For what sets does the Lebesgue Differentiation Theorem hold in one dimension?











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Lebesgue's differentiation theorem states that if $x$ is a point in $mathbb{R}^n$ and $f:mathbb{R}^nrightarrowmathbb{R}$ is a Lebesgue integrable function, then the limit of $frac{int_B f dlambda}{lambda(B)}$ over all balls $B$ centered at $x$ as the diameter of $B$ goes to $0$ is equal almost everywhere to $f(x)$. But if you replace balls with other kinds of set with diameter going to $0$, this need not be true. For instance it need not be true if you replace balls with rectangles.



But my question is, what is known about what collections of sets the Lebesgue Differentiation Theorem holds for in one dimension? We know it holds for bounded intervals. Does it hold for finite unions of bounded open intervals? What about bounded Borel sets? What about bounded Lebesgue measurable sets in general?



Or are all these open problems?










share|cite|improve this question




























    up vote
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    down vote

    favorite












    Lebesgue's differentiation theorem states that if $x$ is a point in $mathbb{R}^n$ and $f:mathbb{R}^nrightarrowmathbb{R}$ is a Lebesgue integrable function, then the limit of $frac{int_B f dlambda}{lambda(B)}$ over all balls $B$ centered at $x$ as the diameter of $B$ goes to $0$ is equal almost everywhere to $f(x)$. But if you replace balls with other kinds of set with diameter going to $0$, this need not be true. For instance it need not be true if you replace balls with rectangles.



    But my question is, what is known about what collections of sets the Lebesgue Differentiation Theorem holds for in one dimension? We know it holds for bounded intervals. Does it hold for finite unions of bounded open intervals? What about bounded Borel sets? What about bounded Lebesgue measurable sets in general?



    Or are all these open problems?










    share|cite|improve this question


























      up vote
      7
      down vote

      favorite









      up vote
      7
      down vote

      favorite











      Lebesgue's differentiation theorem states that if $x$ is a point in $mathbb{R}^n$ and $f:mathbb{R}^nrightarrowmathbb{R}$ is a Lebesgue integrable function, then the limit of $frac{int_B f dlambda}{lambda(B)}$ over all balls $B$ centered at $x$ as the diameter of $B$ goes to $0$ is equal almost everywhere to $f(x)$. But if you replace balls with other kinds of set with diameter going to $0$, this need not be true. For instance it need not be true if you replace balls with rectangles.



      But my question is, what is known about what collections of sets the Lebesgue Differentiation Theorem holds for in one dimension? We know it holds for bounded intervals. Does it hold for finite unions of bounded open intervals? What about bounded Borel sets? What about bounded Lebesgue measurable sets in general?



      Or are all these open problems?










      share|cite|improve this question















      Lebesgue's differentiation theorem states that if $x$ is a point in $mathbb{R}^n$ and $f:mathbb{R}^nrightarrowmathbb{R}$ is a Lebesgue integrable function, then the limit of $frac{int_B f dlambda}{lambda(B)}$ over all balls $B$ centered at $x$ as the diameter of $B$ goes to $0$ is equal almost everywhere to $f(x)$. But if you replace balls with other kinds of set with diameter going to $0$, this need not be true. For instance it need not be true if you replace balls with rectangles.



      But my question is, what is known about what collections of sets the Lebesgue Differentiation Theorem holds for in one dimension? We know it holds for bounded intervals. Does it hold for finite unions of bounded open intervals? What about bounded Borel sets? What about bounded Lebesgue measurable sets in general?



      Or are all these open problems?







      real-analysis measure-theory integration open-problems lebesgue-measure






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      share|cite|improve this question













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      share|cite|improve this question








      edited Nov 24 at 8:07









      YCor

      27k380132




      27k380132










      asked Nov 19 at 7:39









      Keshav Srinivasan

      1,52811228




      1,52811228






















          3 Answers
          3






          active

          oldest

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          up vote
          4
          down vote













          For bounded unions of intervals it of course does not hold: take two disjoint mesaurable subsets $A,B$ of $mathbb{R}$ such that $lambda(Acap E)$ and $lambda (Bcap E)$ are positive for any interval $E$. Then for $ain A$ any interval $delta ni a$ contains a density point $bin B$ and you may take the union of an interval $delta_1subset delta$ consisting mostly of points of $B$ and define $B=delta_1 cup (a-lambda(delta_1)/5,a+lambda(delta_1)/5)$. This set is a small union of at most two intervals, contains $a$, but contains many points from $B$. So for $f=chi_A$ the Lebesgue ratio is far from 1.






          share|cite|improve this answer






























            up vote
            3
            down vote













            Perhaps Doob's martingale convergence theorem can help. Suppose that $newcommand{bR}{mathbb{R}}$ $Csubset bR^n$ is a cube and $fin L^1 (C)$. Suppose that $newcommand{eP}{mathscr{P}}$
            $$
            eP_1prec eP_2prec cdots
            $$

            is a sequence of finite partitions of $C$ into Borel subsets such that for any $BineP_{i+1}$ there exists $B'in eP_i$, $B'supset B$. Assume that the $sigma$-algebra generated by
            $$
            bigcup_{igeq 1}eP_i
            $$

            coincides with the Borel $sigma$-algebra of $C$. For example this happens if for any open subset $Usubset bR^n$ and any $xin Ccap U$ there exists $igeq 1$ and $B_iineP_i$ such that
            $$
            xin B_isubset Ucap C.
            $$

            For any $xin C$ there exists a unique sequence of Borel sets $B_i(x)ineP_i$ such that $B_i(x)ni x$, $forall i$. Then Doob's martingale convergence theorem implies that for almost any $xin C$ we have
            $$
            f(x) =lim_{ito infty}frac{1}{{rm vol};(B_i(x))}int_{B_i(x)} f(y)dy.
            $$

            A bit more is true. If we define
            $$
            f_i(x)=sum_{Bin eP_i}frac{1}{{rm vol};(B)}left(int_B f(y) dyright) I_B(x),
            $$

            where $I_B$ denotes the indicator functio of the subset $B$, then $f_ito f$ in $L^1$ as $itoinfty$.






            share|cite|improve this answer






























              up vote
              0
              down vote













              The classic reference on differeniation theorems (and, yes, martingales as Liviu mentions):



              Hayes, C. A.; Pauc, C. Y., Derivation and martingales, Ergebnisse der Mathematik und ihrer Grenzgebiete. 49. Berlin-Heidelberg-New York: Springer-Verlag. VII, 203 p. (1970). ZBL0192.40604.



              a plug

              Some more modern results may be found in



              G. A. Edgar & L. Sucheston, Stopping Times and Directed Processes, Encyclopedia of Mathematics and Its Applications 47, Cambridge University Press, 1992






              share|cite|improve this answer























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                3 Answers
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                active

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                up vote
                4
                down vote













                For bounded unions of intervals it of course does not hold: take two disjoint mesaurable subsets $A,B$ of $mathbb{R}$ such that $lambda(Acap E)$ and $lambda (Bcap E)$ are positive for any interval $E$. Then for $ain A$ any interval $delta ni a$ contains a density point $bin B$ and you may take the union of an interval $delta_1subset delta$ consisting mostly of points of $B$ and define $B=delta_1 cup (a-lambda(delta_1)/5,a+lambda(delta_1)/5)$. This set is a small union of at most two intervals, contains $a$, but contains many points from $B$. So for $f=chi_A$ the Lebesgue ratio is far from 1.






                share|cite|improve this answer



























                  up vote
                  4
                  down vote













                  For bounded unions of intervals it of course does not hold: take two disjoint mesaurable subsets $A,B$ of $mathbb{R}$ such that $lambda(Acap E)$ and $lambda (Bcap E)$ are positive for any interval $E$. Then for $ain A$ any interval $delta ni a$ contains a density point $bin B$ and you may take the union of an interval $delta_1subset delta$ consisting mostly of points of $B$ and define $B=delta_1 cup (a-lambda(delta_1)/5,a+lambda(delta_1)/5)$. This set is a small union of at most two intervals, contains $a$, but contains many points from $B$. So for $f=chi_A$ the Lebesgue ratio is far from 1.






                  share|cite|improve this answer

























                    up vote
                    4
                    down vote










                    up vote
                    4
                    down vote









                    For bounded unions of intervals it of course does not hold: take two disjoint mesaurable subsets $A,B$ of $mathbb{R}$ such that $lambda(Acap E)$ and $lambda (Bcap E)$ are positive for any interval $E$. Then for $ain A$ any interval $delta ni a$ contains a density point $bin B$ and you may take the union of an interval $delta_1subset delta$ consisting mostly of points of $B$ and define $B=delta_1 cup (a-lambda(delta_1)/5,a+lambda(delta_1)/5)$. This set is a small union of at most two intervals, contains $a$, but contains many points from $B$. So for $f=chi_A$ the Lebesgue ratio is far from 1.






                    share|cite|improve this answer














                    For bounded unions of intervals it of course does not hold: take two disjoint mesaurable subsets $A,B$ of $mathbb{R}$ such that $lambda(Acap E)$ and $lambda (Bcap E)$ are positive for any interval $E$. Then for $ain A$ any interval $delta ni a$ contains a density point $bin B$ and you may take the union of an interval $delta_1subset delta$ consisting mostly of points of $B$ and define $B=delta_1 cup (a-lambda(delta_1)/5,a+lambda(delta_1)/5)$. This set is a small union of at most two intervals, contains $a$, but contains many points from $B$. So for $f=chi_A$ the Lebesgue ratio is far from 1.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 23 at 23:38

























                    answered Nov 19 at 10:58









                    Fedor Petrov

                    46.7k5110217




                    46.7k5110217






















                        up vote
                        3
                        down vote













                        Perhaps Doob's martingale convergence theorem can help. Suppose that $newcommand{bR}{mathbb{R}}$ $Csubset bR^n$ is a cube and $fin L^1 (C)$. Suppose that $newcommand{eP}{mathscr{P}}$
                        $$
                        eP_1prec eP_2prec cdots
                        $$

                        is a sequence of finite partitions of $C$ into Borel subsets such that for any $BineP_{i+1}$ there exists $B'in eP_i$, $B'supset B$. Assume that the $sigma$-algebra generated by
                        $$
                        bigcup_{igeq 1}eP_i
                        $$

                        coincides with the Borel $sigma$-algebra of $C$. For example this happens if for any open subset $Usubset bR^n$ and any $xin Ccap U$ there exists $igeq 1$ and $B_iineP_i$ such that
                        $$
                        xin B_isubset Ucap C.
                        $$

                        For any $xin C$ there exists a unique sequence of Borel sets $B_i(x)ineP_i$ such that $B_i(x)ni x$, $forall i$. Then Doob's martingale convergence theorem implies that for almost any $xin C$ we have
                        $$
                        f(x) =lim_{ito infty}frac{1}{{rm vol};(B_i(x))}int_{B_i(x)} f(y)dy.
                        $$

                        A bit more is true. If we define
                        $$
                        f_i(x)=sum_{Bin eP_i}frac{1}{{rm vol};(B)}left(int_B f(y) dyright) I_B(x),
                        $$

                        where $I_B$ denotes the indicator functio of the subset $B$, then $f_ito f$ in $L^1$ as $itoinfty$.






                        share|cite|improve this answer



























                          up vote
                          3
                          down vote













                          Perhaps Doob's martingale convergence theorem can help. Suppose that $newcommand{bR}{mathbb{R}}$ $Csubset bR^n$ is a cube and $fin L^1 (C)$. Suppose that $newcommand{eP}{mathscr{P}}$
                          $$
                          eP_1prec eP_2prec cdots
                          $$

                          is a sequence of finite partitions of $C$ into Borel subsets such that for any $BineP_{i+1}$ there exists $B'in eP_i$, $B'supset B$. Assume that the $sigma$-algebra generated by
                          $$
                          bigcup_{igeq 1}eP_i
                          $$

                          coincides with the Borel $sigma$-algebra of $C$. For example this happens if for any open subset $Usubset bR^n$ and any $xin Ccap U$ there exists $igeq 1$ and $B_iineP_i$ such that
                          $$
                          xin B_isubset Ucap C.
                          $$

                          For any $xin C$ there exists a unique sequence of Borel sets $B_i(x)ineP_i$ such that $B_i(x)ni x$, $forall i$. Then Doob's martingale convergence theorem implies that for almost any $xin C$ we have
                          $$
                          f(x) =lim_{ito infty}frac{1}{{rm vol};(B_i(x))}int_{B_i(x)} f(y)dy.
                          $$

                          A bit more is true. If we define
                          $$
                          f_i(x)=sum_{Bin eP_i}frac{1}{{rm vol};(B)}left(int_B f(y) dyright) I_B(x),
                          $$

                          where $I_B$ denotes the indicator functio of the subset $B$, then $f_ito f$ in $L^1$ as $itoinfty$.






                          share|cite|improve this answer

























                            up vote
                            3
                            down vote










                            up vote
                            3
                            down vote









                            Perhaps Doob's martingale convergence theorem can help. Suppose that $newcommand{bR}{mathbb{R}}$ $Csubset bR^n$ is a cube and $fin L^1 (C)$. Suppose that $newcommand{eP}{mathscr{P}}$
                            $$
                            eP_1prec eP_2prec cdots
                            $$

                            is a sequence of finite partitions of $C$ into Borel subsets such that for any $BineP_{i+1}$ there exists $B'in eP_i$, $B'supset B$. Assume that the $sigma$-algebra generated by
                            $$
                            bigcup_{igeq 1}eP_i
                            $$

                            coincides with the Borel $sigma$-algebra of $C$. For example this happens if for any open subset $Usubset bR^n$ and any $xin Ccap U$ there exists $igeq 1$ and $B_iineP_i$ such that
                            $$
                            xin B_isubset Ucap C.
                            $$

                            For any $xin C$ there exists a unique sequence of Borel sets $B_i(x)ineP_i$ such that $B_i(x)ni x$, $forall i$. Then Doob's martingale convergence theorem implies that for almost any $xin C$ we have
                            $$
                            f(x) =lim_{ito infty}frac{1}{{rm vol};(B_i(x))}int_{B_i(x)} f(y)dy.
                            $$

                            A bit more is true. If we define
                            $$
                            f_i(x)=sum_{Bin eP_i}frac{1}{{rm vol};(B)}left(int_B f(y) dyright) I_B(x),
                            $$

                            where $I_B$ denotes the indicator functio of the subset $B$, then $f_ito f$ in $L^1$ as $itoinfty$.






                            share|cite|improve this answer














                            Perhaps Doob's martingale convergence theorem can help. Suppose that $newcommand{bR}{mathbb{R}}$ $Csubset bR^n$ is a cube and $fin L^1 (C)$. Suppose that $newcommand{eP}{mathscr{P}}$
                            $$
                            eP_1prec eP_2prec cdots
                            $$

                            is a sequence of finite partitions of $C$ into Borel subsets such that for any $BineP_{i+1}$ there exists $B'in eP_i$, $B'supset B$. Assume that the $sigma$-algebra generated by
                            $$
                            bigcup_{igeq 1}eP_i
                            $$

                            coincides with the Borel $sigma$-algebra of $C$. For example this happens if for any open subset $Usubset bR^n$ and any $xin Ccap U$ there exists $igeq 1$ and $B_iineP_i$ such that
                            $$
                            xin B_isubset Ucap C.
                            $$

                            For any $xin C$ there exists a unique sequence of Borel sets $B_i(x)ineP_i$ such that $B_i(x)ni x$, $forall i$. Then Doob's martingale convergence theorem implies that for almost any $xin C$ we have
                            $$
                            f(x) =lim_{ito infty}frac{1}{{rm vol};(B_i(x))}int_{B_i(x)} f(y)dy.
                            $$

                            A bit more is true. If we define
                            $$
                            f_i(x)=sum_{Bin eP_i}frac{1}{{rm vol};(B)}left(int_B f(y) dyright) I_B(x),
                            $$

                            where $I_B$ denotes the indicator functio of the subset $B$, then $f_ito f$ in $L^1$ as $itoinfty$.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Nov 19 at 13:07

























                            answered Nov 19 at 10:24









                            Liviu Nicolaescu

                            25.1k256109




                            25.1k256109






















                                up vote
                                0
                                down vote













                                The classic reference on differeniation theorems (and, yes, martingales as Liviu mentions):



                                Hayes, C. A.; Pauc, C. Y., Derivation and martingales, Ergebnisse der Mathematik und ihrer Grenzgebiete. 49. Berlin-Heidelberg-New York: Springer-Verlag. VII, 203 p. (1970). ZBL0192.40604.



                                a plug

                                Some more modern results may be found in



                                G. A. Edgar & L. Sucheston, Stopping Times and Directed Processes, Encyclopedia of Mathematics and Its Applications 47, Cambridge University Press, 1992






                                share|cite|improve this answer



























                                  up vote
                                  0
                                  down vote













                                  The classic reference on differeniation theorems (and, yes, martingales as Liviu mentions):



                                  Hayes, C. A.; Pauc, C. Y., Derivation and martingales, Ergebnisse der Mathematik und ihrer Grenzgebiete. 49. Berlin-Heidelberg-New York: Springer-Verlag. VII, 203 p. (1970). ZBL0192.40604.



                                  a plug

                                  Some more modern results may be found in



                                  G. A. Edgar & L. Sucheston, Stopping Times and Directed Processes, Encyclopedia of Mathematics and Its Applications 47, Cambridge University Press, 1992






                                  share|cite|improve this answer

























                                    up vote
                                    0
                                    down vote










                                    up vote
                                    0
                                    down vote









                                    The classic reference on differeniation theorems (and, yes, martingales as Liviu mentions):



                                    Hayes, C. A.; Pauc, C. Y., Derivation and martingales, Ergebnisse der Mathematik und ihrer Grenzgebiete. 49. Berlin-Heidelberg-New York: Springer-Verlag. VII, 203 p. (1970). ZBL0192.40604.



                                    a plug

                                    Some more modern results may be found in



                                    G. A. Edgar & L. Sucheston, Stopping Times and Directed Processes, Encyclopedia of Mathematics and Its Applications 47, Cambridge University Press, 1992






                                    share|cite|improve this answer














                                    The classic reference on differeniation theorems (and, yes, martingales as Liviu mentions):



                                    Hayes, C. A.; Pauc, C. Y., Derivation and martingales, Ergebnisse der Mathematik und ihrer Grenzgebiete. 49. Berlin-Heidelberg-New York: Springer-Verlag. VII, 203 p. (1970). ZBL0192.40604.



                                    a plug

                                    Some more modern results may be found in



                                    G. A. Edgar & L. Sucheston, Stopping Times and Directed Processes, Encyclopedia of Mathematics and Its Applications 47, Cambridge University Press, 1992







                                    share|cite|improve this answer














                                    share|cite|improve this answer



                                    share|cite|improve this answer








                                    edited Nov 24 at 12:38

























                                    answered Nov 24 at 11:47









                                    Gerald Edgar

                                    27.7k374154




                                    27.7k374154






























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