Mixing
Function almost always periodic for every period is constant
$begingroup$
Let be $(X, mathcal{A}, mu)$ a measure space and $f: X to [0, +infty]$ a measurable function such that for every $x in X$ we have $f(t + x) = f(t)$ for almost every $t in X$. Is true that $f$ is constant? or constant almost everywhere?
Thanks in advance.
measure-theory almost-everywhere measurable-functions
asked Jan 7 at 10:18
Marco All-in NervoMarco All-in Nervo
36118
$endgroup$
add a comment |
$begingroup$
Let be $(X, mathcal{A}, mu)$ a measure space and $f: X to [0, +infty]$ a measurable function such that for every $x in X$ we have $f(t + x) = f(t)$ for almost every $t in X$. Is true that $f$ is constant? or constant almost everywhere?
Thanks in advance.
measure-theory almost-everywhere measurable-functions
asked Jan 7 at 10:18
Marco All-in NervoMarco All-in Nervo
36118
$endgroup$
add a comment |
$begingroup$
Let be $(X, mathcal{A}, mu)$ a measure space and $f: X to [0, +infty]$ a measurable function such that for every $x in X$ we have $f(t + x) = f(t)$ for almost every $t in X$. Is true that $f$ is constant? or constant almost everywhere?
Thanks in advance.
measure-theory almost-everywhere measurable-functions
asked Jan 7 at 10:18
Marco All-in NervoMarco All-in Nervo
36118
$endgroup$
Let be $(X, mathcal{A}, mu)$ a measure space and $f: X to [0, +infty]$ a measurable function such that for every $x in X$ we have $f(t + x) = f(t)$ for almost every $t in X$. Is true that $f$ is constant? or constant almost everywhere?
Thanks in advance.
measure-theory almost-everywhere measurable-functions
measure-theory almost-everywhere measurable-functions
asked Jan 7 at 10:18
Marco All-in NervoMarco All-in Nervo
36118
asked Jan 7 at 10:18
Marco All-in NervoMarco All-in Nervo
36118
asked Jan 7 at 10:18
Marco All-in NervoMarco All-in Nervo
36118
asked Jan 7 at 10:18
Marco All-in NervoMarco All-in Nervo
36118
asked Jan 7 at 10:18
Marco All-in NervoMarco All-in Nervo
36118
36118
add a comment |
add a comment |
1 Answer
1
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Hints: as a simple consequence of Fubini's Theorem, for almost all $t$, the equation $f(x+t)=f(t)$ holds for almost all $x$. [Use the fact that $int int |f(x+t)-f(t)| dx dt =int int |f(x+t)-f(t)| dt dx$ ]. In particular this holds for one value of $t$ and hence $f$ is constant almost everywhere. Can you see that $f$ need not be constant? [Take characteristic function of a non-empty set of measure $0$]. Will be happy to provide a detailed answer if needed.
answered Jan 7 at 10:33
Kavi Rama MurthyKavi Rama Murthy
56.1k42158
$endgroup$
$begingroup$
Thanks! Really cool and helpful. Only the last step is not clear to me: we need that $mu$ is translation-invariant to conclude? Because I got $exists t_0$ such that for almost all x $f(x + t_0) = f(t_0)$ and I want to conclude that $exists t_0$ such that for almost all x $f(x) = f(t_0)$. Is translation-invariance needed?
$endgroup$
– Marco All-in Nervo
Jan 7 at 20:11
$begingroup$
(i.e that if we translate the null set it remains a null set)
$endgroup$
– Marco All-in Nervo
Jan 7 at 20:42
$begingroup$
@MarcoAll-inNervo Precisely! I am using translation invariance of Lebesgue measure in the last part.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 23:14
add a comment |
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1 Answer
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1 Answer
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active
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votes
$begingroup$
Hints: as a simple consequence of Fubini's Theorem, for almost all $t$, the equation $f(x+t)=f(t)$ holds for almost all $x$. [Use the fact that $int int |f(x+t)-f(t)| dx dt =int int |f(x+t)-f(t)| dt dx$ ]. In particular this holds for one value of $t$ and hence $f$ is constant almost everywhere. Can you see that $f$ need not be constant? [Take characteristic function of a non-empty set of measure $0$]. Will be happy to provide a detailed answer if needed.
answered Jan 7 at 10:33
Kavi Rama MurthyKavi Rama Murthy
56.1k42158
$endgroup$
$begingroup$
Thanks! Really cool and helpful. Only the last step is not clear to me: we need that $mu$ is translation-invariant to conclude? Because I got $exists t_0$ such that for almost all x $f(x + t_0) = f(t_0)$ and I want to conclude that $exists t_0$ such that for almost all x $f(x) = f(t_0)$. Is translation-invariance needed?
$endgroup$
– Marco All-in Nervo
Jan 7 at 20:11
$begingroup$
(i.e that if we translate the null set it remains a null set)
$endgroup$
– Marco All-in Nervo
Jan 7 at 20:42
$begingroup$
@MarcoAll-inNervo Precisely! I am using translation invariance of Lebesgue measure in the last part.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 23:14
add a comment |
$begingroup$
Hints: as a simple consequence of Fubini's Theorem, for almost all $t$, the equation $f(x+t)=f(t)$ holds for almost all $x$. [Use the fact that $int int |f(x+t)-f(t)| dx dt =int int |f(x+t)-f(t)| dt dx$ ]. In particular this holds for one value of $t$ and hence $f$ is constant almost everywhere. Can you see that $f$ need not be constant? [Take characteristic function of a non-empty set of measure $0$]. Will be happy to provide a detailed answer if needed.
answered Jan 7 at 10:33
Kavi Rama MurthyKavi Rama Murthy
56.1k42158
$endgroup$
$begingroup$
Thanks! Really cool and helpful. Only the last step is not clear to me: we need that $mu$ is translation-invariant to conclude? Because I got $exists t_0$ such that for almost all x $f(x + t_0) = f(t_0)$ and I want to conclude that $exists t_0$ such that for almost all x $f(x) = f(t_0)$. Is translation-invariance needed?
$endgroup$
– Marco All-in Nervo
Jan 7 at 20:11
$begingroup$
(i.e that if we translate the null set it remains a null set)
$endgroup$
– Marco All-in Nervo
Jan 7 at 20:42
$begingroup$
@MarcoAll-inNervo Precisely! I am using translation invariance of Lebesgue measure in the last part.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 23:14
add a comment |
$begingroup$
Hints: as a simple consequence of Fubini's Theorem, for almost all $t$, the equation $f(x+t)=f(t)$ holds for almost all $x$. [Use the fact that $int int |f(x+t)-f(t)| dx dt =int int |f(x+t)-f(t)| dt dx$ ]. In particular this holds for one value of $t$ and hence $f$ is constant almost everywhere. Can you see that $f$ need not be constant? [Take characteristic function of a non-empty set of measure $0$]. Will be happy to provide a detailed answer if needed.
answered Jan 7 at 10:33
Kavi Rama MurthyKavi Rama Murthy
56.1k42158
$endgroup$
Hints: as a simple consequence of Fubini's Theorem, for almost all $t$, the equation $f(x+t)=f(t)$ holds for almost all $x$. [Use the fact that $int int |f(x+t)-f(t)| dx dt =int int |f(x+t)-f(t)| dt dx$ ]. In particular this holds for one value of $t$ and hence $f$ is constant almost everywhere. Can you see that $f$ need not be constant? [Take characteristic function of a non-empty set of measure $0$]. Will be happy to provide a detailed answer if needed.
answered Jan 7 at 10:33
Kavi Rama MurthyKavi Rama Murthy
56.1k42158
edited Jan 7 at 11:52
answered Jan 7 at 10:33
Kavi Rama MurthyKavi Rama Murthy
56.1k42158
answered Jan 7 at 10:33
Kavi Rama MurthyKavi Rama Murthy
56.1k42158
answered Jan 7 at 10:33
Kavi Rama MurthyKavi Rama Murthy
56.1k42158
56.1k42158
$begingroup$
Thanks! Really cool and helpful. Only the last step is not clear to me: we need that $mu$ is translation-invariant to conclude? Because I got $exists t_0$ such that for almost all x $f(x + t_0) = f(t_0)$ and I want to conclude that $exists t_0$ such that for almost all x $f(x) = f(t_0)$. Is translation-invariance needed?
$endgroup$
– Marco All-in Nervo
Jan 7 at 20:11
$begingroup$
(i.e that if we translate the null set it remains a null set)
$endgroup$
– Marco All-in Nervo
Jan 7 at 20:42
$begingroup$
@MarcoAll-inNervo Precisely! I am using translation invariance of Lebesgue measure in the last part.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 23:14
add a comment |
$begingroup$
Thanks! Really cool and helpful. Only the last step is not clear to me: we need that $mu$ is translation-invariant to conclude? Because I got $exists t_0$ such that for almost all x $f(x + t_0) = f(t_0)$ and I want to conclude that $exists t_0$ such that for almost all x $f(x) = f(t_0)$. Is translation-invariance needed?
$endgroup$
– Marco All-in Nervo
Jan 7 at 20:11
$begingroup$
(i.e that if we translate the null set it remains a null set)
$endgroup$
– Marco All-in Nervo
Jan 7 at 20:42
$begingroup$
@MarcoAll-inNervo Precisely! I am using translation invariance of Lebesgue measure in the last part.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 23:14
$begingroup$
Thanks! Really cool and helpful. Only the last step is not clear to me: we need that $mu$ is translation-invariant to conclude? Because I got $exists t_0$ such that for almost all x $f(x + t_0) = f(t_0)$ and I want to conclude that $exists t_0$ such that for almost all x $f(x) = f(t_0)$. Is translation-invariance needed?
$endgroup$
– Marco All-in Nervo
Jan 7 at 20:11
$begingroup$
Thanks! Really cool and helpful. Only the last step is not clear to me: we need that $mu$ is translation-invariant to conclude? Because I got $exists t_0$ such that for almost all x $f(x + t_0) = f(t_0)$ and I want to conclude that $exists t_0$ such that for almost all x $f(x) = f(t_0)$. Is translation-invariance needed?
$endgroup$
– Marco All-in Nervo
Jan 7 at 20:11
$begingroup$
(i.e that if we translate the null set it remains a null set)
$endgroup$
– Marco All-in Nervo
Jan 7 at 20:42
$begingroup$
(i.e that if we translate the null set it remains a null set)
$endgroup$
– Marco All-in Nervo
Jan 7 at 20:42
$begingroup$
@MarcoAll-inNervo Precisely! I am using translation invariance of Lebesgue measure in the last part.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 23:14
$begingroup$
@MarcoAll-inNervo Precisely! I am using translation invariance of Lebesgue measure in the last part.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 23:14
add a comment |
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