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Bounded, measurable and supported on a set of finite measure function
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Suppose f is a bounded and measurable function on R and supported on a set of finite measure. Prove that for every $epsilon gt 0$ there exists a simple function $s$ such that $int |f-s|dx$ $lt epsilon$
I have question: Can I let this simple function be supported on a set of finite measure? Then this simple function will be bounded and measurable and supported, then I can use the linearity of integral and the definition of the integral of this kind of functions to prove that.
I am wondering if I am right. I really appreciate your help if you would like to give me another proof.
Thanks
real-analysis
asked Dec 8 '13 at 17:28
YangYang
206417
$endgroup$
add a comment |
$begingroup$
Suppose f is a bounded and measurable function on R and supported on a set of finite measure. Prove that for every $epsilon gt 0$ there exists a simple function $s$ such that $int |f-s|dx$ $lt epsilon$
I have question: Can I let this simple function be supported on a set of finite measure? Then this simple function will be bounded and measurable and supported, then I can use the linearity of integral and the definition of the integral of this kind of functions to prove that.
I am wondering if I am right. I really appreciate your help if you would like to give me another proof.
Thanks
real-analysis
asked Dec 8 '13 at 17:28
YangYang
206417
$endgroup$
add a comment |
$begingroup$
Suppose f is a bounded and measurable function on R and supported on a set of finite measure. Prove that for every $epsilon gt 0$ there exists a simple function $s$ such that $int |f-s|dx$ $lt epsilon$
I have question: Can I let this simple function be supported on a set of finite measure? Then this simple function will be bounded and measurable and supported, then I can use the linearity of integral and the definition of the integral of this kind of functions to prove that.
I am wondering if I am right. I really appreciate your help if you would like to give me another proof.
Thanks
real-analysis
asked Dec 8 '13 at 17:28
YangYang
206417
$endgroup$
Suppose f is a bounded and measurable function on R and supported on a set of finite measure. Prove that for every $epsilon gt 0$ there exists a simple function $s$ such that $int |f-s|dx$ $lt epsilon$
I have question: Can I let this simple function be supported on a set of finite measure? Then this simple function will be bounded and measurable and supported, then I can use the linearity of integral and the definition of the integral of this kind of functions to prove that.
I am wondering if I am right. I really appreciate your help if you would like to give me another proof.
Thanks
real-analysis
real-analysis
asked Dec 8 '13 at 17:28
YangYang
206417
asked Dec 8 '13 at 17:28
YangYang
206417
asked Dec 8 '13 at 17:28
YangYang
206417
asked Dec 8 '13 at 17:28
YangYang
206417
asked Dec 8 '13 at 17:28
YangYang
206417
206417
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You can choose any simple function that satisfies the condition whether it is supported by finite set or not. In fact, simple function is always bounded and measurable without any condition.
answered Dec 8 '13 at 18:13
Worawit TepsanWorawit Tepsan
333212
$endgroup$
$begingroup$
Can you give me another proof of this question, Wit@Worawit Tepsan
$endgroup$
– Yang
Dec 8 '13 at 20:10
add a comment |
$begingroup$
We assume that $f$ is supported on $E$ which $m(E)=L<infty.$ So we can consider a function $f$ on E.
Since $f$ is bounded, it must have a and b such that
$a leq f(x) leq b$ for all $xin E$. Let $N > frac{L(b-a)}{varepsilon}.$
We define $E_i = {x in E : a+i(b-a)/N leq f(x) < a+(i+1)(b-a)/N}$.
Choose $x_iin E_i$ and define
$ s(x) = sum_{i=0}^{N} f(x_i) lambda(E_i)$.
Then you can show that
$|f(x)-s(x)|leq frac{varepsilon}{L}.$
And so on.
answered Dec 8 '13 at 20:49
Worawit TepsanWorawit Tepsan
333212
$endgroup$
$begingroup$
the question is that the domain is R which has infinity measure@Worawit Tepsan
$endgroup$
– Yang
Dec 8 '13 at 21:14
$begingroup$
I edited the proof.
$endgroup$
– Worawit Tepsan
Dec 8 '13 at 21:41
1
$begingroup$
@Worawit Tepsan, can you please elaborate on how you can show that $|f(x) - s(x)|leq frac{epsilon}{L}$?
$endgroup$
– ALannister
Nov 18 '15 at 15:17
$begingroup$
I did typo. I should choose $N>frac{L(b-a)}{varepsilon}.$
$endgroup$
– Worawit Tepsan
Nov 29 '15 at 3:33
add a comment |
$begingroup$
Let $X$ be the support of $f$ and $mathbb{1}_X$ its characteristic function. Now observe that $f=fcdot mathbb{1}_X$ and that $s cdot mathbb{1}_X$ approximate $f$.
answered Jan 19 at 12:20
lzralbulzralbu
640512
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can choose any simple function that satisfies the condition whether it is supported by finite set or not. In fact, simple function is always bounded and measurable without any condition.
answered Dec 8 '13 at 18:13
Worawit TepsanWorawit Tepsan
333212
$endgroup$
$begingroup$
Can you give me another proof of this question, Wit@Worawit Tepsan
$endgroup$
– Yang
Dec 8 '13 at 20:10
add a comment |
$begingroup$
You can choose any simple function that satisfies the condition whether it is supported by finite set or not. In fact, simple function is always bounded and measurable without any condition.
answered Dec 8 '13 at 18:13
Worawit TepsanWorawit Tepsan
333212
$endgroup$
$begingroup$
Can you give me another proof of this question, Wit@Worawit Tepsan
$endgroup$
– Yang
Dec 8 '13 at 20:10
add a comment |
$begingroup$
You can choose any simple function that satisfies the condition whether it is supported by finite set or not. In fact, simple function is always bounded and measurable without any condition.
answered Dec 8 '13 at 18:13
Worawit TepsanWorawit Tepsan
333212
$endgroup$
You can choose any simple function that satisfies the condition whether it is supported by finite set or not. In fact, simple function is always bounded and measurable without any condition.
answered Dec 8 '13 at 18:13
Worawit TepsanWorawit Tepsan
333212
answered Dec 8 '13 at 18:13
Worawit TepsanWorawit Tepsan
333212
answered Dec 8 '13 at 18:13
Worawit TepsanWorawit Tepsan
333212
answered Dec 8 '13 at 18:13
Worawit TepsanWorawit Tepsan
333212
333212
$begingroup$
Can you give me another proof of this question, Wit@Worawit Tepsan
$endgroup$
– Yang
Dec 8 '13 at 20:10
add a comment |
$begingroup$
Can you give me another proof of this question, Wit@Worawit Tepsan
$endgroup$
– Yang
Dec 8 '13 at 20:10
$begingroup$
Can you give me another proof of this question, Wit@Worawit Tepsan
$endgroup$
– Yang
Dec 8 '13 at 20:10
$begingroup$
Can you give me another proof of this question, Wit@Worawit Tepsan
$endgroup$
– Yang
Dec 8 '13 at 20:10
add a comment |
$begingroup$
We assume that $f$ is supported on $E$ which $m(E)=L<infty.$ So we can consider a function $f$ on E.
Since $f$ is bounded, it must have a and b such that
$a leq f(x) leq b$ for all $xin E$. Let $N > frac{L(b-a)}{varepsilon}.$
We define $E_i = {x in E : a+i(b-a)/N leq f(x) < a+(i+1)(b-a)/N}$.
Choose $x_iin E_i$ and define
$ s(x) = sum_{i=0}^{N} f(x_i) lambda(E_i)$.
Then you can show that
$|f(x)-s(x)|leq frac{varepsilon}{L}.$
And so on.
answered Dec 8 '13 at 20:49
Worawit TepsanWorawit Tepsan
333212
$endgroup$
$begingroup$
the question is that the domain is R which has infinity measure@Worawit Tepsan
$endgroup$
– Yang
Dec 8 '13 at 21:14
$begingroup$
I edited the proof.
$endgroup$
– Worawit Tepsan
Dec 8 '13 at 21:41
1
$begingroup$
@Worawit Tepsan, can you please elaborate on how you can show that $|f(x) - s(x)|leq frac{epsilon}{L}$?
$endgroup$
– ALannister
Nov 18 '15 at 15:17
$begingroup$
I did typo. I should choose $N>frac{L(b-a)}{varepsilon}.$
$endgroup$
– Worawit Tepsan
Nov 29 '15 at 3:33
add a comment |
$begingroup$
We assume that $f$ is supported on $E$ which $m(E)=L<infty.$ So we can consider a function $f$ on E.
Since $f$ is bounded, it must have a and b such that
$a leq f(x) leq b$ for all $xin E$. Let $N > frac{L(b-a)}{varepsilon}.$
We define $E_i = {x in E : a+i(b-a)/N leq f(x) < a+(i+1)(b-a)/N}$.
Choose $x_iin E_i$ and define
$ s(x) = sum_{i=0}^{N} f(x_i) lambda(E_i)$.
Then you can show that
$|f(x)-s(x)|leq frac{varepsilon}{L}.$
And so on.
answered Dec 8 '13 at 20:49
Worawit TepsanWorawit Tepsan
333212
$endgroup$
$begingroup$
the question is that the domain is R which has infinity measure@Worawit Tepsan
$endgroup$
– Yang
Dec 8 '13 at 21:14
$begingroup$
I edited the proof.
$endgroup$
– Worawit Tepsan
Dec 8 '13 at 21:41
1
$begingroup$
@Worawit Tepsan, can you please elaborate on how you can show that $|f(x) - s(x)|leq frac{epsilon}{L}$?
$endgroup$
– ALannister
Nov 18 '15 at 15:17
$begingroup$
I did typo. I should choose $N>frac{L(b-a)}{varepsilon}.$
$endgroup$
– Worawit Tepsan
Nov 29 '15 at 3:33
add a comment |
$begingroup$
We assume that $f$ is supported on $E$ which $m(E)=L<infty.$ So we can consider a function $f$ on E.
Since $f$ is bounded, it must have a and b such that
$a leq f(x) leq b$ for all $xin E$. Let $N > frac{L(b-a)}{varepsilon}.$
We define $E_i = {x in E : a+i(b-a)/N leq f(x) < a+(i+1)(b-a)/N}$.
Choose $x_iin E_i$ and define
$ s(x) = sum_{i=0}^{N} f(x_i) lambda(E_i)$.
Then you can show that
$|f(x)-s(x)|leq frac{varepsilon}{L}.$
And so on.
answered Dec 8 '13 at 20:49
Worawit TepsanWorawit Tepsan
333212
$endgroup$
We assume that $f$ is supported on $E$ which $m(E)=L<infty.$ So we can consider a function $f$ on E.
Since $f$ is bounded, it must have a and b such that
$a leq f(x) leq b$ for all $xin E$. Let $N > frac{L(b-a)}{varepsilon}.$
We define $E_i = {x in E : a+i(b-a)/N leq f(x) < a+(i+1)(b-a)/N}$.
Choose $x_iin E_i$ and define
$ s(x) = sum_{i=0}^{N} f(x_i) lambda(E_i)$.
Then you can show that
$|f(x)-s(x)|leq frac{varepsilon}{L}.$
And so on.
answered Dec 8 '13 at 20:49
Worawit TepsanWorawit Tepsan
333212
edited Nov 29 '15 at 3:34
answered Dec 8 '13 at 20:49
Worawit TepsanWorawit Tepsan
333212
answered Dec 8 '13 at 20:49
Worawit TepsanWorawit Tepsan
333212
answered Dec 8 '13 at 20:49
Worawit TepsanWorawit Tepsan
333212
333212
$begingroup$
the question is that the domain is R which has infinity measure@Worawit Tepsan
$endgroup$
– Yang
Dec 8 '13 at 21:14
$begingroup$
I edited the proof.
$endgroup$
– Worawit Tepsan
Dec 8 '13 at 21:41
1
$begingroup$
@Worawit Tepsan, can you please elaborate on how you can show that $|f(x) - s(x)|leq frac{epsilon}{L}$?
$endgroup$
– ALannister
Nov 18 '15 at 15:17
$begingroup$
I did typo. I should choose $N>frac{L(b-a)}{varepsilon}.$
$endgroup$
– Worawit Tepsan
Nov 29 '15 at 3:33
add a comment |
$begingroup$
the question is that the domain is R which has infinity measure@Worawit Tepsan
$endgroup$
– Yang
Dec 8 '13 at 21:14
$begingroup$
I edited the proof.
$endgroup$
– Worawit Tepsan
Dec 8 '13 at 21:41
1
$begingroup$
@Worawit Tepsan, can you please elaborate on how you can show that $|f(x) - s(x)|leq frac{epsilon}{L}$?
$endgroup$
– ALannister
Nov 18 '15 at 15:17
$begingroup$
I did typo. I should choose $N>frac{L(b-a)}{varepsilon}.$
$endgroup$
– Worawit Tepsan
Nov 29 '15 at 3:33
$begingroup$
the question is that the domain is R which has infinity measure@Worawit Tepsan
$endgroup$
– Yang
Dec 8 '13 at 21:14
$begingroup$
the question is that the domain is R which has infinity measure@Worawit Tepsan
$endgroup$
– Yang
Dec 8 '13 at 21:14
$begingroup$
I edited the proof.
$endgroup$
– Worawit Tepsan
Dec 8 '13 at 21:41
$begingroup$
I edited the proof.
$endgroup$
– Worawit Tepsan
Dec 8 '13 at 21:41
1
1
$begingroup$
@Worawit Tepsan, can you please elaborate on how you can show that $|f(x) - s(x)|leq frac{epsilon}{L}$?
$endgroup$
– ALannister
Nov 18 '15 at 15:17
$begingroup$
@Worawit Tepsan, can you please elaborate on how you can show that $|f(x) - s(x)|leq frac{epsilon}{L}$?
$endgroup$
– ALannister
Nov 18 '15 at 15:17
$begingroup$
I did typo. I should choose $N>frac{L(b-a)}{varepsilon}.$
$endgroup$
– Worawit Tepsan
Nov 29 '15 at 3:33
$begingroup$
I did typo. I should choose $N>frac{L(b-a)}{varepsilon}.$
$endgroup$
– Worawit Tepsan
Nov 29 '15 at 3:33
add a comment |
$begingroup$
Let $X$ be the support of $f$ and $mathbb{1}_X$ its characteristic function. Now observe that $f=fcdot mathbb{1}_X$ and that $s cdot mathbb{1}_X$ approximate $f$.
answered Jan 19 at 12:20
lzralbulzralbu
640512
$endgroup$
add a comment |
$begingroup$
Let $X$ be the support of $f$ and $mathbb{1}_X$ its characteristic function. Now observe that $f=fcdot mathbb{1}_X$ and that $s cdot mathbb{1}_X$ approximate $f$.
answered Jan 19 at 12:20
lzralbulzralbu
640512
$endgroup$
add a comment |
$begingroup$
Let $X$ be the support of $f$ and $mathbb{1}_X$ its characteristic function. Now observe that $f=fcdot mathbb{1}_X$ and that $s cdot mathbb{1}_X$ approximate $f$.
answered Jan 19 at 12:20
lzralbulzralbu
640512
$endgroup$
Let $X$ be the support of $f$ and $mathbb{1}_X$ its characteristic function. Now observe that $f=fcdot mathbb{1}_X$ and that $s cdot mathbb{1}_X$ approximate $f$.
answered Jan 19 at 12:20
lzralbulzralbu
640512
answered Jan 19 at 12:20
lzralbulzralbu
640512
answered Jan 19 at 12:20
lzralbulzralbu
640512
answered Jan 19 at 12:20
lzralbulzralbu
640512
640512
add a comment |
add a comment |
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