showing that $|f|_1=sup {int_{[a,b]}tau(x)dx mid tau text{ step function and } taule f} $
$begingroup$
Let $a,binmathbb{R}$ such that $a<b$ and $fcolon [a,b]to mathbb{R}$ a non-negative function. Is then
$$|f|_1=sup {int_{[a,b]}tau(x)dx mid tau text{ step function and } taule f} ?$$
I have seen it to be true under additionally assuming that $f$ is Riemann-integrable. Therefore, I initially tried to find a counterexample (with a non Riemann-integrable function). But then I came across here Lebesgue integrable implies Riemann integrable?, so that the equality probably is true (note that simple functions are more general as step functions!). And now I don't know whether it's true or not (I tend to vote for wrong, but I still don't have a counterexample)) ..
I appreciate any help.
integration measure-theory lebesgue-integral step-function
asked Dec 29 '18 at 15:16
hettyhetty
1656
$endgroup$
add a comment |
$begingroup$
Let $a,binmathbb{R}$ such that $a<b$ and $fcolon [a,b]to mathbb{R}$ a non-negative function. Is then
$$|f|_1=sup {int_{[a,b]}tau(x)dx mid tau text{ step function and } taule f} ?$$
I have seen it to be true under additionally assuming that $f$ is Riemann-integrable. Therefore, I initially tried to find a counterexample (with a non Riemann-integrable function). But then I came across here Lebesgue integrable implies Riemann integrable?, so that the equality probably is true (note that simple functions are more general as step functions!). And now I don't know whether it's true or not (I tend to vote for wrong, but I still don't have a counterexample)) ..
I appreciate any help.
integration measure-theory lebesgue-integral step-function
asked Dec 29 '18 at 15:16
hettyhetty
1656
$endgroup$
$begingroup$
That's actually close to the definition of integral (in the Lebesgue sense) of a non-negative measurable function, where you consider simple functions in place of step functions, see here.
$endgroup$
– Anguepa
Dec 29 '18 at 15:42
$begingroup$
It would be good if we are given a definition of $|f|_1.$ Is it $|f|_1 = int_{[a,b]}|f(x)|,dx?$
$endgroup$
– Idonknow
Dec 29 '18 at 15:53
$begingroup$
@Idonknow Oh sorry. yes, it is
$endgroup$
– hetty
Dec 29 '18 at 15:57
add a comment |
$begingroup$
Let $a,binmathbb{R}$ such that $a<b$ and $fcolon [a,b]to mathbb{R}$ a non-negative function. Is then
$$|f|_1=sup {int_{[a,b]}tau(x)dx mid tau text{ step function and } taule f} ?$$
I have seen it to be true under additionally assuming that $f$ is Riemann-integrable. Therefore, I initially tried to find a counterexample (with a non Riemann-integrable function). But then I came across here Lebesgue integrable implies Riemann integrable?, so that the equality probably is true (note that simple functions are more general as step functions!). And now I don't know whether it's true or not (I tend to vote for wrong, but I still don't have a counterexample)) ..
I appreciate any help.
integration measure-theory lebesgue-integral step-function
asked Dec 29 '18 at 15:16
hettyhetty
1656
$endgroup$
Let $a,binmathbb{R}$ such that $a<b$ and $fcolon [a,b]to mathbb{R}$ a non-negative function. Is then
$$|f|_1=sup {int_{[a,b]}tau(x)dx mid tau text{ step function and } taule f} ?$$
I have seen it to be true under additionally assuming that $f$ is Riemann-integrable. Therefore, I initially tried to find a counterexample (with a non Riemann-integrable function). But then I came across here Lebesgue integrable implies Riemann integrable?, so that the equality probably is true (note that simple functions are more general as step functions!). And now I don't know whether it's true or not (I tend to vote for wrong, but I still don't have a counterexample)) ..
I appreciate any help.
integration measure-theory lebesgue-integral step-function
integration measure-theory lebesgue-integral step-function
asked Dec 29 '18 at 15:16
hettyhetty
1656
asked Dec 29 '18 at 15:16
hettyhetty
1656
asked Dec 29 '18 at 15:16
hettyhetty
1656
asked Dec 29 '18 at 15:16
hettyhetty
1656
asked Dec 29 '18 at 15:16
hettyhetty
1656
1656
$begingroup$
That's actually close to the definition of integral (in the Lebesgue sense) of a non-negative measurable function, where you consider simple functions in place of step functions, see here.
$endgroup$
– Anguepa
Dec 29 '18 at 15:42
$begingroup$
It would be good if we are given a definition of $|f|_1.$ Is it $|f|_1 = int_{[a,b]}|f(x)|,dx?$
$endgroup$
– Idonknow
Dec 29 '18 at 15:53
$begingroup$
@Idonknow Oh sorry. yes, it is
$endgroup$
– hetty
Dec 29 '18 at 15:57
add a comment |
$begingroup$
That's actually close to the definition of integral (in the Lebesgue sense) of a non-negative measurable function, where you consider simple functions in place of step functions, see here.
$endgroup$
– Anguepa
Dec 29 '18 at 15:42
$begingroup$
It would be good if we are given a definition of $|f|_1.$ Is it $|f|_1 = int_{[a,b]}|f(x)|,dx?$
$endgroup$
– Idonknow
Dec 29 '18 at 15:53
$begingroup$
@Idonknow Oh sorry. yes, it is
$endgroup$
– hetty
Dec 29 '18 at 15:57
$begingroup$
That's actually close to the definition of integral (in the Lebesgue sense) of a non-negative measurable function, where you consider simple functions in place of step functions, see here.
$endgroup$
– Anguepa
Dec 29 '18 at 15:42
$begingroup$
That's actually close to the definition of integral (in the Lebesgue sense) of a non-negative measurable function, where you consider simple functions in place of step functions, see here.
$endgroup$
– Anguepa
Dec 29 '18 at 15:42
$begingroup$
It would be good if we are given a definition of $|f|_1.$ Is it $|f|_1 = int_{[a,b]}|f(x)|,dx?$
$endgroup$
– Idonknow
Dec 29 '18 at 15:53
$begingroup$
It would be good if we are given a definition of $|f|_1.$ Is it $|f|_1 = int_{[a,b]}|f(x)|,dx?$
$endgroup$
– Idonknow
Dec 29 '18 at 15:53
$begingroup$
@Idonknow Oh sorry. yes, it is
$endgroup$
– hetty
Dec 29 '18 at 15:57
$begingroup$
@Idonknow Oh sorry. yes, it is
$endgroup$
– hetty
Dec 29 '18 at 15:57
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I'm afraid this is not true for all (Lebesgue) integrable functions.
Consider the indicator function on $[a,b]setminus mathbb{Q}$, namely $mathbb{I}_{[a,b]setminus mathbb{Q}} : [a,b]rightarrow {0,1}$ where $mathbb{I}_{[a,b]setminus mathbb{Q}}(x)=1$ for $xin [a,b]setminus mathbb{Q}$ and $0$ otherwise.
This function only differs from the indicator function on $[a,b]$ on countably many points, so
$$
int mathbb{I}_{[a,b]setminus mathbb{Q}} = int mathbb{I}_{[a,b]}=b-a.
$$
However clearly any step function $s:[a,b]rightarrow mathbb{R}$ with $sleq mathbb{I}_{[a,b]setminus mathbb{Q}}$ maps into $(-infty, 0]$.
It is true however for continuous functions on $[a,b]$, since these are Riemann-integrable.
answered Dec 29 '18 at 15:52
AnguepaAnguepa
1,391819
$endgroup$
$begingroup$
I see, thank you.
$endgroup$
– hetty
Dec 29 '18 at 16:17
add a comment |
Your Answer
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
I'm afraid this is not true for all (Lebesgue) integrable functions.
Consider the indicator function on $[a,b]setminus mathbb{Q}$, namely $mathbb{I}_{[a,b]setminus mathbb{Q}} : [a,b]rightarrow {0,1}$ where $mathbb{I}_{[a,b]setminus mathbb{Q}}(x)=1$ for $xin [a,b]setminus mathbb{Q}$ and $0$ otherwise.
This function only differs from the indicator function on $[a,b]$ on countably many points, so
$$
int mathbb{I}_{[a,b]setminus mathbb{Q}} = int mathbb{I}_{[a,b]}=b-a.
$$
However clearly any step function $s:[a,b]rightarrow mathbb{R}$ with $sleq mathbb{I}_{[a,b]setminus mathbb{Q}}$ maps into $(-infty, 0]$.
It is true however for continuous functions on $[a,b]$, since these are Riemann-integrable.
answered Dec 29 '18 at 15:52
AnguepaAnguepa
1,391819
$endgroup$
$begingroup$
I see, thank you.
$endgroup$
– hetty
Dec 29 '18 at 16:17
add a comment |
$begingroup$
I'm afraid this is not true for all (Lebesgue) integrable functions.
Consider the indicator function on $[a,b]setminus mathbb{Q}$, namely $mathbb{I}_{[a,b]setminus mathbb{Q}} : [a,b]rightarrow {0,1}$ where $mathbb{I}_{[a,b]setminus mathbb{Q}}(x)=1$ for $xin [a,b]setminus mathbb{Q}$ and $0$ otherwise.
This function only differs from the indicator function on $[a,b]$ on countably many points, so
$$
int mathbb{I}_{[a,b]setminus mathbb{Q}} = int mathbb{I}_{[a,b]}=b-a.
$$
However clearly any step function $s:[a,b]rightarrow mathbb{R}$ with $sleq mathbb{I}_{[a,b]setminus mathbb{Q}}$ maps into $(-infty, 0]$.
It is true however for continuous functions on $[a,b]$, since these are Riemann-integrable.
answered Dec 29 '18 at 15:52
AnguepaAnguepa
1,391819
$endgroup$
$begingroup$
I see, thank you.
$endgroup$
– hetty
Dec 29 '18 at 16:17
add a comment |
$begingroup$
I'm afraid this is not true for all (Lebesgue) integrable functions.
Consider the indicator function on $[a,b]setminus mathbb{Q}$, namely $mathbb{I}_{[a,b]setminus mathbb{Q}} : [a,b]rightarrow {0,1}$ where $mathbb{I}_{[a,b]setminus mathbb{Q}}(x)=1$ for $xin [a,b]setminus mathbb{Q}$ and $0$ otherwise.
This function only differs from the indicator function on $[a,b]$ on countably many points, so
$$
int mathbb{I}_{[a,b]setminus mathbb{Q}} = int mathbb{I}_{[a,b]}=b-a.
$$
However clearly any step function $s:[a,b]rightarrow mathbb{R}$ with $sleq mathbb{I}_{[a,b]setminus mathbb{Q}}$ maps into $(-infty, 0]$.
It is true however for continuous functions on $[a,b]$, since these are Riemann-integrable.
answered Dec 29 '18 at 15:52
AnguepaAnguepa
1,391819
$endgroup$
I'm afraid this is not true for all (Lebesgue) integrable functions.
Consider the indicator function on $[a,b]setminus mathbb{Q}$, namely $mathbb{I}_{[a,b]setminus mathbb{Q}} : [a,b]rightarrow {0,1}$ where $mathbb{I}_{[a,b]setminus mathbb{Q}}(x)=1$ for $xin [a,b]setminus mathbb{Q}$ and $0$ otherwise.
This function only differs from the indicator function on $[a,b]$ on countably many points, so
$$
int mathbb{I}_{[a,b]setminus mathbb{Q}} = int mathbb{I}_{[a,b]}=b-a.
$$
However clearly any step function $s:[a,b]rightarrow mathbb{R}$ with $sleq mathbb{I}_{[a,b]setminus mathbb{Q}}$ maps into $(-infty, 0]$.
It is true however for continuous functions on $[a,b]$, since these are Riemann-integrable.
answered Dec 29 '18 at 15:52
AnguepaAnguepa
1,391819
edited Dec 29 '18 at 21:27
answered Dec 29 '18 at 15:52
AnguepaAnguepa
1,391819
answered Dec 29 '18 at 15:52
AnguepaAnguepa
1,391819
answered Dec 29 '18 at 15:52
AnguepaAnguepa
1,391819
1,391819
$begingroup$
I see, thank you.
$endgroup$
– hetty
Dec 29 '18 at 16:17
add a comment |
$begingroup$
I see, thank you.
$endgroup$
– hetty
Dec 29 '18 at 16:17
$begingroup$
I see, thank you.
$endgroup$
– hetty
Dec 29 '18 at 16:17
$begingroup$
I see, thank you.
$endgroup$
– hetty
Dec 29 '18 at 16:17
add a comment |
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$begingroup$
That's actually close to the definition of integral (in the Lebesgue sense) of a non-negative measurable function, where you consider simple functions in place of step functions, see here.
$endgroup$
– Anguepa
Dec 29 '18 at 15:42
$begingroup$
It would be good if we are given a definition of $|f|_1.$ Is it $|f|_1 = int_{[a,b]}|f(x)|,dx?$
$endgroup$
– Idonknow
Dec 29 '18 at 15:53
$begingroup$
@Idonknow Oh sorry. yes, it is
$endgroup$
– hetty
Dec 29 '18 at 15:57