An Application of Convergence theorem
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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
edited Dec 24 '18 at 20:41
Clement C.
51.2k34093
asked Dec 24 '18 at 19:44
InfinityInfinity
394113
$endgroup$
add a comment |
$begingroup$
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
edited Dec 24 '18 at 20:41
Clement C.
51.2k34093
asked Dec 24 '18 at 19:44
InfinityInfinity
394113
$endgroup$
$begingroup$
the integrand is bounded above by 1, why cannot you use DCT?
$endgroup$
– ablmf
Dec 24 '18 at 20:13
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@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
add a comment |
$begingroup$
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
edited Dec 24 '18 at 20:41
Clement C.
51.2k34093
asked Dec 24 '18 at 19:44
InfinityInfinity
394113
$endgroup$
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
edited Dec 24 '18 at 20:41
Clement C.
51.2k34093
asked Dec 24 '18 at 19:44
InfinityInfinity
394113
edited Dec 24 '18 at 20:41
Clement C.
51.2k34093
asked Dec 24 '18 at 19:44
InfinityInfinity
394113
edited Dec 24 '18 at 20:41
Clement C.
51.2k34093
edited Dec 24 '18 at 20:41
Clement C.
51.2k34093
edited Dec 24 '18 at 20:41
Clement C.
51.2k34093
51.2k34093
asked Dec 24 '18 at 19:44
InfinityInfinity
394113
asked Dec 24 '18 at 19:44
InfinityInfinity
394113
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
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Hint: Let $f_n$ be the $n$th integrand. Then $0le f_1le f_2 le cdots $
answered Dec 24 '18 at 20:28
zhw.zhw.
75.4k43275
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add a comment |
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1 Answer
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$begingroup$
Hint: Let $f_n$ be the $n$th integrand. Then $0le f_1le f_2 le cdots $
answered Dec 24 '18 at 20:28
zhw.zhw.
75.4k43275
$endgroup$
add a comment |
$begingroup$
Hint: Let $f_n$ be the $n$th integrand. Then $0le f_1le f_2 le cdots $
answered Dec 24 '18 at 20:28
zhw.zhw.
75.4k43275
$endgroup$
add a comment |
$begingroup$
Hint: Let $f_n$ be the $n$th integrand. Then $0le f_1le f_2 le cdots $
answered Dec 24 '18 at 20:28
zhw.zhw.
75.4k43275
$endgroup$
Hint: Let $f_n$ be the $n$th integrand. Then $0le f_1le f_2 le cdots $
answered Dec 24 '18 at 20:28
zhw.zhw.
75.4k43275
answered Dec 24 '18 at 20:28
zhw.zhw.
75.4k43275
answered Dec 24 '18 at 20:28
zhw.zhw.
75.4k43275
answered Dec 24 '18 at 20:28
zhw.zhw.
75.4k43275
75.4k43275
add a comment |
add a comment |
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$begingroup$
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