Simple function formula
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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 at 13:32
Anas BOUALII
253
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up vote
0
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favorite
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 at 13:32
Anas BOUALII
253
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 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 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 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 at 13:43
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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 at 13:32
Anas BOUALII
253
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 at 13:32
Anas BOUALII
253
asked Nov 20 at 13:32
Anas BOUALII
253
asked Nov 20 at 13:32
Anas BOUALII
253
asked Nov 20 at 13:32
Anas BOUALII
253
asked Nov 20 at 13:32
Anas BOUALII
253
253
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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 at 13:43
add a comment |
1 Answer
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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 at 13:39
Federico
2,629510
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
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 at 13:39
Federico
2,629510
add a comment |
up vote
0
down vote
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 at 13:39
Federico
2,629510
add a comment |
up vote
0
down vote
up vote
0
down vote
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 at 13:39
Federico
2,629510
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 at 13:39
Federico
2,629510
answered Nov 20 at 13:39
Federico
2,629510
answered Nov 20 at 13:39
Federico
2,629510
answered Nov 20 at 13:39
Federico
2,629510
2,629510
add a comment |
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 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 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 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 at 13:43