$left{x:f(x)neq 0right}$ decomposes into a numerable union of sets of finite measurement. with $f$ complex...
0
$begingroup$
Let $f$ function with valued in $mathbb{C}$ and $fin L^1(mathbb{R})$. The set $left{x:f(x)neq 0right}$ decomposes into a numerable union of sets of
Finite measurement.
I want
$left{x:f(x)neq 0right}=bigcup_{nin mathbb{N}} left{x:|f(x)|>1/nright} $
But I do not know if this applies in this case with complex values ...
real-analysis complex-analysis measure-theory
asked Jan 10 at 18:28
eraldcoileraldcoil
395211
$endgroup$
|
show 1 more comment
0
$begingroup$
Let $f$ function with valued in $mathbb{C}$ and $fin L^1(mathbb{R})$. The set $left{x:f(x)neq 0right}$ decomposes into a numerable union of sets of
Finite measurement.
I want
$left{x:f(x)neq 0right}=bigcup_{nin mathbb{N}} left{x:|f(x)|>1/nright} $
But I do not know if this applies in this case with complex values ...
real-analysis complex-analysis measure-theory
asked Jan 10 at 18:28
eraldcoileraldcoil
395211
$endgroup$
$begingroup$
I have already corrected it
$endgroup$
– eraldcoil
Jan 10 at 18:34
1
$begingroup$
In which case the result is true, simply because $f(x)=0$ iff there existed $ninmathbb N$ such that $|f| > 1/n $ (even for complex valued $f$)
$endgroup$
– Calvin Khor
Jan 10 at 18:36
1
$begingroup$
(...should be $neq 0$ instead of $=0$)
$endgroup$
– Calvin Khor
Jan 10 at 18:56
$begingroup$
In mi notes, say $f(x)neq 0$...
$endgroup$
– eraldcoil
Jan 10 at 19:02
1
$begingroup$
I was correcting my comment that was too old for me to edit. Do you understand ?
$endgroup$
– Calvin Khor
Jan 10 at 19:03
|
show 1 more comment
0
0
0
$begingroup$
Let $f$ function with valued in $mathbb{C}$ and $fin L^1(mathbb{R})$. The set $left{x:f(x)neq 0right}$ decomposes into a numerable union of sets of
Finite measurement.
I want
$left{x:f(x)neq 0right}=bigcup_{nin mathbb{N}} left{x:|f(x)|>1/nright} $
But I do not know if this applies in this case with complex values ...
real-analysis complex-analysis measure-theory
asked Jan 10 at 18:28
eraldcoileraldcoil
395211
$endgroup$
Let $f$ function with valued in $mathbb{C}$ and $fin L^1(mathbb{R})$. The set $left{x:f(x)neq 0right}$ decomposes into a numerable union of sets of
Finite measurement.
I want
$left{x:f(x)neq 0right}=bigcup_{nin mathbb{N}} left{x:|f(x)|>1/nright} $
But I do not know if this applies in this case with complex values ...
real-analysis complex-analysis measure-theory
real-analysis complex-analysis measure-theory
asked Jan 10 at 18:28
eraldcoileraldcoil
395211
asked Jan 10 at 18:28
eraldcoileraldcoil
395211
edited Jan 10 at 18:34
eraldcoil
asked Jan 10 at 18:28
eraldcoileraldcoil
395211
asked Jan 10 at 18:28
eraldcoileraldcoil
395211
asked Jan 10 at 18:28
eraldcoileraldcoil
395211
395211
$begingroup$
I have already corrected it
$endgroup$
– eraldcoil
Jan 10 at 18:34
1
$begingroup$
In which case the result is true, simply because $f(x)=0$ iff there existed $ninmathbb N$ such that $|f| > 1/n $ (even for complex valued $f$)
$endgroup$
– Calvin Khor
Jan 10 at 18:36
1
$begingroup$
(...should be $neq 0$ instead of $=0$)
$endgroup$
– Calvin Khor
Jan 10 at 18:56
$begingroup$
In mi notes, say $f(x)neq 0$...
$endgroup$
– eraldcoil
Jan 10 at 19:02
1
$begingroup$
I was correcting my comment that was too old for me to edit. Do you understand ?
$endgroup$
– Calvin Khor
Jan 10 at 19:03
|
show 1 more comment
$begingroup$
I have already corrected it
$endgroup$
– eraldcoil
Jan 10 at 18:34
1
$begingroup$
In which case the result is true, simply because $f(x)=0$ iff there existed $ninmathbb N$ such that $|f| > 1/n $ (even for complex valued $f$)
$endgroup$
– Calvin Khor
Jan 10 at 18:36
1
$begingroup$
(...should be $neq 0$ instead of $=0$)
$endgroup$
– Calvin Khor
Jan 10 at 18:56
$begingroup$
In mi notes, say $f(x)neq 0$...
$endgroup$
– eraldcoil
Jan 10 at 19:02
1
$begingroup$
I was correcting my comment that was too old for me to edit. Do you understand ?
$endgroup$
– Calvin Khor
Jan 10 at 19:03
$begingroup$
I have already corrected it
$endgroup$
– eraldcoil
Jan 10 at 18:34
$begingroup$
I have already corrected it
$endgroup$
– eraldcoil
Jan 10 at 18:34
1
1
$begingroup$
In which case the result is true, simply because $f(x)=0$ iff there existed $ninmathbb N$ such that $|f| > 1/n $ (even for complex valued $f$)
$endgroup$
– Calvin Khor
Jan 10 at 18:36
$begingroup$
In which case the result is true, simply because $f(x)=0$ iff there existed $ninmathbb N$ such that $|f| > 1/n $ (even for complex valued $f$)
$endgroup$
– Calvin Khor
Jan 10 at 18:36
1
1
$begingroup$
(...should be $neq 0$ instead of $=0$)
$endgroup$
– Calvin Khor
Jan 10 at 18:56
$begingroup$
(...should be $neq 0$ instead of $=0$)
$endgroup$
– Calvin Khor
Jan 10 at 18:56
$begingroup$
In mi notes, say $f(x)neq 0$...
$endgroup$
– eraldcoil
Jan 10 at 19:02
$begingroup$
In mi notes, say $f(x)neq 0$...
$endgroup$
– eraldcoil
Jan 10 at 19:02
1
1
$begingroup$
I was correcting my comment that was too old for me to edit. Do you understand ?
$endgroup$
– Calvin Khor
Jan 10 at 19:03
$begingroup$
I was correcting my comment that was too old for me to edit. Do you understand ?
$endgroup$
– Calvin Khor
Jan 10 at 19:03
|
show 1 more comment
0
active
oldest
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$begingroup$
I have already corrected it
$endgroup$
– eraldcoil
Jan 10 at 18:34
1
$begingroup$
In which case the result is true, simply because $f(x)=0$ iff there existed $ninmathbb N$ such that $|f| > 1/n $ (even for complex valued $f$)
$endgroup$
– Calvin Khor
Jan 10 at 18:36
1
$begingroup$
(...should be $neq 0$ instead of $=0$)
$endgroup$
– Calvin Khor
Jan 10 at 18:56
$begingroup$
In mi notes, say $f(x)neq 0$...
$endgroup$
– eraldcoil
Jan 10 at 19:02
1
$begingroup$
I was correcting my comment that was too old for me to edit. Do you understand ?
$endgroup$
– Calvin Khor
Jan 10 at 19:03