Every section null implies null in product measure
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Let $Asubset[0,1]^2$ be a set such that every section $A_x={y:(x,y)in A}$ is a null set in $[0,1]$. Can we conclude that $A$ is a null set in $[0,1]^2$?
Some context: It is a standard fact that if $A$ is measurable in the product measure, then each section $A_x$ is measurable. In the converse direction, it seems that each $A_x$ measurable does not imply that $A$ is measurable (even in the 2D Lebesgue sense). The question above is about what happens if we replace measurable by null.
lebesgue-measure product-measure
asked Nov 27 at 16:42
timur
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Let $Asubset[0,1]^2$ be a set such that every section $A_x={y:(x,y)in A}$ is a null set in $[0,1]$. Can we conclude that $A$ is a null set in $[0,1]^2$?
Some context: It is a standard fact that if $A$ is measurable in the product measure, then each section $A_x$ is measurable. In the converse direction, it seems that each $A_x$ measurable does not imply that $A$ is measurable (even in the 2D Lebesgue sense). The question above is about what happens if we replace measurable by null.
lebesgue-measure product-measure
asked Nov 27 at 16:42
timur
11.5k1943
@bof: You are right, if we take $A$ to be Lebesgue measurable in $mathbb{R}^2$. On the other hand, if we consider the product of 1-dimensional Lebesgue measures in $mathbb{R}^2$, then every section would be measurable.
– timur
6 hours ago
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up vote
0
down vote
favorite
Let $Asubset[0,1]^2$ be a set such that every section $A_x={y:(x,y)in A}$ is a null set in $[0,1]$. Can we conclude that $A$ is a null set in $[0,1]^2$?
Some context: It is a standard fact that if $A$ is measurable in the product measure, then each section $A_x$ is measurable. In the converse direction, it seems that each $A_x$ measurable does not imply that $A$ is measurable (even in the 2D Lebesgue sense). The question above is about what happens if we replace measurable by null.
lebesgue-measure product-measure
asked Nov 27 at 16:42
timur
11.5k1943
Let $Asubset[0,1]^2$ be a set such that every section $A_x={y:(x,y)in A}$ is a null set in $[0,1]$. Can we conclude that $A$ is a null set in $[0,1]^2$?
Some context: It is a standard fact that if $A$ is measurable in the product measure, then each section $A_x$ is measurable. In the converse direction, it seems that each $A_x$ measurable does not imply that $A$ is measurable (even in the 2D Lebesgue sense). The question above is about what happens if we replace measurable by null.
lebesgue-measure product-measure
lebesgue-measure product-measure
asked Nov 27 at 16:42
timur
11.5k1943
asked Nov 27 at 16:42
timur
11.5k1943
edited 6 hours ago
asked Nov 27 at 16:42
timur
11.5k1943
asked Nov 27 at 16:42
timur
11.5k1943
asked Nov 27 at 16:42
timur
11.5k1943
11.5k1943
@bof: You are right, if we take $A$ to be Lebesgue measurable in $mathbb{R}^2$. On the other hand, if we consider the product of 1-dimensional Lebesgue measures in $mathbb{R}^2$, then every section would be measurable.
– timur
6 hours ago
add a comment |
@bof: You are right, if we take $A$ to be Lebesgue measurable in $mathbb{R}^2$. On the other hand, if we consider the product of 1-dimensional Lebesgue measures in $mathbb{R}^2$, then every section would be measurable.
– timur
6 hours ago
@bof: You are right, if we take $A$ to be Lebesgue measurable in $mathbb{R}^2$. On the other hand, if we consider the product of 1-dimensional Lebesgue measures in $mathbb{R}^2$, then every section would be measurable.
– timur
6 hours ago
@bof: You are right, if we take $A$ to be Lebesgue measurable in $mathbb{R}^2$. On the other hand, if we consider the product of 1-dimensional Lebesgue measures in $mathbb{R}^2$, then every section would be measurable.
– timur
6 hours ago
add a comment |
1 Answer
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Assuming the continuum hypothesis, $mathbb R^2$ can be partitioned into two (nonmeasurable) sets, one with all vertical sections countable, the other with all horizontal sections countable. If you just want the sections to be Lebesgue null sets instead of countable sets, the continuum hypothesis can be replaced by the weaker assumption, that every set of real numbers of cardinality less than $2^{aleph_0}$ is a Lebesgue null set.
This has nothing to do with measure theory, it is just basic set theory: given a well-ordering
of $mathbb R$, we can define a partition of $mathbb R^2$ into two sets $A$ and $B$ such that every vertical section of $A$, and every horizontal section of $B$, has cardinality less than the continuum.
answered 21 hours ago
bof
48.9k451116
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Assuming the continuum hypothesis, $mathbb R^2$ can be partitioned into two (nonmeasurable) sets, one with all vertical sections countable, the other with all horizontal sections countable. If you just want the sections to be Lebesgue null sets instead of countable sets, the continuum hypothesis can be replaced by the weaker assumption, that every set of real numbers of cardinality less than $2^{aleph_0}$ is a Lebesgue null set.
This has nothing to do with measure theory, it is just basic set theory: given a well-ordering
of $mathbb R$, we can define a partition of $mathbb R^2$ into two sets $A$ and $B$ such that every vertical section of $A$, and every horizontal section of $B$, has cardinality less than the continuum.
answered 21 hours ago
bof
48.9k451116
add a comment |
up vote
1
down vote
accepted
Assuming the continuum hypothesis, $mathbb R^2$ can be partitioned into two (nonmeasurable) sets, one with all vertical sections countable, the other with all horizontal sections countable. If you just want the sections to be Lebesgue null sets instead of countable sets, the continuum hypothesis can be replaced by the weaker assumption, that every set of real numbers of cardinality less than $2^{aleph_0}$ is a Lebesgue null set.
This has nothing to do with measure theory, it is just basic set theory: given a well-ordering
of $mathbb R$, we can define a partition of $mathbb R^2$ into two sets $A$ and $B$ such that every vertical section of $A$, and every horizontal section of $B$, has cardinality less than the continuum.
answered 21 hours ago
bof
48.9k451116
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Assuming the continuum hypothesis, $mathbb R^2$ can be partitioned into two (nonmeasurable) sets, one with all vertical sections countable, the other with all horizontal sections countable. If you just want the sections to be Lebesgue null sets instead of countable sets, the continuum hypothesis can be replaced by the weaker assumption, that every set of real numbers of cardinality less than $2^{aleph_0}$ is a Lebesgue null set.
This has nothing to do with measure theory, it is just basic set theory: given a well-ordering
of $mathbb R$, we can define a partition of $mathbb R^2$ into two sets $A$ and $B$ such that every vertical section of $A$, and every horizontal section of $B$, has cardinality less than the continuum.
answered 21 hours ago
bof
48.9k451116
Assuming the continuum hypothesis, $mathbb R^2$ can be partitioned into two (nonmeasurable) sets, one with all vertical sections countable, the other with all horizontal sections countable. If you just want the sections to be Lebesgue null sets instead of countable sets, the continuum hypothesis can be replaced by the weaker assumption, that every set of real numbers of cardinality less than $2^{aleph_0}$ is a Lebesgue null set.
This has nothing to do with measure theory, it is just basic set theory: given a well-ordering
of $mathbb R$, we can define a partition of $mathbb R^2$ into two sets $A$ and $B$ such that every vertical section of $A$, and every horizontal section of $B$, has cardinality less than the continuum.
answered 21 hours ago
bof
48.9k451116
edited 6 hours ago
answered 21 hours ago
bof
48.9k451116
answered 21 hours ago
bof
48.9k451116
answered 21 hours ago
bof
48.9k451116
48.9k451116
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@bof: You are right, if we take $A$ to be Lebesgue measurable in $mathbb{R}^2$. On the other hand, if we consider the product of 1-dimensional Lebesgue measures in $mathbb{R}^2$, then every section would be measurable.
– timur
6 hours ago