Why is the indicator function of rational numbers a simple function?
1
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I understand the simple function is one that is measurable and takes on finitely many values. However, the amount of values in the set Q are infinite. Thanks
real-analysis measure-theory simple-functions
asked Jan 14 at 21:08
Thomas Mc DonaldThomas Mc Donald
63
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|
show 1 more comment
1
$begingroup$
I understand the simple function is one that is measurable and takes on finitely many values. However, the amount of values in the set Q are infinite. Thanks
real-analysis measure-theory simple-functions
asked Jan 14 at 21:08
Thomas Mc DonaldThomas Mc Donald
63
$endgroup$
$begingroup$
The indicator function, any indicator function, takes on at most two values.
$endgroup$
– lulu
Jan 14 at 21:09
$begingroup$
The indicator function of the rationals is $fcolon mathbb R to {0,1}$ with $f(x)=1$ iff $xin mathbb Q$. It only takes two values.
$endgroup$
– Ittay Weiss
Jan 14 at 21:09
2
$begingroup$
Is it because it only takes on the value 0 and 1?
$endgroup$
– Thomas Mc Donald
Jan 14 at 21:09
$begingroup$
Ah I see, very stupid, thanks!
$endgroup$
– Thomas Mc Donald
Jan 14 at 21:10
$begingroup$
Yes! Well, you also have to argue that $mathbb Q$ is measurable, but as it is countable it has measure $0$.
$endgroup$
– lulu
Jan 14 at 21:10
|
show 1 more comment
1
1
1
$begingroup$
I understand the simple function is one that is measurable and takes on finitely many values. However, the amount of values in the set Q are infinite. Thanks
real-analysis measure-theory simple-functions
asked Jan 14 at 21:08
Thomas Mc DonaldThomas Mc Donald
63
$endgroup$
I understand the simple function is one that is measurable and takes on finitely many values. However, the amount of values in the set Q are infinite. Thanks
real-analysis measure-theory simple-functions
real-analysis measure-theory simple-functions
asked Jan 14 at 21:08
Thomas Mc DonaldThomas Mc Donald
63
asked Jan 14 at 21:08
Thomas Mc DonaldThomas Mc Donald
63
asked Jan 14 at 21:08
Thomas Mc DonaldThomas Mc Donald
63
asked Jan 14 at 21:08
Thomas Mc DonaldThomas Mc Donald
63
asked Jan 14 at 21:08
Thomas Mc DonaldThomas Mc Donald
63
63
$begingroup$
The indicator function, any indicator function, takes on at most two values.
$endgroup$
– lulu
Jan 14 at 21:09
$begingroup$
The indicator function of the rationals is $fcolon mathbb R to {0,1}$ with $f(x)=1$ iff $xin mathbb Q$. It only takes two values.
$endgroup$
– Ittay Weiss
Jan 14 at 21:09
2
$begingroup$
Is it because it only takes on the value 0 and 1?
$endgroup$
– Thomas Mc Donald
Jan 14 at 21:09
$begingroup$
Ah I see, very stupid, thanks!
$endgroup$
– Thomas Mc Donald
Jan 14 at 21:10
$begingroup$
Yes! Well, you also have to argue that $mathbb Q$ is measurable, but as it is countable it has measure $0$.
$endgroup$
– lulu
Jan 14 at 21:10
|
show 1 more comment
$begingroup$
The indicator function, any indicator function, takes on at most two values.
$endgroup$
– lulu
Jan 14 at 21:09
$begingroup$
The indicator function of the rationals is $fcolon mathbb R to {0,1}$ with $f(x)=1$ iff $xin mathbb Q$. It only takes two values.
$endgroup$
– Ittay Weiss
Jan 14 at 21:09
2
$begingroup$
Is it because it only takes on the value 0 and 1?
$endgroup$
– Thomas Mc Donald
Jan 14 at 21:09
$begingroup$
Ah I see, very stupid, thanks!
$endgroup$
– Thomas Mc Donald
Jan 14 at 21:10
$begingroup$
Yes! Well, you also have to argue that $mathbb Q$ is measurable, but as it is countable it has measure $0$.
$endgroup$
– lulu
Jan 14 at 21:10
$begingroup$
The indicator function, any indicator function, takes on at most two values.
$endgroup$
– lulu
Jan 14 at 21:09
$begingroup$
The indicator function, any indicator function, takes on at most two values.
$endgroup$
– lulu
Jan 14 at 21:09
$begingroup$
The indicator function of the rationals is $fcolon mathbb R to {0,1}$ with $f(x)=1$ iff $xin mathbb Q$. It only takes two values.
$endgroup$
– Ittay Weiss
Jan 14 at 21:09
$begingroup$
The indicator function of the rationals is $fcolon mathbb R to {0,1}$ with $f(x)=1$ iff $xin mathbb Q$. It only takes two values.
$endgroup$
– Ittay Weiss
Jan 14 at 21:09
2
2
$begingroup$
Is it because it only takes on the value 0 and 1?
$endgroup$
– Thomas Mc Donald
Jan 14 at 21:09
$begingroup$
Is it because it only takes on the value 0 and 1?
$endgroup$
– Thomas Mc Donald
Jan 14 at 21:09
$begingroup$
Ah I see, very stupid, thanks!
$endgroup$
– Thomas Mc Donald
Jan 14 at 21:10
$begingroup$
Ah I see, very stupid, thanks!
$endgroup$
– Thomas Mc Donald
Jan 14 at 21:10
$begingroup$
Yes! Well, you also have to argue that $mathbb Q$ is measurable, but as it is countable it has measure $0$.
$endgroup$
– lulu
Jan 14 at 21:10
$begingroup$
Yes! Well, you also have to argue that $mathbb Q$ is measurable, but as it is countable it has measure $0$.
$endgroup$
– lulu
Jan 14 at 21:10
|
show 1 more comment
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$begingroup$
The indicator function, any indicator function, takes on at most two values.
$endgroup$
– lulu
Jan 14 at 21:09
$begingroup$
The indicator function of the rationals is $fcolon mathbb R to {0,1}$ with $f(x)=1$ iff $xin mathbb Q$. It only takes two values.
$endgroup$
– Ittay Weiss
Jan 14 at 21:09
2
$begingroup$
Is it because it only takes on the value 0 and 1?
$endgroup$
– Thomas Mc Donald
Jan 14 at 21:09
$begingroup$
Ah I see, very stupid, thanks!
$endgroup$
– Thomas Mc Donald
Jan 14 at 21:10
$begingroup$
Yes! Well, you also have to argue that $mathbb Q$ is measurable, but as it is countable it has measure $0$.
$endgroup$
– lulu
Jan 14 at 21:10