Let $f_n:[1,infty) rightarrow mathbb{R}$ be defined by $f_n(x)=frac{1}{x}chi_{[n,infty)}(x)$. Does $int f_n...
$begingroup$
Let $f_n:[1,infty) rightarrow mathbb{R}$ be defined by $f_n(x)=frac{1}{x}chi_{[n,infty)}(x)$. Does $int f_n to int f$?
I know it is an application of DCT but I really don't know to come up with a bound that would satisfy the hypothesis of DCT. Something like $frac{1}{x^2}$ should work but $frac{1}{x} > frac{1}{x^2}$.
real-analysis limits measure-theory convergence lebesgue-integral
edited Dec 14 '18 at 22:20
Batominovski
33.1k33293
asked Dec 14 '18 at 22:09
InfinityInfinity
346112
$endgroup$
add a comment |
$begingroup$
Let $f_n:[1,infty) rightarrow mathbb{R}$ be defined by $f_n(x)=frac{1}{x}chi_{[n,infty)}(x)$. Does $int f_n to int f$?
I know it is an application of DCT but I really don't know to come up with a bound that would satisfy the hypothesis of DCT. Something like $frac{1}{x^2}$ should work but $frac{1}{x} > frac{1}{x^2}$.
real-analysis limits measure-theory convergence lebesgue-integral
edited Dec 14 '18 at 22:20
Batominovski
33.1k33293
asked Dec 14 '18 at 22:09
InfinityInfinity
346112
$endgroup$
$begingroup$
As defined, we have $int f_n = +infty$ but $f = 0$ so $int f = 0$.
$endgroup$
– mechanodroid
Dec 14 '18 at 22:13
$begingroup$
$int f_n = infty$ for all $n$. On the other side $f=0$ so $int f = 0$.
$endgroup$
– Kolja
Dec 14 '18 at 22:13
$begingroup$
what is an integrable 'g' which will bound $f_n$ so that I can apply DCT @mechanodroid
$endgroup$
– Infinity
Dec 14 '18 at 22:20
$begingroup$
We have $int f_n notto int f$ so you cannot apply DCT, the integrable function $g$ doesn't exist.
$endgroup$
– mechanodroid
Dec 14 '18 at 22:23
$begingroup$
Furthermore, the sequence is monotone decreasing. It is also an example showing why you need monotone increasing for the Monotone Convergence Theorem.
$endgroup$
– GEdgar
Dec 14 '18 at 22:41
add a comment |
$begingroup$
Let $f_n:[1,infty) rightarrow mathbb{R}$ be defined by $f_n(x)=frac{1}{x}chi_{[n,infty)}(x)$. Does $int f_n to int f$?
I know it is an application of DCT but I really don't know to come up with a bound that would satisfy the hypothesis of DCT. Something like $frac{1}{x^2}$ should work but $frac{1}{x} > frac{1}{x^2}$.
real-analysis limits measure-theory convergence lebesgue-integral
edited Dec 14 '18 at 22:20
Batominovski
33.1k33293
asked Dec 14 '18 at 22:09
InfinityInfinity
346112
$endgroup$
Let $f_n:[1,infty) rightarrow mathbb{R}$ be defined by $f_n(x)=frac{1}{x}chi_{[n,infty)}(x)$. Does $int f_n to int f$?
I know it is an application of DCT but I really don't know to come up with a bound that would satisfy the hypothesis of DCT. Something like $frac{1}{x^2}$ should work but $frac{1}{x} > frac{1}{x^2}$.
real-analysis limits measure-theory convergence lebesgue-integral
real-analysis limits measure-theory convergence lebesgue-integral
edited Dec 14 '18 at 22:20
Batominovski
33.1k33293
asked Dec 14 '18 at 22:09
InfinityInfinity
346112
edited Dec 14 '18 at 22:20
Batominovski
33.1k33293
asked Dec 14 '18 at 22:09
InfinityInfinity
346112
edited Dec 14 '18 at 22:20
Batominovski
33.1k33293
edited Dec 14 '18 at 22:20
Batominovski
33.1k33293
edited Dec 14 '18 at 22:20
Batominovski
33.1k33293
33.1k33293
asked Dec 14 '18 at 22:09
InfinityInfinity
346112
asked Dec 14 '18 at 22:09
InfinityInfinity
346112
asked Dec 14 '18 at 22:09
InfinityInfinity
346112
346112
$begingroup$
As defined, we have $int f_n = +infty$ but $f = 0$ so $int f = 0$.
$endgroup$
– mechanodroid
Dec 14 '18 at 22:13
$begingroup$
$int f_n = infty$ for all $n$. On the other side $f=0$ so $int f = 0$.
$endgroup$
– Kolja
Dec 14 '18 at 22:13
$begingroup$
what is an integrable 'g' which will bound $f_n$ so that I can apply DCT @mechanodroid
$endgroup$
– Infinity
Dec 14 '18 at 22:20
$begingroup$
We have $int f_n notto int f$ so you cannot apply DCT, the integrable function $g$ doesn't exist.
$endgroup$
– mechanodroid
Dec 14 '18 at 22:23
$begingroup$
Furthermore, the sequence is monotone decreasing. It is also an example showing why you need monotone increasing for the Monotone Convergence Theorem.
$endgroup$
– GEdgar
Dec 14 '18 at 22:41
add a comment |
$begingroup$
As defined, we have $int f_n = +infty$ but $f = 0$ so $int f = 0$.
$endgroup$
– mechanodroid
Dec 14 '18 at 22:13
$begingroup$
$int f_n = infty$ for all $n$. On the other side $f=0$ so $int f = 0$.
$endgroup$
– Kolja
Dec 14 '18 at 22:13
$begingroup$
what is an integrable 'g' which will bound $f_n$ so that I can apply DCT @mechanodroid
$endgroup$
– Infinity
Dec 14 '18 at 22:20
$begingroup$
We have $int f_n notto int f$ so you cannot apply DCT, the integrable function $g$ doesn't exist.
$endgroup$
– mechanodroid
Dec 14 '18 at 22:23
$begingroup$
Furthermore, the sequence is monotone decreasing. It is also an example showing why you need monotone increasing for the Monotone Convergence Theorem.
$endgroup$
– GEdgar
Dec 14 '18 at 22:41
$begingroup$
As defined, we have $int f_n = +infty$ but $f = 0$ so $int f = 0$.
$endgroup$
– mechanodroid
Dec 14 '18 at 22:13
$begingroup$
As defined, we have $int f_n = +infty$ but $f = 0$ so $int f = 0$.
$endgroup$
– mechanodroid
Dec 14 '18 at 22:13
$begingroup$
$int f_n = infty$ for all $n$. On the other side $f=0$ so $int f = 0$.
$endgroup$
– Kolja
Dec 14 '18 at 22:13
$begingroup$
$int f_n = infty$ for all $n$. On the other side $f=0$ so $int f = 0$.
$endgroup$
– Kolja
Dec 14 '18 at 22:13
$begingroup$
what is an integrable 'g' which will bound $f_n$ so that I can apply DCT @mechanodroid
$endgroup$
– Infinity
Dec 14 '18 at 22:20
$begingroup$
what is an integrable 'g' which will bound $f_n$ so that I can apply DCT @mechanodroid
$endgroup$
– Infinity
Dec 14 '18 at 22:20
$begingroup$
We have $int f_n notto int f$ so you cannot apply DCT, the integrable function $g$ doesn't exist.
$endgroup$
– mechanodroid
Dec 14 '18 at 22:23
$begingroup$
We have $int f_n notto int f$ so you cannot apply DCT, the integrable function $g$ doesn't exist.
$endgroup$
– mechanodroid
Dec 14 '18 at 22:23
$begingroup$
Furthermore, the sequence is monotone decreasing. It is also an example showing why you need monotone increasing for the Monotone Convergence Theorem.
$endgroup$
– GEdgar
Dec 14 '18 at 22:41
$begingroup$
Furthermore, the sequence is monotone decreasing. It is also an example showing why you need monotone increasing for the Monotone Convergence Theorem.
$endgroup$
– GEdgar
Dec 14 '18 at 22:41
add a comment |
1 Answer
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as stated in the comments, the first thing to see is that $f_nto 0$ pointwise everywhere as $ntoinfty$. Also $int_{[n,infty)}f_n=+infty$ no matter what is $nin[1,infty)$ so if you suppose that the DCT holds true, then by intechanging limit and integrals you would obtain $limint_{[n,infty)}f_n=+infty=int_{mathbb{R}}lim f_n=0$ which is not possibly true.
answered Dec 14 '18 at 22:36
MalikMalik
1239
$endgroup$
add a comment |
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$begingroup$
as stated in the comments, the first thing to see is that $f_nto 0$ pointwise everywhere as $ntoinfty$. Also $int_{[n,infty)}f_n=+infty$ no matter what is $nin[1,infty)$ so if you suppose that the DCT holds true, then by intechanging limit and integrals you would obtain $limint_{[n,infty)}f_n=+infty=int_{mathbb{R}}lim f_n=0$ which is not possibly true.
answered Dec 14 '18 at 22:36
MalikMalik
1239
$endgroup$
add a comment |
$begingroup$
as stated in the comments, the first thing to see is that $f_nto 0$ pointwise everywhere as $ntoinfty$. Also $int_{[n,infty)}f_n=+infty$ no matter what is $nin[1,infty)$ so if you suppose that the DCT holds true, then by intechanging limit and integrals you would obtain $limint_{[n,infty)}f_n=+infty=int_{mathbb{R}}lim f_n=0$ which is not possibly true.
answered Dec 14 '18 at 22:36
MalikMalik
1239
$endgroup$
add a comment |
$begingroup$
as stated in the comments, the first thing to see is that $f_nto 0$ pointwise everywhere as $ntoinfty$. Also $int_{[n,infty)}f_n=+infty$ no matter what is $nin[1,infty)$ so if you suppose that the DCT holds true, then by intechanging limit and integrals you would obtain $limint_{[n,infty)}f_n=+infty=int_{mathbb{R}}lim f_n=0$ which is not possibly true.
answered Dec 14 '18 at 22:36
MalikMalik
1239
$endgroup$
as stated in the comments, the first thing to see is that $f_nto 0$ pointwise everywhere as $ntoinfty$. Also $int_{[n,infty)}f_n=+infty$ no matter what is $nin[1,infty)$ so if you suppose that the DCT holds true, then by intechanging limit and integrals you would obtain $limint_{[n,infty)}f_n=+infty=int_{mathbb{R}}lim f_n=0$ which is not possibly true.
answered Dec 14 '18 at 22:36
MalikMalik
1239
answered Dec 14 '18 at 22:36
MalikMalik
1239
answered Dec 14 '18 at 22:36
MalikMalik
1239
answered Dec 14 '18 at 22:36
MalikMalik
1239
1239
add a comment |
add a comment |
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$begingroup$
As defined, we have $int f_n = +infty$ but $f = 0$ so $int f = 0$.
$endgroup$
– mechanodroid
Dec 14 '18 at 22:13
$begingroup$
$int f_n = infty$ for all $n$. On the other side $f=0$ so $int f = 0$.
$endgroup$
– Kolja
Dec 14 '18 at 22:13
$begingroup$
what is an integrable 'g' which will bound $f_n$ so that I can apply DCT @mechanodroid
$endgroup$
– Infinity
Dec 14 '18 at 22:20
$begingroup$
We have $int f_n notto int f$ so you cannot apply DCT, the integrable function $g$ doesn't exist.
$endgroup$
– mechanodroid
Dec 14 '18 at 22:23
$begingroup$
Furthermore, the sequence is monotone decreasing. It is also an example showing why you need monotone increasing for the Monotone Convergence Theorem.
$endgroup$
– GEdgar
Dec 14 '18 at 22:41