at 16:20






  • 3




    @user7534 Then you should write $int fdmu$ or $sum_{i} f(x_i)mu({x_i})$. Not a mixup of both.
    – drhab
    Jan 1 at 16:29
















  • 4




    I don't think that $sum f(x_i)dmu (x_i) $ is a standard notation
    – Jakobian
    Jan 1 at 16:16










  • @Jakobian it is any counting measure with weight 1 on each point.
    – user7534
    Jan 1 at 16:20






  • 3




    @user7534 Then you should write $int fdmu$ or $sum_{i} f(x_i)mu({x_i})$. Not a mixup of both.
    – drhab
    Jan 1 at 16:29










4




4




I don't think that $sum f(x_i)dmu (x_i) $ is a standard notation
– Jakobian
Jan 1 at 16:16




I don't think that $sum f(x_i)dmu (x_i) $ is a standard notation
– Jakobian
Jan 1 at 16:16












@Jakobian it is any counting measure with weight 1 on each point.
– user7534
Jan 1 at 16:20




@Jakobian it is any counting measure with weight 1 on each point.
– user7534
Jan 1 at 16:20




3




3




@user7534 Then you should write $int fdmu$ or $sum_{i} f(x_i)mu({x_i})$. Not a mixup of both.
– drhab
Jan 1 at 16:29






@user7534 Then you should write $int fdmu$ or $sum_{i} f(x_i)mu({x_i})$. Not a mixup of both.
– drhab
Jan 1 at 16:29












3 Answers
3






active

oldest

votes


















7














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
    Jan 1 at 16:33






  • 1




    @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
    Jan 1 at 16:38





















2














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














    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





















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      7














      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
        Jan 1 at 16:33






      • 1




        @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
        Jan 1 at 16:38


















      7














      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
        Jan 1 at 16:33






      • 1




        @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
        Jan 1 at 16:38
















      7












      7








      7






      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 Jan 1 at 16:43

























      answered Jan 1 at 16:24









      Ben W

      1,725514




      1,725514












      • Thanks, you mind clarify the last part with a lower bound on the increment?
        – user7534
        Jan 1 at 16:33






      • 1




        @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
        Jan 1 at 16:38




















      • Thanks, you mind clarify the last part with a lower bound on the increment?
        – user7534
        Jan 1 at 16:33






      • 1




        @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
        Jan 1 at 16:38


















      Thanks, you mind clarify the last part with a lower bound on the increment?
      – user7534
      Jan 1 at 16:33




      Thanks, you mind clarify the last part with a lower bound on the increment?
      – user7534
      Jan 1 at 16:33




      1




      1




      @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
      Jan 1 at 16:38






      @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
      Jan 1 at 16:38













      2














      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


























        2














        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
























          2












          2








          2






          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 Jan 1 at 16:21









          drhab

          98.1k544129




          98.1k544129























              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 Jan 1 at 16:19









                  Ethan Bolker

                  41.7k547110




                  41.7k547110






























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