An Application of Convergence theorem












1












$begingroup$



Let $(X,M.mu)$ be a measure space and let $fcolon X rightarrow mathbb{R}$ be a measurable function. Show that $$lim_{n to infty} int_X left(1-left(frac{2}{e^{f(x)}+e^{-f(x)}}right)^n right)dmu=mu({x in X:f(x)neq0})$$




My attempt:



I think its an application of Dominated or Monotone Convergence Theorem followed by Chebyshev's inequality. I didn't able to dominate the function inside the integral with an integrable function(so, i guess its not DCT) and even if I try to use MCT, I can't use it. The thing inside the integral is $left(1-frac{1}{cosh(f(x))}right)^n$ and I only know that $cosh$ is increasing from $[0,infty)$ but its decreasing from $(-infty,0]$.



Any help is appreciated!






real-analysis measure-theory lebesgue-integral







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$endgroup$












  • $begingroup$
    the integrand is bounded above by 1, why cannot you use DCT?
    $endgroup$
    – ablmf
    Dec 24 '18 at 20:13










  • $begingroup$
    @ablmf Well, is the constant $1$ function integrable? That may not fly when $mu(X)=infty$.
    $endgroup$
    – Clement C.
    Dec 24 '18 at 20:43










  • $begingroup$
    I am, however, very confused as to why and where Chebyshev's inequality would come into play.
    $endgroup$
    – Clement C.
    Dec 24 '18 at 20:48
















1












$begingroup$



Let $(X,M.mu)$ be a measure space and let $fcolon X rightarrow mathbb{R}$ be a measurable function. Show that $$lim_{n to infty} int_X left(1-left(frac{2}{e^{f(x)}+e^{-f(x)}}right)^n right)dmu=mu({x in X:f(x)neq0})$$




My attempt:



I think its an application of Dominated or Monotone Convergence Theorem followed by Chebyshev's inequality. I didn't able to dominate the function inside the integral with an integrable function(so, i guess its not DCT) and even if I try to use MCT, I can't use it. The thing inside the integral is $left(1-frac{1}{cosh(f(x))}right)^n$ and I only know that $cosh$ is increasing from $[0,infty)$ but its decreasing from $(-infty,0]$.



Any help is appreciated!






real-analysis measure-theory lebesgue-integral







share|cite|improve this question











$endgroup$












  • $begingroup$
    the integrand is bounded above by 1, why cannot you use DCT?
    $endgroup$
    – ablmf
    Dec 24 '18 at 20:13










  • $begingroup$
    @ablmf Well, is the constant $1$ function integrable? That may not fly when $mu(X)=infty$.
    $endgroup$
    – Clement C.
    Dec 24 '18 at 20:43










  • $begingroup$
    I am, however, very confused as to why and where Chebyshev's inequality would come into play.
    $endgroup$
    – Clement C.
    Dec 24 '18 at 20:48














1












1








1





$begingroup$



Let $(X,M.mu)$ be a measure space and let $fcolon X rightarrow mathbb{R}$ be a measurable function. Show that $$lim_{n to infty} int_X left(1-left(frac{2}{e^{f(x)}+e^{-f(x)}}right)^n right)dmu=mu({x in X:f(x)neq0})$$




My attempt:



I think its an application of Dominated or Monotone Convergence Theorem followed by Chebyshev's inequality. I didn't able to dominate the function inside the integral with an integrable function(so, i guess its not DCT) and even if I try to use MCT, I can't use it. The thing inside the integral is $left(1-frac{1}{cosh(f(x))}right)^n$ and I only know that $cosh$ is increasing from $[0,infty)$ but its decreasing from $(-infty,0]$.



Any help is appreciated!






real-analysis measure-theory lebesgue-integral







share|cite|improve this question











$endgroup$





Let $(X,M.mu)$ be a measure space and let $fcolon X rightarrow mathbb{R}$ be a measurable function. Show that $$lim_{n to infty} int_X left(1-left(frac{2}{e^{f(x)}+e^{-f(x)}}right)^n right)dmu=mu({x in X:f(x)neq0})$$




My attempt:



I think its an application of Dominated or Monotone Convergence Theorem followed by Chebyshev's inequality. I didn't able to dominate the function inside the integral with an integrable function(so, i guess its not DCT) and even if I try to use MCT, I can't use it. The thing inside the integral is $left(1-frac{1}{cosh(f(x))}right)^n$ and I only know that $cosh$ is increasing from $[0,infty)$ but its decreasing from $(-infty,0]$.



Any help is appreciated!






real-analysis measure-theory lebesgue-integral




real-analysis measure-theory lebesgue-integral






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 24 '18 at 20:41









Clement C.

51.2k34093




51.2k34093










asked Dec 24 '18 at 19:44









InfinityInfinity

394113




394113












  • $begingroup$
    the integrand is bounded above by 1, why cannot you use DCT?
    $endgroup$
    – ablmf
    Dec 24 '18 at 20:13










  • $begingroup$
    @ablmf Well, is the constant $1$ function integrable? That may not fly when $mu(X)=infty$.
    $endgroup$
    – Clement C.
    Dec 24 '18 at 20:43










  • $begingroup$
    I am, however, very confused as to why and where Chebyshev's inequality would come into play.
    $endgroup$
    – Clement C.
    Dec 24 '18 at 20:48


















  • $begingroup$
    the integrand is bounded above by 1, why cannot you use DCT?
    $endgroup$
    – ablmf
    Dec 24 '18 at 20:13










  • $begingroup$
    @ablmf Well, is the constant $1$ function integrable? That may not fly when $mu(X)=infty$.
    $endgroup$
    – Clement C.
    Dec 24 '18 at 20:43










  • $begingroup$
    I am, however, very confused as to why and where Chebyshev's inequality would come into play.
    $endgroup$
    – Clement C.
    Dec 24 '18 at 20:48
















$begingroup$
the integrand is bounded above by 1, why cannot you use DCT?
$endgroup$
– ablmf
Dec 24 '18 at 20:13




$begingroup$
the integrand is bounded above by 1, why cannot you use DCT?
$endgroup$
– ablmf
Dec 24 '18 at 20:13












$begingroup$
@ablmf Well, is the constant $1$ function integrable? That may not fly when $mu(X)=infty$.
$endgroup$
– Clement C.
Dec 24 '18 at 20:43




$begingroup$
@ablmf Well, is the constant $1$ function integrable? That may not fly when $mu(X)=infty$.
$endgroup$
– Clement C.
Dec 24 '18 at 20:43












$begingroup$
I am, however, very confused as to why and where Chebyshev's inequality would come into play.
$endgroup$
– Clement C.
Dec 24 '18 at 20:48




$begingroup$
I am, however, very confused as to why and where Chebyshev's inequality would come into play.
$endgroup$
– Clement C.
Dec 24 '18 at 20:48










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Hint: Let $f_n$ be the $n$th integrand. Then $0le f_1le f_2 le cdots $






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    $begingroup$

    Hint: Let $f_n$ be the $n$th integrand. Then $0le f_1le f_2 le cdots $






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Hint: Let $f_n$ be the $n$th integrand. Then $0le f_1le f_2 le cdots $






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Hint: Let $f_n$ be the $n$th integrand. Then $0le f_1le f_2 le cdots $






        share|cite|improve this answer









        $endgroup$



        Hint: Let $f_n$ be the $n$th integrand. Then $0le f_1le f_2 le cdots $







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 24 '18 at 20:28









        zhw.zhw.

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