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For what sets does the Lebesgue Diffferentiation 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?
integration open-problems lebesgue-measure
asked yesterday
Keshav Srinivasan
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up vote
7
down vote
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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?
integration open-problems lebesgue-measure
asked yesterday
Keshav Srinivasan
1,49311227
add a comment |
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?
integration open-problems lebesgue-measure
asked yesterday
Keshav Srinivasan
1,49311227
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?
integration open-problems lebesgue-measure
integration open-problems lebesgue-measure
asked yesterday
Keshav Srinivasan
1,49311227
asked yesterday
Keshav Srinivasan
1,49311227
asked yesterday
Keshav Srinivasan
1,49311227
asked yesterday
Keshav Srinivasan
1,49311227
asked yesterday
Keshav Srinivasan
1,49311227
1,49311227
add a comment |
add a comment |
2 Answers
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3
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It does not hold for rectangles, but holds for rectangles with prescribed directions of edges (that is, for the sets of the form $B=prod_{k=1}^n [a_k,b_k]$). 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.
answered 22 hours ago
Fedor Petrov
46.5k5109217
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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 $itoinfty$.
answered 22 hours ago
Liviu Nicolaescu
25k256108
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
It does not hold for rectangles, but holds for rectangles with prescribed directions of edges (that is, for the sets of the form $B=prod_{k=1}^n [a_k,b_k]$). 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.
answered 22 hours ago
Fedor Petrov
46.5k5109217
add a comment |
up vote
3
down vote
It does not hold for rectangles, but holds for rectangles with prescribed directions of edges (that is, for the sets of the form $B=prod_{k=1}^n [a_k,b_k]$). 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.
answered 22 hours ago
Fedor Petrov
46.5k5109217
add a comment |
up vote
3
down vote
up vote
3
down vote
It does not hold for rectangles, but holds for rectangles with prescribed directions of edges (that is, for the sets of the form $B=prod_{k=1}^n [a_k,b_k]$). 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.
answered 22 hours ago
Fedor Petrov
46.5k5109217
It does not hold for rectangles, but holds for rectangles with prescribed directions of edges (that is, for the sets of the form $B=prod_{k=1}^n [a_k,b_k]$). 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.
answered 22 hours ago
Fedor Petrov
46.5k5109217
answered 22 hours ago
Fedor Petrov
46.5k5109217
answered 22 hours ago
Fedor Petrov
46.5k5109217
answered 22 hours ago
Fedor Petrov
46.5k5109217
46.5k5109217
add a comment |
add a comment |
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 $itoinfty$.
answered 22 hours ago
Liviu Nicolaescu
25k256108
add a comment |
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 $itoinfty$.
answered 22 hours ago
Liviu Nicolaescu
25k256108
add a comment |
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 $itoinfty$.
answered 22 hours ago
Liviu Nicolaescu
25k256108
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 $itoinfty$.
answered 22 hours ago
Liviu Nicolaescu
25k256108
edited 20 hours ago
answered 22 hours ago
Liviu Nicolaescu
25k256108
answered 22 hours ago
Liviu Nicolaescu
25k256108
answered 22 hours ago
Liviu Nicolaescu
25k256108
25k256108
add a comment |
add a comment |
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