Mixing
Simple function formula
0
Let $f$ be a simple function in $(mathbb{R},B(mathbb{R}),mu)$.
Can we write $f=sum_{k=0}^nalpha_k.1_{[a_k,b_k]}$ instead of $f=sum_{k=0}^nalpha_k.1_{A_k}$ with $A_kin B(mathbb{R})$?
integration lebesgue-integral lebesgue-measure borel-sets
asked Nov 20 '18 at 13:32
Anas BOUALII
887
add a comment |
0
Let $f$ be a simple function in $(mathbb{R},B(mathbb{R}),mu)$.
Can we write $f=sum_{k=0}^nalpha_k.1_{[a_k,b_k]}$ instead of $f=sum_{k=0}^nalpha_k.1_{A_k}$ with $A_kin B(mathbb{R})$?
integration lebesgue-integral lebesgue-measure borel-sets
asked Nov 20 '18 at 13:32
Anas BOUALII
887
Not in general. There are Borel sets which are not intervals. Take for instance the Cantor set (which is closed, hence Borel). How would you decompose it into closed intervals?
– Federico
Nov 20 '18 at 13:36
I see, but in the proof of Reimann Lebesgue Lemma. they choosed a simple function but they wrote it with intervals instead of just measurable sets
– Anas BOUALII
Nov 20 '18 at 13:39
Functions of the form $f=sum_{k=0}^n c_k 1_{[a_k,b_k]}$ are usually called step functions, as opposed to simple functions. Depending on the applications, it might be sufficient to consider only them.
– Federico
Nov 20 '18 at 13:40
So using the approximation theorm in $L^1$. there exists a step function $f_k$ such that the limit is $fin L^1$. ?
– Anas BOUALII
Nov 20 '18 at 13:43
add a comment |
0
0
0
Let $f$ be a simple function in $(mathbb{R},B(mathbb{R}),mu)$.
Can we write $f=sum_{k=0}^nalpha_k.1_{[a_k,b_k]}$ instead of $f=sum_{k=0}^nalpha_k.1_{A_k}$ with $A_kin B(mathbb{R})$?
integration lebesgue-integral lebesgue-measure borel-sets
asked Nov 20 '18 at 13:32
Anas BOUALII
887
Let $f$ be a simple function in $(mathbb{R},B(mathbb{R}),mu)$.
Can we write $f=sum_{k=0}^nalpha_k.1_{[a_k,b_k]}$ instead of $f=sum_{k=0}^nalpha_k.1_{A_k}$ with $A_kin B(mathbb{R})$?
integration lebesgue-integral lebesgue-measure borel-sets
integration lebesgue-integral lebesgue-measure borel-sets
asked Nov 20 '18 at 13:32
Anas BOUALII
887
asked Nov 20 '18 at 13:32
Anas BOUALII
887
asked Nov 20 '18 at 13:32
Anas BOUALII
887
asked Nov 20 '18 at 13:32
Anas BOUALII
887
asked Nov 20 '18 at 13:32
Anas BOUALII
887
887
Not in general. There are Borel sets which are not intervals. Take for instance the Cantor set (which is closed, hence Borel). How would you decompose it into closed intervals?
– Federico
Nov 20 '18 at 13:36
I see, but in the proof of Reimann Lebesgue Lemma. they choosed a simple function but they wrote it with intervals instead of just measurable sets
– Anas BOUALII
Nov 20 '18 at 13:39
Functions of the form $f=sum_{k=0}^n c_k 1_{[a_k,b_k]}$ are usually called step functions, as opposed to simple functions. Depending on the applications, it might be sufficient to consider only them.
– Federico
Nov 20 '18 at 13:40
So using the approximation theorm in $L^1$. there exists a step function $f_k$ such that the limit is $fin L^1$. ?
– Anas BOUALII
Nov 20 '18 at 13:43
add a comment |
Not in general. There are Borel sets which are not intervals. Take for instance the Cantor set (which is closed, hence Borel). How would you decompose it into closed intervals?
– Federico
Nov 20 '18 at 13:36
I see, but in the proof of Reimann Lebesgue Lemma. they choosed a simple function but they wrote it with intervals instead of just measurable sets
– Anas BOUALII
Nov 20 '18 at 13:39
Functions of the form $f=sum_{k=0}^n c_k 1_{[a_k,b_k]}$ are usually called step functions, as opposed to simple functions. Depending on the applications, it might be sufficient to consider only them.
– Federico
Nov 20 '18 at 13:40
So using the approximation theorm in $L^1$. there exists a step function $f_k$ such that the limit is $fin L^1$. ?
– Anas BOUALII
Nov 20 '18 at 13:43
Not in general. There are Borel sets which are not intervals. Take for instance the Cantor set (which is closed, hence Borel). How would you decompose it into closed intervals?
– Federico
Nov 20 '18 at 13:36
Not in general. There are Borel sets which are not intervals. Take for instance the Cantor set (which is closed, hence Borel). How would you decompose it into closed intervals?
– Federico
Nov 20 '18 at 13:36
I see, but in the proof of Reimann Lebesgue Lemma. they choosed a simple function but they wrote it with intervals instead of just measurable sets
– Anas BOUALII
Nov 20 '18 at 13:39
I see, but in the proof of Reimann Lebesgue Lemma. they choosed a simple function but they wrote it with intervals instead of just measurable sets
– Anas BOUALII
Nov 20 '18 at 13:39
Functions of the form $f=sum_{k=0}^n c_k 1_{[a_k,b_k]}$ are usually called step functions, as opposed to simple functions. Depending on the applications, it might be sufficient to consider only them.
– Federico
Nov 20 '18 at 13:40
Functions of the form $f=sum_{k=0}^n c_k 1_{[a_k,b_k]}$ are usually called step functions, as opposed to simple functions. Depending on the applications, it might be sufficient to consider only them.
– Federico
Nov 20 '18 at 13:40
So using the approximation theorm in $L^1$. there exists a step function $f_k$ such that the limit is $fin L^1$. ?
– Anas BOUALII
Nov 20 '18 at 13:43
So using the approximation theorm in $L^1$. there exists a step function $f_k$ such that the limit is $fin L^1$. ?
– Anas BOUALII
Nov 20 '18 at 13:43
add a comment |
1 Answer
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No. Let $A_0=mathbb Z$. Then $f=1_{A_0}$ has no representation of the form $f=sum_{k=0}^n c_k 1_{[a_k,b_k]}$.
answered Nov 20 '18 at 13:39
Federico
4,679514
add a comment |
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1 Answer
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1 Answer
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No. Let $A_0=mathbb Z$. Then $f=1_{A_0}$ has no representation of the form $f=sum_{k=0}^n c_k 1_{[a_k,b_k]}$.
answered Nov 20 '18 at 13:39
Federico
4,679514
add a comment |
0
No. Let $A_0=mathbb Z$. Then $f=1_{A_0}$ has no representation of the form $f=sum_{k=0}^n c_k 1_{[a_k,b_k]}$.
answered Nov 20 '18 at 13:39
Federico
4,679514
add a comment |
0
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No. Let $A_0=mathbb Z$. Then $f=1_{A_0}$ has no representation of the form $f=sum_{k=0}^n c_k 1_{[a_k,b_k]}$.
answered Nov 20 '18 at 13:39
Federico
4,679514
No. Let $A_0=mathbb Z$. Then $f=1_{A_0}$ has no representation of the form $f=sum_{k=0}^n c_k 1_{[a_k,b_k]}$.
answered Nov 20 '18 at 13:39
Federico
4,679514
answered Nov 20 '18 at 13:39
Federico
4,679514
answered Nov 20 '18 at 13:39
Federico
4,679514
answered Nov 20 '18 at 13:39
Federico
4,679514
4,679514
add a comment |
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Not in general. There are Borel sets which are not intervals. Take for instance the Cantor set (which is closed, hence Borel). How would you decompose it into closed intervals?
– Federico
Nov 20 '18 at 13:36
I see, but in the proof of Reimann Lebesgue Lemma. they choosed a simple function but they wrote it with intervals instead of just measurable sets
– Anas BOUALII
Nov 20 '18 at 13:39
Functions of the form $f=sum_{k=0}^n c_k 1_{[a_k,b_k]}$ are usually called step functions, as opposed to simple functions. Depending on the applications, it might be sufficient to consider only them.
– Federico
Nov 20 '18 at 13:40
So using the approximation theorm in $L^1$. there exists a step function $f_k$ such that the limit is $fin L^1$. ?
– Anas BOUALII
Nov 20 '18 at 13:43