$f$ is $mu$-measurable $Rightarrow$ $textrm{sgn}f$ is $mu$-measurable
$begingroup$
I want to prove the following claim
Claim: Let $(Omega, Sigma, mu)$ be a measure space. If $f : Omega to mathbb{C}$ is $mu$-measurable, then $textrm{f}$ is $mu$-measurable.
Here a function $f$ is called $mu$-measurable if there exists a
sequence of simple functions on $Sigma$ $(varphi_n)$ such that
$varphi_n to f$ almost everywhere.We define $textrm{sgn}(z) := z/|z|$ for $z neq 0$ and $textrm{sgn}(z) := 0$
for $z = 0$.
This was just mentioned off-the-cuff in lecture as part of another proof. So I imagine it's not too difficult to see. Still, I haven't been able to prove it to myself or find anything on Stack Overflow about it.
My work so far:
I think I want to show the following:
Want to show: Let $(varphi_n)$ be a sequence of simple functions such that $varphi_n to f$ almost everywhere, with $f : Omega to mathbb{C}$. Then $textrm{sgn}(varphi_n) to textrm{sgn}(f)$ almost everywhere.
We let $varphi_n to f$ pointwise on $Omega setminus N$ with $mu(N) = 0$ and take an $omega in Omega setminus N$.
It suffices to show that $|textrm{sgn}(varphi_n(omega)) - textrm{sgn}(f(omega))| to 0$.
Case 1: $f(omega) neq 0$.
$|varphi_n(omega) - f(omega)| to 0 Rightarrow Big||varphi_n(omega)| - |f(omega)|Big| to 0 Rightarrow Big| frac{|varphi_n(omega)|}{f(omega)} - text{sgn}(f(omega)) Big| to 0$.
Not sure where to go from here. I imagine I take a subsequence $(varphi_{k(n)})$ that with no zero elements and show that $frac{|varphi_{k(n)}(omega)|}{varphi_{k(n)}(omega)} to frac{|varphi_n(omega)|}{f(omega)}$ and then use this to get the convergence I want.
Case 2: $f(omega) = 0$. Here I need to show $text{sgn}(varphi_n(omega)) to 0$. If $exists M in mathbb{N}$ such that $varphi_n(omega) = 0$ $forall n geq M$, then I'm done. Otherwise, I'm not quite sure. Maybe I need to take a subsequence of nonzero $(varphi_{k(n)}(omega))$ and show $text{sgn}(varphi_{k(n)}(omega)) to 0$.
real-analysis measure-theory convergence
asked Dec 23 '18 at 14:17
zxmknzxmkn
340213
$endgroup$
add a comment |
$begingroup$
I want to prove the following claim
Claim: Let $(Omega, Sigma, mu)$ be a measure space. If $f : Omega to mathbb{C}$ is $mu$-measurable, then $textrm{f}$ is $mu$-measurable.
Here a function $f$ is called $mu$-measurable if there exists a
sequence of simple functions on $Sigma$ $(varphi_n)$ such that
$varphi_n to f$ almost everywhere.We define $textrm{sgn}(z) := z/|z|$ for $z neq 0$ and $textrm{sgn}(z) := 0$
for $z = 0$.
This was just mentioned off-the-cuff in lecture as part of another proof. So I imagine it's not too difficult to see. Still, I haven't been able to prove it to myself or find anything on Stack Overflow about it.
My work so far:
I think I want to show the following:
Want to show: Let $(varphi_n)$ be a sequence of simple functions such that $varphi_n to f$ almost everywhere, with $f : Omega to mathbb{C}$. Then $textrm{sgn}(varphi_n) to textrm{sgn}(f)$ almost everywhere.
We let $varphi_n to f$ pointwise on $Omega setminus N$ with $mu(N) = 0$ and take an $omega in Omega setminus N$.
It suffices to show that $|textrm{sgn}(varphi_n(omega)) - textrm{sgn}(f(omega))| to 0$.
Case 1: $f(omega) neq 0$.
$|varphi_n(omega) - f(omega)| to 0 Rightarrow Big||varphi_n(omega)| - |f(omega)|Big| to 0 Rightarrow Big| frac{|varphi_n(omega)|}{f(omega)} - text{sgn}(f(omega)) Big| to 0$.
Not sure where to go from here. I imagine I take a subsequence $(varphi_{k(n)})$ that with no zero elements and show that $frac{|varphi_{k(n)}(omega)|}{varphi_{k(n)}(omega)} to frac{|varphi_n(omega)|}{f(omega)}$ and then use this to get the convergence I want.
Case 2: $f(omega) = 0$. Here I need to show $text{sgn}(varphi_n(omega)) to 0$. If $exists M in mathbb{N}$ such that $varphi_n(omega) = 0$ $forall n geq M$, then I'm done. Otherwise, I'm not quite sure. Maybe I need to take a subsequence of nonzero $(varphi_{k(n)}(omega))$ and show $text{sgn}(varphi_{k(n)}(omega)) to 0$.
real-analysis measure-theory convergence
asked Dec 23 '18 at 14:17
zxmknzxmkn
340213
$endgroup$
$begingroup$
Uhm... If the domain is $Bbb C$, then the image of $operatorname{sgn}$ is not ${0,1}$, but ${0}cup S^1$ (and I'd rather define it as $z/lvert zrvert$ for $zne 0$ because it makes more sense). If the domain is $Bbb R$, then the image is ${-1,0,1}$.
$endgroup$
– Saucy O'Path
Dec 23 '18 at 14:42
$begingroup$
Even on $mathbb R$, It's not true that $sgn(f_n)to sgn(f)$. Take $f_n(x)=frac{1}{n}boldsymbol 1_{mathbb Rsetminus mathbb Q}$. Then $f_n(x)to 0=:f$ for all $x$ but $sgn(f_n)to 1$ a.e. $neq sgn(f)=0$
$endgroup$
– Surb
Dec 23 '18 at 14:43
$begingroup$
@SaucyO'Path I corrected the definition of $text{sgn}$. Thanks.
$endgroup$
– zxmkn
Dec 23 '18 at 15:06
$begingroup$
@Surb Oh, yeah. Then my strategy (WTS: ...) was clearly off. Any ideas for showing the claim here?
$endgroup$
– zxmkn
Dec 23 '18 at 15:08
3
$begingroup$
You have that $sgn(f)(z)=frac{f(z)}{|f(z)|}boldsymbol 1_{{f(z)neq 0}}$ that is a product and quotient of measurable function. Then it's measurable.
$endgroup$
– Surb
Dec 23 '18 at 15:25
add a comment |
$begingroup$
I want to prove the following claim
Claim: Let $(Omega, Sigma, mu)$ be a measure space. If $f : Omega to mathbb{C}$ is $mu$-measurable, then $textrm{f}$ is $mu$-measurable.
Here a function $f$ is called $mu$-measurable if there exists a
sequence of simple functions on $Sigma$ $(varphi_n)$ such that
$varphi_n to f$ almost everywhere.We define $textrm{sgn}(z) := z/|z|$ for $z neq 0$ and $textrm{sgn}(z) := 0$
for $z = 0$.
This was just mentioned off-the-cuff in lecture as part of another proof. So I imagine it's not too difficult to see. Still, I haven't been able to prove it to myself or find anything on Stack Overflow about it.
My work so far:
I think I want to show the following:
Want to show: Let $(varphi_n)$ be a sequence of simple functions such that $varphi_n to f$ almost everywhere, with $f : Omega to mathbb{C}$. Then $textrm{sgn}(varphi_n) to textrm{sgn}(f)$ almost everywhere.
We let $varphi_n to f$ pointwise on $Omega setminus N$ with $mu(N) = 0$ and take an $omega in Omega setminus N$.
It suffices to show that $|textrm{sgn}(varphi_n(omega)) - textrm{sgn}(f(omega))| to 0$.
Case 1: $f(omega) neq 0$.
$|varphi_n(omega) - f(omega)| to 0 Rightarrow Big||varphi_n(omega)| - |f(omega)|Big| to 0 Rightarrow Big| frac{|varphi_n(omega)|}{f(omega)} - text{sgn}(f(omega)) Big| to 0$.
Not sure where to go from here. I imagine I take a subsequence $(varphi_{k(n)})$ that with no zero elements and show that $frac{|varphi_{k(n)}(omega)|}{varphi_{k(n)}(omega)} to frac{|varphi_n(omega)|}{f(omega)}$ and then use this to get the convergence I want.
Case 2: $f(omega) = 0$. Here I need to show $text{sgn}(varphi_n(omega)) to 0$. If $exists M in mathbb{N}$ such that $varphi_n(omega) = 0$ $forall n geq M$, then I'm done. Otherwise, I'm not quite sure. Maybe I need to take a subsequence of nonzero $(varphi_{k(n)}(omega))$ and show $text{sgn}(varphi_{k(n)}(omega)) to 0$.
real-analysis measure-theory convergence
asked Dec 23 '18 at 14:17
zxmknzxmkn
340213
$endgroup$
I want to prove the following claim
Claim: Let $(Omega, Sigma, mu)$ be a measure space. If $f : Omega to mathbb{C}$ is $mu$-measurable, then $textrm{f}$ is $mu$-measurable.
Here a function $f$ is called $mu$-measurable if there exists a
sequence of simple functions on $Sigma$ $(varphi_n)$ such that
$varphi_n to f$ almost everywhere.We define $textrm{sgn}(z) := z/|z|$ for $z neq 0$ and $textrm{sgn}(z) := 0$
for $z = 0$.
This was just mentioned off-the-cuff in lecture as part of another proof. So I imagine it's not too difficult to see. Still, I haven't been able to prove it to myself or find anything on Stack Overflow about it.
My work so far:
I think I want to show the following:
Want to show: Let $(varphi_n)$ be a sequence of simple functions such that $varphi_n to f$ almost everywhere, with $f : Omega to mathbb{C}$. Then $textrm{sgn}(varphi_n) to textrm{sgn}(f)$ almost everywhere.
We let $varphi_n to f$ pointwise on $Omega setminus N$ with $mu(N) = 0$ and take an $omega in Omega setminus N$.
It suffices to show that $|textrm{sgn}(varphi_n(omega)) - textrm{sgn}(f(omega))| to 0$.
Case 1: $f(omega) neq 0$.
$|varphi_n(omega) - f(omega)| to 0 Rightarrow Big||varphi_n(omega)| - |f(omega)|Big| to 0 Rightarrow Big| frac{|varphi_n(omega)|}{f(omega)} - text{sgn}(f(omega)) Big| to 0$.
Not sure where to go from here. I imagine I take a subsequence $(varphi_{k(n)})$ that with no zero elements and show that $frac{|varphi_{k(n)}(omega)|}{varphi_{k(n)}(omega)} to frac{|varphi_n(omega)|}{f(omega)}$ and then use this to get the convergence I want.
Case 2: $f(omega) = 0$. Here I need to show $text{sgn}(varphi_n(omega)) to 0$. If $exists M in mathbb{N}$ such that $varphi_n(omega) = 0$ $forall n geq M$, then I'm done. Otherwise, I'm not quite sure. Maybe I need to take a subsequence of nonzero $(varphi_{k(n)}(omega))$ and show $text{sgn}(varphi_{k(n)}(omega)) to 0$.
real-analysis measure-theory convergence
real-analysis measure-theory convergence
asked Dec 23 '18 at 14:17
zxmknzxmkn
340213
asked Dec 23 '18 at 14:17
zxmknzxmkn
340213
edited Dec 23 '18 at 15:04
zxmkn
asked Dec 23 '18 at 14:17
zxmknzxmkn
340213
asked Dec 23 '18 at 14:17
zxmknzxmkn
340213
asked Dec 23 '18 at 14:17
zxmknzxmkn
340213
340213
$begingroup$
Uhm... If the domain is $Bbb C$, then the image of $operatorname{sgn}$ is not ${0,1}$, but ${0}cup S^1$ (and I'd rather define it as $z/lvert zrvert$ for $zne 0$ because it makes more sense). If the domain is $Bbb R$, then the image is ${-1,0,1}$.
$endgroup$
– Saucy O'Path
Dec 23 '18 at 14:42
$begingroup$
Even on $mathbb R$, It's not true that $sgn(f_n)to sgn(f)$. Take $f_n(x)=frac{1}{n}boldsymbol 1_{mathbb Rsetminus mathbb Q}$. Then $f_n(x)to 0=:f$ for all $x$ but $sgn(f_n)to 1$ a.e. $neq sgn(f)=0$
$endgroup$
– Surb
Dec 23 '18 at 14:43
$begingroup$
@SaucyO'Path I corrected the definition of $text{sgn}$. Thanks.
$endgroup$
– zxmkn
Dec 23 '18 at 15:06
$begingroup$
@Surb Oh, yeah. Then my strategy (WTS: ...) was clearly off. Any ideas for showing the claim here?
$endgroup$
– zxmkn
Dec 23 '18 at 15:08
3
$begingroup$
You have that $sgn(f)(z)=frac{f(z)}{|f(z)|}boldsymbol 1_{{f(z)neq 0}}$ that is a product and quotient of measurable function. Then it's measurable.
$endgroup$
– Surb
Dec 23 '18 at 15:25
add a comment |
$begingroup$
Uhm... If the domain is $Bbb C$, then the image of $operatorname{sgn}$ is not ${0,1}$, but ${0}cup S^1$ (and I'd rather define it as $z/lvert zrvert$ for $zne 0$ because it makes more sense). If the domain is $Bbb R$, then the image is ${-1,0,1}$.
$endgroup$
– Saucy O'Path
Dec 23 '18 at 14:42
$begingroup$
Even on $mathbb R$, It's not true that $sgn(f_n)to sgn(f)$. Take $f_n(x)=frac{1}{n}boldsymbol 1_{mathbb Rsetminus mathbb Q}$. Then $f_n(x)to 0=:f$ for all $x$ but $sgn(f_n)to 1$ a.e. $neq sgn(f)=0$
$endgroup$
– Surb
Dec 23 '18 at 14:43
$begingroup$
@SaucyO'Path I corrected the definition of $text{sgn}$. Thanks.
$endgroup$
– zxmkn
Dec 23 '18 at 15:06
$begingroup$
@Surb Oh, yeah. Then my strategy (WTS: ...) was clearly off. Any ideas for showing the claim here?
$endgroup$
– zxmkn
Dec 23 '18 at 15:08
3
$begingroup$
You have that $sgn(f)(z)=frac{f(z)}{|f(z)|}boldsymbol 1_{{f(z)neq 0}}$ that is a product and quotient of measurable function. Then it's measurable.
$endgroup$
– Surb
Dec 23 '18 at 15:25
$begingroup$
Uhm... If the domain is $Bbb C$, then the image of $operatorname{sgn}$ is not ${0,1}$, but ${0}cup S^1$ (and I'd rather define it as $z/lvert zrvert$ for $zne 0$ because it makes more sense). If the domain is $Bbb R$, then the image is ${-1,0,1}$.
$endgroup$
– Saucy O'Path
Dec 23 '18 at 14:42
$begingroup$
Uhm... If the domain is $Bbb C$, then the image of $operatorname{sgn}$ is not ${0,1}$, but ${0}cup S^1$ (and I'd rather define it as $z/lvert zrvert$ for $zne 0$ because it makes more sense). If the domain is $Bbb R$, then the image is ${-1,0,1}$.
$endgroup$
– Saucy O'Path
Dec 23 '18 at 14:42
$begingroup$
Even on $mathbb R$, It's not true that $sgn(f_n)to sgn(f)$. Take $f_n(x)=frac{1}{n}boldsymbol 1_{mathbb Rsetminus mathbb Q}$. Then $f_n(x)to 0=:f$ for all $x$ but $sgn(f_n)to 1$ a.e. $neq sgn(f)=0$
$endgroup$
– Surb
Dec 23 '18 at 14:43
$begingroup$
Even on $mathbb R$, It's not true that $sgn(f_n)to sgn(f)$. Take $f_n(x)=frac{1}{n}boldsymbol 1_{mathbb Rsetminus mathbb Q}$. Then $f_n(x)to 0=:f$ for all $x$ but $sgn(f_n)to 1$ a.e. $neq sgn(f)=0$
$endgroup$
– Surb
Dec 23 '18 at 14:43
$begingroup$
@SaucyO'Path I corrected the definition of $text{sgn}$. Thanks.
$endgroup$
– zxmkn
Dec 23 '18 at 15:06
$begingroup$
@SaucyO'Path I corrected the definition of $text{sgn}$. Thanks.
$endgroup$
– zxmkn
Dec 23 '18 at 15:06
$begingroup$
@Surb Oh, yeah. Then my strategy (WTS: ...) was clearly off. Any ideas for showing the claim here?
$endgroup$
– zxmkn
Dec 23 '18 at 15:08
$begingroup$
@Surb Oh, yeah. Then my strategy (WTS: ...) was clearly off. Any ideas for showing the claim here?
$endgroup$
– zxmkn
Dec 23 '18 at 15:08
3
3
$begingroup$
You have that $sgn(f)(z)=frac{f(z)}{|f(z)|}boldsymbol 1_{{f(z)neq 0}}$ that is a product and quotient of measurable function. Then it's measurable.
$endgroup$
– Surb
Dec 23 '18 at 15:25
$begingroup$
You have that $sgn(f)(z)=frac{f(z)}{|f(z)|}boldsymbol 1_{{f(z)neq 0}}$ that is a product and quotient of measurable function. Then it's measurable.
$endgroup$
– Surb
Dec 23 '18 at 15:25
add a comment |
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$begingroup$
Uhm... If the domain is $Bbb C$, then the image of $operatorname{sgn}$ is not ${0,1}$, but ${0}cup S^1$ (and I'd rather define it as $z/lvert zrvert$ for $zne 0$ because it makes more sense). If the domain is $Bbb R$, then the image is ${-1,0,1}$.
$endgroup$
– Saucy O'Path
Dec 23 '18 at 14:42
$begingroup$
Even on $mathbb R$, It's not true that $sgn(f_n)to sgn(f)$. Take $f_n(x)=frac{1}{n}boldsymbol 1_{mathbb Rsetminus mathbb Q}$. Then $f_n(x)to 0=:f$ for all $x$ but $sgn(f_n)to 1$ a.e. $neq sgn(f)=0$
$endgroup$
– Surb
Dec 23 '18 at 14:43
$begingroup$
@SaucyO'Path I corrected the definition of $text{sgn}$. Thanks.
$endgroup$
– zxmkn
Dec 23 '18 at 15:06
$begingroup$
@Surb Oh, yeah. Then my strategy (WTS: ...) was clearly off. Any ideas for showing the claim here?
$endgroup$
– zxmkn
Dec 23 '18 at 15:08
3
$begingroup$
You have that $sgn(f)(z)=frac{f(z)}{|f(z)|}boldsymbol 1_{{f(z)neq 0}}$ that is a product and quotient of measurable function. Then it's measurable.
$endgroup$
– Surb
Dec 23 '18 at 15:25