The simple function are dense in $L^{infty} (X)$ if $mu (X) lt infty$
$begingroup$
Let $(X,M,mu)$ be a measure space, show that:
The simple function are dense in $L^{infty} (X)$ if $mu (X) lt infty$
This is an exercise in Stein's functional analysis Page 35.
If $f in L^infty$ , then for any $e gt0$ there exist $N$ , $f lt Ne$ , a.e..
So use $E_n={x:-Ne +ne le f lt -Ne +(n+1)e } $ and let $$f_n(x)=sum_{n=0}^{2N} (-Ne+ne)X_{E_n}(x)$$ where $X_{E_n}$ is the character function of $E_n$ , $0 le n le 2N$, then I can get $$vert vert f-f_n vert vert lt e$$
Since I didn't use the assumption that $mu (X) lt infty$ . Did I make something wrong in the proof ?
real-analysis functional-analysis
asked Dec 4 '18 at 17:20
J.GuoJ.Guo
3229
$endgroup$
add a comment |
$begingroup$
Let $(X,M,mu)$ be a measure space, show that:
The simple function are dense in $L^{infty} (X)$ if $mu (X) lt infty$
This is an exercise in Stein's functional analysis Page 35.
If $f in L^infty$ , then for any $e gt0$ there exist $N$ , $f lt Ne$ , a.e..
So use $E_n={x:-Ne +ne le f lt -Ne +(n+1)e } $ and let $$f_n(x)=sum_{n=0}^{2N} (-Ne+ne)X_{E_n}(x)$$ where $X_{E_n}$ is the character function of $E_n$ , $0 le n le 2N$, then I can get $$vert vert f-f_n vert vert lt e$$
Since I didn't use the assumption that $mu (X) lt infty$ . Did I make something wrong in the proof ?
real-analysis functional-analysis
asked Dec 4 '18 at 17:20
J.GuoJ.Guo
3229
$endgroup$
1
$begingroup$
I don't have the book at hand to check for myself, but you might look closely at their definition of simple function. It may require that each set $E_n$ has finite measure.
$endgroup$
– Umberto P.
Dec 4 '18 at 17:38
$begingroup$
Thank you ! I get the point now.
$endgroup$
– J.Guo
Dec 4 '18 at 17:43
$begingroup$
Nope. Generally it is not required for simple functions to have finite (in measure) support. However you proof is wrong. But just slightly. You have notation abuse. Your $f_{n}$ does not depend on $n$. It depends only $N$. So this proof can't be accepted as your choice of $N$ depends on fixed $e$. (However this is right way to think. Soon you'll get the proof. I think so.)
$endgroup$
– kolobokish
Dec 4 '18 at 18:13
add a comment |
$begingroup$
Let $(X,M,mu)$ be a measure space, show that:
The simple function are dense in $L^{infty} (X)$ if $mu (X) lt infty$
This is an exercise in Stein's functional analysis Page 35.
If $f in L^infty$ , then for any $e gt0$ there exist $N$ , $f lt Ne$ , a.e..
So use $E_n={x:-Ne +ne le f lt -Ne +(n+1)e } $ and let $$f_n(x)=sum_{n=0}^{2N} (-Ne+ne)X_{E_n}(x)$$ where $X_{E_n}$ is the character function of $E_n$ , $0 le n le 2N$, then I can get $$vert vert f-f_n vert vert lt e$$
Since I didn't use the assumption that $mu (X) lt infty$ . Did I make something wrong in the proof ?
real-analysis functional-analysis
asked Dec 4 '18 at 17:20
J.GuoJ.Guo
3229
$endgroup$
Let $(X,M,mu)$ be a measure space, show that:
The simple function are dense in $L^{infty} (X)$ if $mu (X) lt infty$
This is an exercise in Stein's functional analysis Page 35.
If $f in L^infty$ , then for any $e gt0$ there exist $N$ , $f lt Ne$ , a.e..
So use $E_n={x:-Ne +ne le f lt -Ne +(n+1)e } $ and let $$f_n(x)=sum_{n=0}^{2N} (-Ne+ne)X_{E_n}(x)$$ where $X_{E_n}$ is the character function of $E_n$ , $0 le n le 2N$, then I can get $$vert vert f-f_n vert vert lt e$$
Since I didn't use the assumption that $mu (X) lt infty$ . Did I make something wrong in the proof ?
real-analysis functional-analysis
real-analysis functional-analysis
asked Dec 4 '18 at 17:20
J.GuoJ.Guo
3229
asked Dec 4 '18 at 17:20
J.GuoJ.Guo
3229
edited Dec 4 '18 at 17:32
J.Guo
asked Dec 4 '18 at 17:20
J.GuoJ.Guo
3229
asked Dec 4 '18 at 17:20
J.GuoJ.Guo
3229
asked Dec 4 '18 at 17:20
J.GuoJ.Guo
3229
3229
1
$begingroup$
I don't have the book at hand to check for myself, but you might look closely at their definition of simple function. It may require that each set $E_n$ has finite measure.
$endgroup$
– Umberto P.
Dec 4 '18 at 17:38
$begingroup$
Thank you ! I get the point now.
$endgroup$
– J.Guo
Dec 4 '18 at 17:43
$begingroup$
Nope. Generally it is not required for simple functions to have finite (in measure) support. However you proof is wrong. But just slightly. You have notation abuse. Your $f_{n}$ does not depend on $n$. It depends only $N$. So this proof can't be accepted as your choice of $N$ depends on fixed $e$. (However this is right way to think. Soon you'll get the proof. I think so.)
$endgroup$
– kolobokish
Dec 4 '18 at 18:13
add a comment |
1
$begingroup$
I don't have the book at hand to check for myself, but you might look closely at their definition of simple function. It may require that each set $E_n$ has finite measure.
$endgroup$
– Umberto P.
Dec 4 '18 at 17:38
$begingroup$
Thank you ! I get the point now.
$endgroup$
– J.Guo
Dec 4 '18 at 17:43
$begingroup$
Nope. Generally it is not required for simple functions to have finite (in measure) support. However you proof is wrong. But just slightly. You have notation abuse. Your $f_{n}$ does not depend on $n$. It depends only $N$. So this proof can't be accepted as your choice of $N$ depends on fixed $e$. (However this is right way to think. Soon you'll get the proof. I think so.)
$endgroup$
– kolobokish
Dec 4 '18 at 18:13
1
1
$begingroup$
I don't have the book at hand to check for myself, but you might look closely at their definition of simple function. It may require that each set $E_n$ has finite measure.
$endgroup$
– Umberto P.
Dec 4 '18 at 17:38
$begingroup$
I don't have the book at hand to check for myself, but you might look closely at their definition of simple function. It may require that each set $E_n$ has finite measure.
$endgroup$
– Umberto P.
Dec 4 '18 at 17:38
$begingroup$
Thank you ! I get the point now.
$endgroup$
– J.Guo
Dec 4 '18 at 17:43
$begingroup$
Thank you ! I get the point now.
$endgroup$
– J.Guo
Dec 4 '18 at 17:43
$begingroup$
Nope. Generally it is not required for simple functions to have finite (in measure) support. However you proof is wrong. But just slightly. You have notation abuse. Your $f_{n}$ does not depend on $n$. It depends only $N$. So this proof can't be accepted as your choice of $N$ depends on fixed $e$. (However this is right way to think. Soon you'll get the proof. I think so.)
$endgroup$
– kolobokish
Dec 4 '18 at 18:13
$begingroup$
Nope. Generally it is not required for simple functions to have finite (in measure) support. However you proof is wrong. But just slightly. You have notation abuse. Your $f_{n}$ does not depend on $n$. It depends only $N$. So this proof can't be accepted as your choice of $N$ depends on fixed $e$. (However this is right way to think. Soon you'll get the proof. I think so.)
$endgroup$
– kolobokish
Dec 4 '18 at 18:13
add a comment |
1 Answer
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It is true that the simple functions that are not necessarily supported on sets of finite measure (simple functions are just measurable functions on $X$ with a finite image) are dense in $L_{infty}(mu)$ regardless of whether $mu$ is finite. Note that you showed it for none negative bounded functions (I think), but it is easily extensible to complex functions in $L_{infty}(mu)$.
answered Dec 4 '18 at 17:48
pitariverpitariver
317112
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add a comment |
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$begingroup$
It is true that the simple functions that are not necessarily supported on sets of finite measure (simple functions are just measurable functions on $X$ with a finite image) are dense in $L_{infty}(mu)$ regardless of whether $mu$ is finite. Note that you showed it for none negative bounded functions (I think), but it is easily extensible to complex functions in $L_{infty}(mu)$.
answered Dec 4 '18 at 17:48
pitariverpitariver
317112
$endgroup$
add a comment |
$begingroup$
It is true that the simple functions that are not necessarily supported on sets of finite measure (simple functions are just measurable functions on $X$ with a finite image) are dense in $L_{infty}(mu)$ regardless of whether $mu$ is finite. Note that you showed it for none negative bounded functions (I think), but it is easily extensible to complex functions in $L_{infty}(mu)$.
answered Dec 4 '18 at 17:48
pitariverpitariver
317112
$endgroup$
add a comment |
$begingroup$
It is true that the simple functions that are not necessarily supported on sets of finite measure (simple functions are just measurable functions on $X$ with a finite image) are dense in $L_{infty}(mu)$ regardless of whether $mu$ is finite. Note that you showed it for none negative bounded functions (I think), but it is easily extensible to complex functions in $L_{infty}(mu)$.
answered Dec 4 '18 at 17:48
pitariverpitariver
317112
$endgroup$
It is true that the simple functions that are not necessarily supported on sets of finite measure (simple functions are just measurable functions on $X$ with a finite image) are dense in $L_{infty}(mu)$ regardless of whether $mu$ is finite. Note that you showed it for none negative bounded functions (I think), but it is easily extensible to complex functions in $L_{infty}(mu)$.
answered Dec 4 '18 at 17:48
pitariverpitariver
317112
answered Dec 4 '18 at 17:48
pitariverpitariver
317112
answered Dec 4 '18 at 17:48
pitariverpitariver
317112
answered Dec 4 '18 at 17:48
pitariverpitariver
317112
317112
add a comment |
add a comment |
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$begingroup$
I don't have the book at hand to check for myself, but you might look closely at their definition of simple function. It may require that each set $E_n$ has finite measure.
$endgroup$
– Umberto P.
Dec 4 '18 at 17:38
$begingroup$
Thank you ! I get the point now.
$endgroup$
– J.Guo
Dec 4 '18 at 17:43
$begingroup$
Nope. Generally it is not required for simple functions to have finite (in measure) support. However you proof is wrong. But just slightly. You have notation abuse. Your $f_{n}$ does not depend on $n$. It depends only $N$. So this proof can't be accepted as your choice of $N$ depends on fixed $e$. (However this is right way to think. Soon you'll get the proof. I think so.)
$endgroup$
– kolobokish
Dec 4 '18 at 18:13