First mean value theorem for integration and Lebesgue measureability
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According to first mean value theorem for integration, if $G : [a,b] to mathbb{R}$ is a continuous function, there exists $x in (a,b)$ such that
$$int_a^b G(t) dt = G(x)(b-a)$$
Assume $G$ is a continuous function defined on $[a,b]$. For $0 < h < frac{b-a}{2}$
$$overline{G} : y mapsto int_{y-h}^{y+h} G(t) dt$$ is defined for $y in [a+h,b-h]$. Applying the first mean value theorem for integration, for all $y in [a+h,b-h]$, there exists $c_y in (y-h,y+h)$ with
$$overline{G}(y)=int_{y-h}^{y+h} G(t) dt = 2 h G(c_y)$$
Taking for $G$ a constant function, $c_y$ can by any point in $(y-h,y+h)$. Hence we can pick up it in a way for which $y mapsto c_y$ is not a Lebesgue measureable function.
Question: can one find a continuous function $G$ for which $c_y$ is uniquely defined for all $y in (a+h,b-h)$ and such that $y mapsto c_y$ is not Lebesgue measureable?
Jean-Pierre (http://www.mathcounterexamples.net)
calculus integration measure-theory
asked Jun 6 '15 at 0:30
mathcounterexamples.netmathcounterexamples.net
26.8k22157
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add a comment |
$begingroup$
According to first mean value theorem for integration, if $G : [a,b] to mathbb{R}$ is a continuous function, there exists $x in (a,b)$ such that
$$int_a^b G(t) dt = G(x)(b-a)$$
Assume $G$ is a continuous function defined on $[a,b]$. For $0 < h < frac{b-a}{2}$
$$overline{G} : y mapsto int_{y-h}^{y+h} G(t) dt$$ is defined for $y in [a+h,b-h]$. Applying the first mean value theorem for integration, for all $y in [a+h,b-h]$, there exists $c_y in (y-h,y+h)$ with
$$overline{G}(y)=int_{y-h}^{y+h} G(t) dt = 2 h G(c_y)$$
Taking for $G$ a constant function, $c_y$ can by any point in $(y-h,y+h)$. Hence we can pick up it in a way for which $y mapsto c_y$ is not a Lebesgue measureable function.
Question: can one find a continuous function $G$ for which $c_y$ is uniquely defined for all $y in (a+h,b-h)$ and such that $y mapsto c_y$ is not Lebesgue measureable?
Jean-Pierre (http://www.mathcounterexamples.net)
calculus integration measure-theory
asked Jun 6 '15 at 0:30
mathcounterexamples.netmathcounterexamples.net
26.8k22157
$endgroup$
$begingroup$
In fact I have the feeling that my question is not so relevant... Is the condition on the unicity of $c_y$ not implying that $G$ is monotonic. Hence $y mapsto c_y$ is also monotonic and measurable?
$endgroup$
– mathcounterexamples.net
Jun 6 '15 at 9:39
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Are you also assuming that $c_y$ does not depend on $h$ ?
$endgroup$
– Charles Madeline
Jun 24 '18 at 7:57
$begingroup$
I can think of a result which might be of some interest to you (as it seems close to the core of your problem): if $f$ is differentiable and for all $x<y$, the set of $zin (x,y)$ such that $f'(z)=frac{f(y)-f(x)}{y-x}$ is an interval, then $f$ or $-f$ is convex (thus $f'$ monotonic)
$endgroup$
– Charles Madeline
Jun 24 '18 at 8:00
1
$begingroup$
I suppose that $h$ is defined and fixed in the question.
$endgroup$
– mathcounterexamples.net
Jun 24 '18 at 8:42
add a comment |
$begingroup$
According to first mean value theorem for integration, if $G : [a,b] to mathbb{R}$ is a continuous function, there exists $x in (a,b)$ such that
$$int_a^b G(t) dt = G(x)(b-a)$$
Assume $G$ is a continuous function defined on $[a,b]$. For $0 < h < frac{b-a}{2}$
$$overline{G} : y mapsto int_{y-h}^{y+h} G(t) dt$$ is defined for $y in [a+h,b-h]$. Applying the first mean value theorem for integration, for all $y in [a+h,b-h]$, there exists $c_y in (y-h,y+h)$ with
$$overline{G}(y)=int_{y-h}^{y+h} G(t) dt = 2 h G(c_y)$$
Taking for $G$ a constant function, $c_y$ can by any point in $(y-h,y+h)$. Hence we can pick up it in a way for which $y mapsto c_y$ is not a Lebesgue measureable function.
Question: can one find a continuous function $G$ for which $c_y$ is uniquely defined for all $y in (a+h,b-h)$ and such that $y mapsto c_y$ is not Lebesgue measureable?
Jean-Pierre (http://www.mathcounterexamples.net)
calculus integration measure-theory
asked Jun 6 '15 at 0:30
mathcounterexamples.netmathcounterexamples.net
26.8k22157
$endgroup$
According to first mean value theorem for integration, if $G : [a,b] to mathbb{R}$ is a continuous function, there exists $x in (a,b)$ such that
$$int_a^b G(t) dt = G(x)(b-a)$$
Assume $G$ is a continuous function defined on $[a,b]$. For $0 < h < frac{b-a}{2}$
$$overline{G} : y mapsto int_{y-h}^{y+h} G(t) dt$$ is defined for $y in [a+h,b-h]$. Applying the first mean value theorem for integration, for all $y in [a+h,b-h]$, there exists $c_y in (y-h,y+h)$ with
$$overline{G}(y)=int_{y-h}^{y+h} G(t) dt = 2 h G(c_y)$$
Taking for $G$ a constant function, $c_y$ can by any point in $(y-h,y+h)$. Hence we can pick up it in a way for which $y mapsto c_y$ is not a Lebesgue measureable function.
Question: can one find a continuous function $G$ for which $c_y$ is uniquely defined for all $y in (a+h,b-h)$ and such that $y mapsto c_y$ is not Lebesgue measureable?
Jean-Pierre (http://www.mathcounterexamples.net)
calculus integration measure-theory
calculus integration measure-theory
asked Jun 6 '15 at 0:30
mathcounterexamples.netmathcounterexamples.net
26.8k22157
asked Jun 6 '15 at 0:30
mathcounterexamples.netmathcounterexamples.net
26.8k22157
asked Jun 6 '15 at 0:30
mathcounterexamples.netmathcounterexamples.net
26.8k22157
asked Jun 6 '15 at 0:30
mathcounterexamples.netmathcounterexamples.net
26.8k22157
asked Jun 6 '15 at 0:30
mathcounterexamples.netmathcounterexamples.net
26.8k22157
26.8k22157
$begingroup$
In fact I have the feeling that my question is not so relevant... Is the condition on the unicity of $c_y$ not implying that $G$ is monotonic. Hence $y mapsto c_y$ is also monotonic and measurable?
$endgroup$
– mathcounterexamples.net
Jun 6 '15 at 9:39
$begingroup$
Are you also assuming that $c_y$ does not depend on $h$ ?
$endgroup$
– Charles Madeline
Jun 24 '18 at 7:57
$begingroup$
I can think of a result which might be of some interest to you (as it seems close to the core of your problem): if $f$ is differentiable and for all $x<y$, the set of $zin (x,y)$ such that $f'(z)=frac{f(y)-f(x)}{y-x}$ is an interval, then $f$ or $-f$ is convex (thus $f'$ monotonic)
$endgroup$
– Charles Madeline
Jun 24 '18 at 8:00
1
$begingroup$
I suppose that $h$ is defined and fixed in the question.
$endgroup$
– mathcounterexamples.net
Jun 24 '18 at 8:42
add a comment |
$begingroup$
In fact I have the feeling that my question is not so relevant... Is the condition on the unicity of $c_y$ not implying that $G$ is monotonic. Hence $y mapsto c_y$ is also monotonic and measurable?
$endgroup$
– mathcounterexamples.net
Jun 6 '15 at 9:39
$begingroup$
Are you also assuming that $c_y$ does not depend on $h$ ?
$endgroup$
– Charles Madeline
Jun 24 '18 at 7:57
$begingroup$
I can think of a result which might be of some interest to you (as it seems close to the core of your problem): if $f$ is differentiable and for all $x<y$, the set of $zin (x,y)$ such that $f'(z)=frac{f(y)-f(x)}{y-x}$ is an interval, then $f$ or $-f$ is convex (thus $f'$ monotonic)
$endgroup$
– Charles Madeline
Jun 24 '18 at 8:00
1
$begingroup$
I suppose that $h$ is defined and fixed in the question.
$endgroup$
– mathcounterexamples.net
Jun 24 '18 at 8:42
$begingroup$
In fact I have the feeling that my question is not so relevant... Is the condition on the unicity of $c_y$ not implying that $G$ is monotonic. Hence $y mapsto c_y$ is also monotonic and measurable?
$endgroup$
– mathcounterexamples.net
Jun 6 '15 at 9:39
$begingroup$
In fact I have the feeling that my question is not so relevant... Is the condition on the unicity of $c_y$ not implying that $G$ is monotonic. Hence $y mapsto c_y$ is also monotonic and measurable?
$endgroup$
– mathcounterexamples.net
Jun 6 '15 at 9:39
$begingroup$
Are you also assuming that $c_y$ does not depend on $h$ ?
$endgroup$
– Charles Madeline
Jun 24 '18 at 7:57
$begingroup$
Are you also assuming that $c_y$ does not depend on $h$ ?
$endgroup$
– Charles Madeline
Jun 24 '18 at 7:57
$begingroup$
I can think of a result which might be of some interest to you (as it seems close to the core of your problem): if $f$ is differentiable and for all $x<y$, the set of $zin (x,y)$ such that $f'(z)=frac{f(y)-f(x)}{y-x}$ is an interval, then $f$ or $-f$ is convex (thus $f'$ monotonic)
$endgroup$
– Charles Madeline
Jun 24 '18 at 8:00
$begingroup$
I can think of a result which might be of some interest to you (as it seems close to the core of your problem): if $f$ is differentiable and for all $x<y$, the set of $zin (x,y)$ such that $f'(z)=frac{f(y)-f(x)}{y-x}$ is an interval, then $f$ or $-f$ is convex (thus $f'$ monotonic)
$endgroup$
– Charles Madeline
Jun 24 '18 at 8:00
1
1
$begingroup$
I suppose that $h$ is defined and fixed in the question.
$endgroup$
– mathcounterexamples.net
Jun 24 '18 at 8:42
$begingroup$
I suppose that $h$ is defined and fixed in the question.
$endgroup$
– mathcounterexamples.net
Jun 24 '18 at 8:42
add a comment |
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$begingroup$
In fact I have the feeling that my question is not so relevant... Is the condition on the unicity of $c_y$ not implying that $G$ is monotonic. Hence $y mapsto c_y$ is also monotonic and measurable?
$endgroup$
– mathcounterexamples.net
Jun 6 '15 at 9:39
$begingroup$
Are you also assuming that $c_y$ does not depend on $h$ ?
$endgroup$
– Charles Madeline
Jun 24 '18 at 7:57
$begingroup$
I can think of a result which might be of some interest to you (as it seems close to the core of your problem): if $f$ is differentiable and for all $x<y$, the set of $zin (x,y)$ such that $f'(z)=frac{f(y)-f(x)}{y-x}$ is an interval, then $f$ or $-f$ is convex (thus $f'$ monotonic)
$endgroup$
– Charles Madeline
Jun 24 '18 at 8:00
1
$begingroup$
I suppose that $h$ is defined and fixed in the question.
$endgroup$
– mathcounterexamples.net
Jun 24 '18 at 8:42