About the Lebesgue points of a product of functions












1












$begingroup$


We say that a measurable function $g:mathbb R to mathbb R$ has a Lebesgue point at $x in mathbb R$ if



$$ frac1{2s}int_{x-s}^{x+s} |g(y)-g(x)| , dy to 0 quad text{as } s to 0.$$



If $g$ is continuous, then every $x in mathbb R$ is a Lebesgue point of $g$. The Lebesgue differentiation theorem states that a (locally) Lebesgue integrable function has Lebesgue points almost everywhere. The function given by



$$ g(x) = sum_{k=0}^infty (-1)^kchi_{(2^{-k-1},2^{-k}]} $$



is an example of a function that does not have a Lebesgue point at $0$.



Now suppose that $g: mathbb R to mathbb R$ and $h: mathbb R to mathbb R$ have Lebesgue points at every $x in mathbb R$. Is it possible that the product $gh$ has non-Lebesgue points?






real-analysis measure-theory lebesgue-integral







share|cite|improve this question









$endgroup$












  • $begingroup$
    Since $gh$ is not-necessary locally-integrable, the question does not make sense. (Take $h=g = sqrt{x}^{-1}$, then $h$ and $g$ are locally-integrable, but not $gh$.)
    $endgroup$
    – p4sch
    Nov 30 '18 at 13:29












  • $begingroup$
    @p4sch Note that $sqrt{x}^{-1}$ does not have a Lebesgue point at $0$.
    $endgroup$
    – cob
    Nov 30 '18 at 14:18










  • $begingroup$
    @mathworker21 If $g=h=0$, then $gh=0$ has a Lebesgue point at every $x in mathbb R$.
    $endgroup$
    – cob
    Nov 30 '18 at 14:21










  • $begingroup$
    @cob hm I thought you had said "must $gh$ have non-Lebesgue points"
    $endgroup$
    – mathworker21
    Nov 30 '18 at 14:38










  • $begingroup$
    If $(q_n)_{n in mathbb{N}}$ is an enumeration of $mathbb{Q}$, we can define $h(x) = g(x) = sum_{n=1}^infty 2^{-n} (max {sqrt{|q_n}|,1})^{-1} sqrt{|x-q_n|}^{-1}$. This function is locally-integrable, but $hcdot g$ is not locally-integrable in any point of $mathbb{R}$.
    $endgroup$
    – p4sch
    Nov 30 '18 at 15:03


















1












$begingroup$


We say that a measurable function $g:mathbb R to mathbb R$ has a Lebesgue point at $x in mathbb R$ if



$$ frac1{2s}int_{x-s}^{x+s} |g(y)-g(x)| , dy to 0 quad text{as } s to 0.$$



If $g$ is continuous, then every $x in mathbb R$ is a Lebesgue point of $g$. The Lebesgue differentiation theorem states that a (locally) Lebesgue integrable function has Lebesgue points almost everywhere. The function given by



$$ g(x) = sum_{k=0}^infty (-1)^kchi_{(2^{-k-1},2^{-k}]} $$



is an example of a function that does not have a Lebesgue point at $0$.



Now suppose that $g: mathbb R to mathbb R$ and $h: mathbb R to mathbb R$ have Lebesgue points at every $x in mathbb R$. Is it possible that the product $gh$ has non-Lebesgue points?






real-analysis measure-theory lebesgue-integral







share|cite|improve this question









$endgroup$












  • $begingroup$
    Since $gh$ is not-necessary locally-integrable, the question does not make sense. (Take $h=g = sqrt{x}^{-1}$, then $h$ and $g$ are locally-integrable, but not $gh$.)
    $endgroup$
    – p4sch
    Nov 30 '18 at 13:29












  • $begingroup$
    @p4sch Note that $sqrt{x}^{-1}$ does not have a Lebesgue point at $0$.
    $endgroup$
    – cob
    Nov 30 '18 at 14:18










  • $begingroup$
    @mathworker21 If $g=h=0$, then $gh=0$ has a Lebesgue point at every $x in mathbb R$.
    $endgroup$
    – cob
    Nov 30 '18 at 14:21










  • $begingroup$
    @cob hm I thought you had said "must $gh$ have non-Lebesgue points"
    $endgroup$
    – mathworker21
    Nov 30 '18 at 14:38










  • $begingroup$
    If $(q_n)_{n in mathbb{N}}$ is an enumeration of $mathbb{Q}$, we can define $h(x) = g(x) = sum_{n=1}^infty 2^{-n} (max {sqrt{|q_n}|,1})^{-1} sqrt{|x-q_n|}^{-1}$. This function is locally-integrable, but $hcdot g$ is not locally-integrable in any point of $mathbb{R}$.
    $endgroup$
    – p4sch
    Nov 30 '18 at 15:03
















1












1








1


1



$begingroup$


We say that a measurable function $g:mathbb R to mathbb R$ has a Lebesgue point at $x in mathbb R$ if



$$ frac1{2s}int_{x-s}^{x+s} |g(y)-g(x)| , dy to 0 quad text{as } s to 0.$$



If $g$ is continuous, then every $x in mathbb R$ is a Lebesgue point of $g$. The Lebesgue differentiation theorem states that a (locally) Lebesgue integrable function has Lebesgue points almost everywhere. The function given by



$$ g(x) = sum_{k=0}^infty (-1)^kchi_{(2^{-k-1},2^{-k}]} $$



is an example of a function that does not have a Lebesgue point at $0$.



Now suppose that $g: mathbb R to mathbb R$ and $h: mathbb R to mathbb R$ have Lebesgue points at every $x in mathbb R$. Is it possible that the product $gh$ has non-Lebesgue points?






real-analysis measure-theory lebesgue-integral







share|cite|improve this question









$endgroup$




We say that a measurable function $g:mathbb R to mathbb R$ has a Lebesgue point at $x in mathbb R$ if



$$ frac1{2s}int_{x-s}^{x+s} |g(y)-g(x)| , dy to 0 quad text{as } s to 0.$$



If $g$ is continuous, then every $x in mathbb R$ is a Lebesgue point of $g$. The Lebesgue differentiation theorem states that a (locally) Lebesgue integrable function has Lebesgue points almost everywhere. The function given by



$$ g(x) = sum_{k=0}^infty (-1)^kchi_{(2^{-k-1},2^{-k}]} $$



is an example of a function that does not have a Lebesgue point at $0$.



Now suppose that $g: mathbb R to mathbb R$ and $h: mathbb R to mathbb R$ have Lebesgue points at every $x in mathbb R$. Is it possible that the product $gh$ has non-Lebesgue points?






real-analysis measure-theory lebesgue-integral




real-analysis measure-theory lebesgue-integral






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 30 '18 at 13:03









cobcob

1063




1063












  • $begingroup$
    Since $gh$ is not-necessary locally-integrable, the question does not make sense. (Take $h=g = sqrt{x}^{-1}$, then $h$ and $g$ are locally-integrable, but not $gh$.)
    $endgroup$
    – p4sch
    Nov 30 '18 at 13:29












  • $begingroup$
    @p4sch Note that $sqrt{x}^{-1}$ does not have a Lebesgue point at $0$.
    $endgroup$
    – cob
    Nov 30 '18 at 14:18










  • $begingroup$
    @mathworker21 If $g=h=0$, then $gh=0$ has a Lebesgue point at every $x in mathbb R$.
    $endgroup$
    – cob
    Nov 30 '18 at 14:21










  • $begingroup$
    @cob hm I thought you had said "must $gh$ have non-Lebesgue points"
    $endgroup$
    – mathworker21
    Nov 30 '18 at 14:38










  • $begingroup$
    If $(q_n)_{n in mathbb{N}}$ is an enumeration of $mathbb{Q}$, we can define $h(x) = g(x) = sum_{n=1}^infty 2^{-n} (max {sqrt{|q_n}|,1})^{-1} sqrt{|x-q_n|}^{-1}$. This function is locally-integrable, but $hcdot g$ is not locally-integrable in any point of $mathbb{R}$.
    $endgroup$
    – p4sch
    Nov 30 '18 at 15:03




















  • $begingroup$
    Since $gh$ is not-necessary locally-integrable, the question does not make sense. (Take $h=g = sqrt{x}^{-1}$, then $h$ and $g$ are locally-integrable, but not $gh$.)
    $endgroup$
    – p4sch
    Nov 30 '18 at 13:29












  • $begingroup$
    @p4sch Note that $sqrt{x}^{-1}$ does not have a Lebesgue point at $0$.
    $endgroup$
    – cob
    Nov 30 '18 at 14:18










  • $begingroup$
    @mathworker21 If $g=h=0$, then $gh=0$ has a Lebesgue point at every $x in mathbb R$.
    $endgroup$
    – cob
    Nov 30 '18 at 14:21










  • $begingroup$
    @cob hm I thought you had said "must $gh$ have non-Lebesgue points"
    $endgroup$
    – mathworker21
    Nov 30 '18 at 14:38










  • $begingroup$
    If $(q_n)_{n in mathbb{N}}$ is an enumeration of $mathbb{Q}$, we can define $h(x) = g(x) = sum_{n=1}^infty 2^{-n} (max {sqrt{|q_n}|,1})^{-1} sqrt{|x-q_n|}^{-1}$. This function is locally-integrable, but $hcdot g$ is not locally-integrable in any point of $mathbb{R}$.
    $endgroup$
    – p4sch
    Nov 30 '18 at 15:03


















$begingroup$
Since $gh$ is not-necessary locally-integrable, the question does not make sense. (Take $h=g = sqrt{x}^{-1}$, then $h$ and $g$ are locally-integrable, but not $gh$.)
$endgroup$
– p4sch
Nov 30 '18 at 13:29






$begingroup$
Since $gh$ is not-necessary locally-integrable, the question does not make sense. (Take $h=g = sqrt{x}^{-1}$, then $h$ and $g$ are locally-integrable, but not $gh$.)
$endgroup$
– p4sch
Nov 30 '18 at 13:29














$begingroup$
@p4sch Note that $sqrt{x}^{-1}$ does not have a Lebesgue point at $0$.
$endgroup$
– cob
Nov 30 '18 at 14:18




$begingroup$
@p4sch Note that $sqrt{x}^{-1}$ does not have a Lebesgue point at $0$.
$endgroup$
– cob
Nov 30 '18 at 14:18












$begingroup$
@mathworker21 If $g=h=0$, then $gh=0$ has a Lebesgue point at every $x in mathbb R$.
$endgroup$
– cob
Nov 30 '18 at 14:21




$begingroup$
@mathworker21 If $g=h=0$, then $gh=0$ has a Lebesgue point at every $x in mathbb R$.
$endgroup$
– cob
Nov 30 '18 at 14:21












$begingroup$
@cob hm I thought you had said "must $gh$ have non-Lebesgue points"
$endgroup$
– mathworker21
Nov 30 '18 at 14:38




$begingroup$
@cob hm I thought you had said "must $gh$ have non-Lebesgue points"
$endgroup$
– mathworker21
Nov 30 '18 at 14:38












$begingroup$
If $(q_n)_{n in mathbb{N}}$ is an enumeration of $mathbb{Q}$, we can define $h(x) = g(x) = sum_{n=1}^infty 2^{-n} (max {sqrt{|q_n}|,1})^{-1} sqrt{|x-q_n|}^{-1}$. This function is locally-integrable, but $hcdot g$ is not locally-integrable in any point of $mathbb{R}$.
$endgroup$
– p4sch
Nov 30 '18 at 15:03






$begingroup$
If $(q_n)_{n in mathbb{N}}$ is an enumeration of $mathbb{Q}$, we can define $h(x) = g(x) = sum_{n=1}^infty 2^{-n} (max {sqrt{|q_n}|,1})^{-1} sqrt{|x-q_n|}^{-1}$. This function is locally-integrable, but $hcdot g$ is not locally-integrable in any point of $mathbb{R}$.
$endgroup$
– p4sch
Nov 30 '18 at 15:03












1 Answer
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Take $$g(x)=sum_{k=0}^infty 2^{k/2} chi_{[2^{-k}-4^{-k},2^{-k}]}.$$ It is easy to see that every $x neq 0$ is a Lebesgue point of $g$. At $x=0$ we have $$frac1{2^{-n+1}}int_{-2^{-n}}^{2^{-n}} |g(y)-g(0)| , dy = 2^{n-1}sum_{k=n}^infty 2^{-3k/2} = 2^{-n/2-1}sum_{k=0}^infty 2^{-3k/2}$$ which goes to $0$ as $n to infty$. This also implies $$frac1{2s}int_{-s}^{s} |g(y)-g(0)| , dy to 0 quad text{as $s to 0$}$$ and therefore $0$ is also a Lebesgue point of $g$.



However, $$g^2(x)=sum_{k=0}^infty 2^k chi_{[2^{-k}-4^{-k},2^{-k}]}$$ does not have a Lebesgue point at $0$, which can be seen by adapting the exponents in the above calculation. In fact, $g^2$ does not have a Lebesgue point at $0$ even after redefining $g^2(0)$. Also note that $g^2$ is integrable at $0$.






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    $begingroup$

    Take $$g(x)=sum_{k=0}^infty 2^{k/2} chi_{[2^{-k}-4^{-k},2^{-k}]}.$$ It is easy to see that every $x neq 0$ is a Lebesgue point of $g$. At $x=0$ we have $$frac1{2^{-n+1}}int_{-2^{-n}}^{2^{-n}} |g(y)-g(0)| , dy = 2^{n-1}sum_{k=n}^infty 2^{-3k/2} = 2^{-n/2-1}sum_{k=0}^infty 2^{-3k/2}$$ which goes to $0$ as $n to infty$. This also implies $$frac1{2s}int_{-s}^{s} |g(y)-g(0)| , dy to 0 quad text{as $s to 0$}$$ and therefore $0$ is also a Lebesgue point of $g$.



    However, $$g^2(x)=sum_{k=0}^infty 2^k chi_{[2^{-k}-4^{-k},2^{-k}]}$$ does not have a Lebesgue point at $0$, which can be seen by adapting the exponents in the above calculation. In fact, $g^2$ does not have a Lebesgue point at $0$ even after redefining $g^2(0)$. Also note that $g^2$ is integrable at $0$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Take $$g(x)=sum_{k=0}^infty 2^{k/2} chi_{[2^{-k}-4^{-k},2^{-k}]}.$$ It is easy to see that every $x neq 0$ is a Lebesgue point of $g$. At $x=0$ we have $$frac1{2^{-n+1}}int_{-2^{-n}}^{2^{-n}} |g(y)-g(0)| , dy = 2^{n-1}sum_{k=n}^infty 2^{-3k/2} = 2^{-n/2-1}sum_{k=0}^infty 2^{-3k/2}$$ which goes to $0$ as $n to infty$. This also implies $$frac1{2s}int_{-s}^{s} |g(y)-g(0)| , dy to 0 quad text{as $s to 0$}$$ and therefore $0$ is also a Lebesgue point of $g$.



      However, $$g^2(x)=sum_{k=0}^infty 2^k chi_{[2^{-k}-4^{-k},2^{-k}]}$$ does not have a Lebesgue point at $0$, which can be seen by adapting the exponents in the above calculation. In fact, $g^2$ does not have a Lebesgue point at $0$ even after redefining $g^2(0)$. Also note that $g^2$ is integrable at $0$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Take $$g(x)=sum_{k=0}^infty 2^{k/2} chi_{[2^{-k}-4^{-k},2^{-k}]}.$$ It is easy to see that every $x neq 0$ is a Lebesgue point of $g$. At $x=0$ we have $$frac1{2^{-n+1}}int_{-2^{-n}}^{2^{-n}} |g(y)-g(0)| , dy = 2^{n-1}sum_{k=n}^infty 2^{-3k/2} = 2^{-n/2-1}sum_{k=0}^infty 2^{-3k/2}$$ which goes to $0$ as $n to infty$. This also implies $$frac1{2s}int_{-s}^{s} |g(y)-g(0)| , dy to 0 quad text{as $s to 0$}$$ and therefore $0$ is also a Lebesgue point of $g$.



        However, $$g^2(x)=sum_{k=0}^infty 2^k chi_{[2^{-k}-4^{-k},2^{-k}]}$$ does not have a Lebesgue point at $0$, which can be seen by adapting the exponents in the above calculation. In fact, $g^2$ does not have a Lebesgue point at $0$ even after redefining $g^2(0)$. Also note that $g^2$ is integrable at $0$.






        share|cite|improve this answer









        $endgroup$



        Take $$g(x)=sum_{k=0}^infty 2^{k/2} chi_{[2^{-k}-4^{-k},2^{-k}]}.$$ It is easy to see that every $x neq 0$ is a Lebesgue point of $g$. At $x=0$ we have $$frac1{2^{-n+1}}int_{-2^{-n}}^{2^{-n}} |g(y)-g(0)| , dy = 2^{n-1}sum_{k=n}^infty 2^{-3k/2} = 2^{-n/2-1}sum_{k=0}^infty 2^{-3k/2}$$ which goes to $0$ as $n to infty$. This also implies $$frac1{2s}int_{-s}^{s} |g(y)-g(0)| , dy to 0 quad text{as $s to 0$}$$ and therefore $0$ is also a Lebesgue point of $g$.



        However, $$g^2(x)=sum_{k=0}^infty 2^k chi_{[2^{-k}-4^{-k},2^{-k}]}$$ does not have a Lebesgue point at $0$, which can be seen by adapting the exponents in the above calculation. In fact, $g^2$ does not have a Lebesgue point at $0$ even after redefining $g^2(0)$. Also note that $g^2$ is integrable at $0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 30 '18 at 17:47









        cobcob

        1063




        1063






























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