Application of DCT, integrability on $[0,infty)$












0












$begingroup$


Let $f$ be integrable on $[0,infty$).



Show that lim$_{n rightarrow infty}$ $int_0^{infty}$ $f(x) frac{x}{n+x}$ $dx$ = $0$.



Ok so I want to say I can take the limit inside, but this is a consequence of DCT or MCT, so to apply DCT, this is what I have come up with so far:



define $f_n$ = $f(x)$ $cdot$ $frac{x}{n+x}$ (my intuition from previous measure theory problems is that I need this step)



then {$f_n}$ is a sequence of measurable functions,



but are the $f_n$ bounded above $f(x)$ ?



*Note by assumption we have $int_0^{infty}$ $f(x)$ $dx < infty$.



Using all of this am I able to somehow deduce that



lim$_{n rightarrow infty}$ $int_0^{infty}$ $f(x) frac{x}{n+x}$ $dx$ = $int_0^{infty}$ $f(x)$ lim$_{n rightarrow infty}$ $frac{x}{n+x}$ $dx$ = $int_0^{infty}$ $f(x)$ $dx$ $int_0^{infty}$ lim$_{n rightarrow infty}$ $frac{x}{n+x}$ $dx$



I am VERY hesitant on my above two equalities, specially the last one, where am I wrong or need input? Thanks in advance and sorry if I am missing something obvious.






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







share|cite|improve this question









$endgroup$












  • $begingroup$
    Are you really asking if $frac{x}{n+x} le 1$ for $x > 0$ and $n ge 1$?
    $endgroup$
    – mathworker21
    Dec 23 '18 at 22:17












  • $begingroup$
    omg duh, $x > 0$. I didn't consider that, thats why I said forgive me If im asking something obvious, Im learning, I don't learn fast. SORRY FOR BEING STUPID.
    $endgroup$
    – Hossien Sahebjame
    Dec 23 '18 at 22:27










  • $begingroup$
    happens to all of us
    $endgroup$
    – mathworker21
    Dec 23 '18 at 22:30










  • $begingroup$
    lol thanks, again, Im new to measure theory, im really trying to get a lot better so feel free to comment if you ever see me ask noob questions.
    $endgroup$
    – Hossien Sahebjame
    Dec 23 '18 at 22:32
















0












$begingroup$


Let $f$ be integrable on $[0,infty$).



Show that lim$_{n rightarrow infty}$ $int_0^{infty}$ $f(x) frac{x}{n+x}$ $dx$ = $0$.



Ok so I want to say I can take the limit inside, but this is a consequence of DCT or MCT, so to apply DCT, this is what I have come up with so far:



define $f_n$ = $f(x)$ $cdot$ $frac{x}{n+x}$ (my intuition from previous measure theory problems is that I need this step)



then {$f_n}$ is a sequence of measurable functions,



but are the $f_n$ bounded above $f(x)$ ?



*Note by assumption we have $int_0^{infty}$ $f(x)$ $dx < infty$.



Using all of this am I able to somehow deduce that



lim$_{n rightarrow infty}$ $int_0^{infty}$ $f(x) frac{x}{n+x}$ $dx$ = $int_0^{infty}$ $f(x)$ lim$_{n rightarrow infty}$ $frac{x}{n+x}$ $dx$ = $int_0^{infty}$ $f(x)$ $dx$ $int_0^{infty}$ lim$_{n rightarrow infty}$ $frac{x}{n+x}$ $dx$



I am VERY hesitant on my above two equalities, specially the last one, where am I wrong or need input? Thanks in advance and sorry if I am missing something obvious.






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







share|cite|improve this question









$endgroup$












  • $begingroup$
    Are you really asking if $frac{x}{n+x} le 1$ for $x > 0$ and $n ge 1$?
    $endgroup$
    – mathworker21
    Dec 23 '18 at 22:17












  • $begingroup$
    omg duh, $x > 0$. I didn't consider that, thats why I said forgive me If im asking something obvious, Im learning, I don't learn fast. SORRY FOR BEING STUPID.
    $endgroup$
    – Hossien Sahebjame
    Dec 23 '18 at 22:27










  • $begingroup$
    happens to all of us
    $endgroup$
    – mathworker21
    Dec 23 '18 at 22:30










  • $begingroup$
    lol thanks, again, Im new to measure theory, im really trying to get a lot better so feel free to comment if you ever see me ask noob questions.
    $endgroup$
    – Hossien Sahebjame
    Dec 23 '18 at 22:32














0












0








0


1



$begingroup$


Let $f$ be integrable on $[0,infty$).



Show that lim$_{n rightarrow infty}$ $int_0^{infty}$ $f(x) frac{x}{n+x}$ $dx$ = $0$.



Ok so I want to say I can take the limit inside, but this is a consequence of DCT or MCT, so to apply DCT, this is what I have come up with so far:



define $f_n$ = $f(x)$ $cdot$ $frac{x}{n+x}$ (my intuition from previous measure theory problems is that I need this step)



then {$f_n}$ is a sequence of measurable functions,



but are the $f_n$ bounded above $f(x)$ ?



*Note by assumption we have $int_0^{infty}$ $f(x)$ $dx < infty$.



Using all of this am I able to somehow deduce that



lim$_{n rightarrow infty}$ $int_0^{infty}$ $f(x) frac{x}{n+x}$ $dx$ = $int_0^{infty}$ $f(x)$ lim$_{n rightarrow infty}$ $frac{x}{n+x}$ $dx$ = $int_0^{infty}$ $f(x)$ $dx$ $int_0^{infty}$ lim$_{n rightarrow infty}$ $frac{x}{n+x}$ $dx$



I am VERY hesitant on my above two equalities, specially the last one, where am I wrong or need input? Thanks in advance and sorry if I am missing something obvious.






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







share|cite|improve this question









$endgroup$




Let $f$ be integrable on $[0,infty$).



Show that lim$_{n rightarrow infty}$ $int_0^{infty}$ $f(x) frac{x}{n+x}$ $dx$ = $0$.



Ok so I want to say I can take the limit inside, but this is a consequence of DCT or MCT, so to apply DCT, this is what I have come up with so far:



define $f_n$ = $f(x)$ $cdot$ $frac{x}{n+x}$ (my intuition from previous measure theory problems is that I need this step)



then {$f_n}$ is a sequence of measurable functions,



but are the $f_n$ bounded above $f(x)$ ?



*Note by assumption we have $int_0^{infty}$ $f(x)$ $dx < infty$.



Using all of this am I able to somehow deduce that



lim$_{n rightarrow infty}$ $int_0^{infty}$ $f(x) frac{x}{n+x}$ $dx$ = $int_0^{infty}$ $f(x)$ lim$_{n rightarrow infty}$ $frac{x}{n+x}$ $dx$ = $int_0^{infty}$ $f(x)$ $dx$ $int_0^{infty}$ lim$_{n rightarrow infty}$ $frac{x}{n+x}$ $dx$



I am VERY hesitant on my above two equalities, specially the last one, where am I wrong or need input? Thanks in advance and sorry if I am missing something obvious.






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 Dec 23 '18 at 22:14









Hossien SahebjameHossien Sahebjame

979




979












  • $begingroup$
    Are you really asking if $frac{x}{n+x} le 1$ for $x > 0$ and $n ge 1$?
    $endgroup$
    – mathworker21
    Dec 23 '18 at 22:17












  • $begingroup$
    omg duh, $x > 0$. I didn't consider that, thats why I said forgive me If im asking something obvious, Im learning, I don't learn fast. SORRY FOR BEING STUPID.
    $endgroup$
    – Hossien Sahebjame
    Dec 23 '18 at 22:27










  • $begingroup$
    happens to all of us
    $endgroup$
    – mathworker21
    Dec 23 '18 at 22:30










  • $begingroup$
    lol thanks, again, Im new to measure theory, im really trying to get a lot better so feel free to comment if you ever see me ask noob questions.
    $endgroup$
    – Hossien Sahebjame
    Dec 23 '18 at 22:32


















  • $begingroup$
    Are you really asking if $frac{x}{n+x} le 1$ for $x > 0$ and $n ge 1$?
    $endgroup$
    – mathworker21
    Dec 23 '18 at 22:17












  • $begingroup$
    omg duh, $x > 0$. I didn't consider that, thats why I said forgive me If im asking something obvious, Im learning, I don't learn fast. SORRY FOR BEING STUPID.
    $endgroup$
    – Hossien Sahebjame
    Dec 23 '18 at 22:27










  • $begingroup$
    happens to all of us
    $endgroup$
    – mathworker21
    Dec 23 '18 at 22:30










  • $begingroup$
    lol thanks, again, Im new to measure theory, im really trying to get a lot better so feel free to comment if you ever see me ask noob questions.
    $endgroup$
    – Hossien Sahebjame
    Dec 23 '18 at 22:32
















$begingroup$
Are you really asking if $frac{x}{n+x} le 1$ for $x > 0$ and $n ge 1$?
$endgroup$
– mathworker21
Dec 23 '18 at 22:17






$begingroup$
Are you really asking if $frac{x}{n+x} le 1$ for $x > 0$ and $n ge 1$?
$endgroup$
– mathworker21
Dec 23 '18 at 22:17














$begingroup$
omg duh, $x > 0$. I didn't consider that, thats why I said forgive me If im asking something obvious, Im learning, I don't learn fast. SORRY FOR BEING STUPID.
$endgroup$
– Hossien Sahebjame
Dec 23 '18 at 22:27




$begingroup$
omg duh, $x > 0$. I didn't consider that, thats why I said forgive me If im asking something obvious, Im learning, I don't learn fast. SORRY FOR BEING STUPID.
$endgroup$
– Hossien Sahebjame
Dec 23 '18 at 22:27












$begingroup$
happens to all of us
$endgroup$
– mathworker21
Dec 23 '18 at 22:30




$begingroup$
happens to all of us
$endgroup$
– mathworker21
Dec 23 '18 at 22:30












$begingroup$
lol thanks, again, Im new to measure theory, im really trying to get a lot better so feel free to comment if you ever see me ask noob questions.
$endgroup$
– Hossien Sahebjame
Dec 23 '18 at 22:32




$begingroup$
lol thanks, again, Im new to measure theory, im really trying to get a lot better so feel free to comment if you ever see me ask noob questions.
$endgroup$
– Hossien Sahebjame
Dec 23 '18 at 22:32










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$begingroup$

I will do this from scratch. First, integrability reads $int |f(x)|<infty$. Now, set $f_n(x)= f(x)frac{x}{n+x}$, and observe that, $|f_n(x)|leq |f(x)|$ point wise, as $frac{x}{n+x}leq 1$ holds always, whenever $xgeq 0$, and $ngeq 1$.



Now, also note that, for any fixed $x$, $lim_{ntoinfty}f_n(x)=0$, point wise. Hence, by DCT, $lim_n int f_n(x); dx=int lim_{ntoinfty}f_n(x); dx = 0$, as $lim_n f_n(x)=0$.






share|cite|improve this answer









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    $begingroup$

    I will do this from scratch. First, integrability reads $int |f(x)|<infty$. Now, set $f_n(x)= f(x)frac{x}{n+x}$, and observe that, $|f_n(x)|leq |f(x)|$ point wise, as $frac{x}{n+x}leq 1$ holds always, whenever $xgeq 0$, and $ngeq 1$.



    Now, also note that, for any fixed $x$, $lim_{ntoinfty}f_n(x)=0$, point wise. Hence, by DCT, $lim_n int f_n(x); dx=int lim_{ntoinfty}f_n(x); dx = 0$, as $lim_n f_n(x)=0$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      I will do this from scratch. First, integrability reads $int |f(x)|<infty$. Now, set $f_n(x)= f(x)frac{x}{n+x}$, and observe that, $|f_n(x)|leq |f(x)|$ point wise, as $frac{x}{n+x}leq 1$ holds always, whenever $xgeq 0$, and $ngeq 1$.



      Now, also note that, for any fixed $x$, $lim_{ntoinfty}f_n(x)=0$, point wise. Hence, by DCT, $lim_n int f_n(x); dx=int lim_{ntoinfty}f_n(x); dx = 0$, as $lim_n f_n(x)=0$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        I will do this from scratch. First, integrability reads $int |f(x)|<infty$. Now, set $f_n(x)= f(x)frac{x}{n+x}$, and observe that, $|f_n(x)|leq |f(x)|$ point wise, as $frac{x}{n+x}leq 1$ holds always, whenever $xgeq 0$, and $ngeq 1$.



        Now, also note that, for any fixed $x$, $lim_{ntoinfty}f_n(x)=0$, point wise. Hence, by DCT, $lim_n int f_n(x); dx=int lim_{ntoinfty}f_n(x); dx = 0$, as $lim_n f_n(x)=0$.






        share|cite|improve this answer









        $endgroup$



        I will do this from scratch. First, integrability reads $int |f(x)|<infty$. Now, set $f_n(x)= f(x)frac{x}{n+x}$, and observe that, $|f_n(x)|leq |f(x)|$ point wise, as $frac{x}{n+x}leq 1$ holds always, whenever $xgeq 0$, and $ngeq 1$.



        Now, also note that, for any fixed $x$, $lim_{ntoinfty}f_n(x)=0$, point wise. Hence, by DCT, $lim_n int f_n(x); dx=int lim_{ntoinfty}f_n(x); dx = 0$, as $lim_n f_n(x)=0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 23 '18 at 22:19









        AaronAaron

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