Prove Lusin's Theorem with characteristic function and simple function.
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Lusin's Theorem: Suppose $f$ is a complex measurable on $X$, $mu(A)<infty$, $f(x)=0$ if $xnotin A$ and $epsilon >0$. Then there exists a $g in C_c(X)$ such that $mu ({x:f(x)neq g(x)}) <epsilon$.
In Rudin page 55-56, author proves with the general case of $f$.
Here, I want to prove theorem with $f$ is a characteristic function, $f$ is a simple function because Rudin's proof is long and complex.
Please hint me!
Thank you!
measure-theory
edited Nov 19 at 14:31
Davide Giraudo
124k16150259
asked Apr 11 '13 at 16:09
Đặng Phước nhật
262
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up vote
2
down vote
favorite
Lusin's Theorem: Suppose $f$ is a complex measurable on $X$, $mu(A)<infty$, $f(x)=0$ if $xnotin A$ and $epsilon >0$. Then there exists a $g in C_c(X)$ such that $mu ({x:f(x)neq g(x)}) <epsilon$.
In Rudin page 55-56, author proves with the general case of $f$.
Here, I want to prove theorem with $f$ is a characteristic function, $f$ is a simple function because Rudin's proof is long and complex.
Please hint me!
Thank you!
measure-theory
edited Nov 19 at 14:31
Davide Giraudo
124k16150259
asked Apr 11 '13 at 16:09
Đặng Phước nhật
262
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Lusin's Theorem: Suppose $f$ is a complex measurable on $X$, $mu(A)<infty$, $f(x)=0$ if $xnotin A$ and $epsilon >0$. Then there exists a $g in C_c(X)$ such that $mu ({x:f(x)neq g(x)}) <epsilon$.
In Rudin page 55-56, author proves with the general case of $f$.
Here, I want to prove theorem with $f$ is a characteristic function, $f$ is a simple function because Rudin's proof is long and complex.
Please hint me!
Thank you!
measure-theory
edited Nov 19 at 14:31
Davide Giraudo
124k16150259
asked Apr 11 '13 at 16:09
Đặng Phước nhật
262
Lusin's Theorem: Suppose $f$ is a complex measurable on $X$, $mu(A)<infty$, $f(x)=0$ if $xnotin A$ and $epsilon >0$. Then there exists a $g in C_c(X)$ such that $mu ({x:f(x)neq g(x)}) <epsilon$.
In Rudin page 55-56, author proves with the general case of $f$.
Here, I want to prove theorem with $f$ is a characteristic function, $f$ is a simple function because Rudin's proof is long and complex.
Please hint me!
Thank you!
measure-theory
measure-theory
edited Nov 19 at 14:31
Davide Giraudo
124k16150259
asked Apr 11 '13 at 16:09
Đặng Phước nhật
262
edited Nov 19 at 14:31
Davide Giraudo
124k16150259
asked Apr 11 '13 at 16:09
Đặng Phước nhật
262
edited Nov 19 at 14:31
Davide Giraudo
124k16150259
edited Nov 19 at 14:31
Davide Giraudo
124k16150259
edited Nov 19 at 14:31
Davide Giraudo
124k16150259
124k16150259
asked Apr 11 '13 at 16:09
Đặng Phước nhật
262
asked Apr 11 '13 at 16:09
Đặng Phước nhật
262
asked Apr 11 '13 at 16:09
Đặng Phước nhật
262
262
add a comment |
add a comment |
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