If $A, B in [m,n] subseteq mathbb{R}$ are sets of measure zero, then is $AB = {ab | a in A, b in B}$ measure...
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If $A, B in [m,n]$, where $[m,n] subseteq mathbb{R}$ are sets of measure zero, then is $AB = {ab ; | ; a in A, b in B}$ measure zero?
real-analysis analysis measure-theory lebesgue-measure outer-measure
asked Nov 23 at 8:24
Ralph Xu
566
closed as off-topic by Brahadeesh, supinf, Christopher, Gibbs, Eckhard Nov 23 at 21:00
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If $A, B in [m,n]$, where $[m,n] subseteq mathbb{R}$ are sets of measure zero, then is $AB = {ab ; | ; a in A, b in B}$ measure zero?
real-analysis analysis measure-theory lebesgue-measure outer-measure
asked Nov 23 at 8:24
Ralph Xu
566
closed as off-topic by Brahadeesh, supinf, Christopher, Gibbs, Eckhard Nov 23 at 21:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, supinf, Christopher, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If $A, B in [m,n]$, where $[m,n] subseteq mathbb{R}$ are sets of measure zero, then is $AB = {ab ; | ; a in A, b in B}$ measure zero?
real-analysis analysis measure-theory lebesgue-measure outer-measure
asked Nov 23 at 8:24
Ralph Xu
566
If $A, B in [m,n]$, where $[m,n] subseteq mathbb{R}$ are sets of measure zero, then is $AB = {ab ; | ; a in A, b in B}$ measure zero?
real-analysis analysis measure-theory lebesgue-measure outer-measure
real-analysis analysis measure-theory lebesgue-measure outer-measure
asked Nov 23 at 8:24
Ralph Xu
566
asked Nov 23 at 8:24
Ralph Xu
566
asked Nov 23 at 8:24
Ralph Xu
566
asked Nov 23 at 8:24
Ralph Xu
566
asked Nov 23 at 8:24
Ralph Xu
566
566
closed as off-topic by Brahadeesh, supinf, Christopher, Gibbs, Eckhard Nov 23 at 21:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, supinf, Christopher, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Brahadeesh, supinf, Christopher, Gibbs, Eckhard Nov 23 at 21:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, supinf, Christopher, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
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1 Answer
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If $C$ is the Cantor set and $A=B={e^{x}:xin C}$ then $A$ and $B$ have measure $0$ but $AB$ contains all numbers of the form $e^{x+y}$ with $x,y in C$. Since $C+C=[0,2]$ it follows that $AB$ does not have measure $0$. [I have used the following facts: $e^{x}$ is absolutely continuous on $[0,1]$ and any absolutely continuous function maps sets of measure $0$ to sets of measure $0$. Hence $A$ and $B$ have measure $0$]. Note also that $AB$ contains the entire interval $[e^{0},e^{2}]$ so it has positive measure.
answered Nov 23 at 8:37
Kavi Rama Murthy
43k31751
I followed your argument, but I was thinking about the following, but I'm not able to figure out where went wrong: Let $E = A cup B$. Then $AB subseteq E^2$. We know $E$ has measure zero. Since the function $f(x) = x^2$ is Lipschitz, it maps sets of measure zero to sets of measure zero, so $E^2$ has measure zero, which means its subset $AB$ is also measure zero.
– Ralph Xu
Nov 23 at 8:58
$AB$ is the image of $C+C=[0,1]$ under the absolutely continuous function $e^x$. We know that every absolutely continuous function sends a set of measure zero to a set of measure zero. But can we say that it sends a set of positive measure to a set of positive measure?
– Anupam
Nov 23 at 9:03
@Anupam No you cannot say that any absolutely continuous function maps a set of positive measure to a set of positive measure. (An obvious counterexample is given by a constant function). But in my argument $AB$ contains all the numbers from $e^{0}$ to $e^{2}$ so it has positive measure.
– Kavi Rama Murthy
Nov 23 at 9:12
Ok. Sorry, I got your argument. May be $C+C=[0,2]$ and $AB$ contains all elements from $e^0$ to $e^4$.
– Anupam
Nov 23 at 9:16
1
@RalphXu You are getting confused with notations. It is not true that $AB$ is contained in the image of $E$ under $f(x)=x^{2}$. The image consists of the points $t^{2}$, with $t in E$ not products of any two elements of $E$.
– Kavi Rama Murthy
Nov 23 at 9:16
|
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
If $C$ is the Cantor set and $A=B={e^{x}:xin C}$ then $A$ and $B$ have measure $0$ but $AB$ contains all numbers of the form $e^{x+y}$ with $x,y in C$. Since $C+C=[0,2]$ it follows that $AB$ does not have measure $0$. [I have used the following facts: $e^{x}$ is absolutely continuous on $[0,1]$ and any absolutely continuous function maps sets of measure $0$ to sets of measure $0$. Hence $A$ and $B$ have measure $0$]. Note also that $AB$ contains the entire interval $[e^{0},e^{2}]$ so it has positive measure.
answered Nov 23 at 8:37
Kavi Rama Murthy
43k31751
I followed your argument, but I was thinking about the following, but I'm not able to figure out where went wrong: Let $E = A cup B$. Then $AB subseteq E^2$. We know $E$ has measure zero. Since the function $f(x) = x^2$ is Lipschitz, it maps sets of measure zero to sets of measure zero, so $E^2$ has measure zero, which means its subset $AB$ is also measure zero.
– Ralph Xu
Nov 23 at 8:58
$AB$ is the image of $C+C=[0,1]$ under the absolutely continuous function $e^x$. We know that every absolutely continuous function sends a set of measure zero to a set of measure zero. But can we say that it sends a set of positive measure to a set of positive measure?
– Anupam
Nov 23 at 9:03
@Anupam No you cannot say that any absolutely continuous function maps a set of positive measure to a set of positive measure. (An obvious counterexample is given by a constant function). But in my argument $AB$ contains all the numbers from $e^{0}$ to $e^{2}$ so it has positive measure.
– Kavi Rama Murthy
Nov 23 at 9:12
Ok. Sorry, I got your argument. May be $C+C=[0,2]$ and $AB$ contains all elements from $e^0$ to $e^4$.
– Anupam
Nov 23 at 9:16
1
@RalphXu You are getting confused with notations. It is not true that $AB$ is contained in the image of $E$ under $f(x)=x^{2}$. The image consists of the points $t^{2}$, with $t in E$ not products of any two elements of $E$.
– Kavi Rama Murthy
Nov 23 at 9:16
|
show 2 more comments
up vote
4
down vote
accepted
If $C$ is the Cantor set and $A=B={e^{x}:xin C}$ then $A$ and $B$ have measure $0$ but $AB$ contains all numbers of the form $e^{x+y}$ with $x,y in C$. Since $C+C=[0,2]$ it follows that $AB$ does not have measure $0$. [I have used the following facts: $e^{x}$ is absolutely continuous on $[0,1]$ and any absolutely continuous function maps sets of measure $0$ to sets of measure $0$. Hence $A$ and $B$ have measure $0$]. Note also that $AB$ contains the entire interval $[e^{0},e^{2}]$ so it has positive measure.
answered Nov 23 at 8:37
Kavi Rama Murthy
43k31751
I followed your argument, but I was thinking about the following, but I'm not able to figure out where went wrong: Let $E = A cup B$. Then $AB subseteq E^2$. We know $E$ has measure zero. Since the function $f(x) = x^2$ is Lipschitz, it maps sets of measure zero to sets of measure zero, so $E^2$ has measure zero, which means its subset $AB$ is also measure zero.
– Ralph Xu
Nov 23 at 8:58
$AB$ is the image of $C+C=[0,1]$ under the absolutely continuous function $e^x$. We know that every absolutely continuous function sends a set of measure zero to a set of measure zero. But can we say that it sends a set of positive measure to a set of positive measure?
– Anupam
Nov 23 at 9:03
@Anupam No you cannot say that any absolutely continuous function maps a set of positive measure to a set of positive measure. (An obvious counterexample is given by a constant function). But in my argument $AB$ contains all the numbers from $e^{0}$ to $e^{2}$ so it has positive measure.
– Kavi Rama Murthy
Nov 23 at 9:12
Ok. Sorry, I got your argument. May be $C+C=[0,2]$ and $AB$ contains all elements from $e^0$ to $e^4$.
– Anupam
Nov 23 at 9:16
1
@RalphXu You are getting confused with notations. It is not true that $AB$ is contained in the image of $E$ under $f(x)=x^{2}$. The image consists of the points $t^{2}$, with $t in E$ not products of any two elements of $E$.
– Kavi Rama Murthy
Nov 23 at 9:16
|
show 2 more comments
up vote
4
down vote
accepted
up vote
4
down vote
accepted
If $C$ is the Cantor set and $A=B={e^{x}:xin C}$ then $A$ and $B$ have measure $0$ but $AB$ contains all numbers of the form $e^{x+y}$ with $x,y in C$. Since $C+C=[0,2]$ it follows that $AB$ does not have measure $0$. [I have used the following facts: $e^{x}$ is absolutely continuous on $[0,1]$ and any absolutely continuous function maps sets of measure $0$ to sets of measure $0$. Hence $A$ and $B$ have measure $0$]. Note also that $AB$ contains the entire interval $[e^{0},e^{2}]$ so it has positive measure.
answered Nov 23 at 8:37
Kavi Rama Murthy
43k31751
If $C$ is the Cantor set and $A=B={e^{x}:xin C}$ then $A$ and $B$ have measure $0$ but $AB$ contains all numbers of the form $e^{x+y}$ with $x,y in C$. Since $C+C=[0,2]$ it follows that $AB$ does not have measure $0$. [I have used the following facts: $e^{x}$ is absolutely continuous on $[0,1]$ and any absolutely continuous function maps sets of measure $0$ to sets of measure $0$. Hence $A$ and $B$ have measure $0$]. Note also that $AB$ contains the entire interval $[e^{0},e^{2}]$ so it has positive measure.
answered Nov 23 at 8:37
Kavi Rama Murthy
43k31751
edited Nov 23 at 9:32
answered Nov 23 at 8:37
Kavi Rama Murthy
43k31751
answered Nov 23 at 8:37
Kavi Rama Murthy
43k31751
answered Nov 23 at 8:37
Kavi Rama Murthy
43k31751
43k31751
I followed your argument, but I was thinking about the following, but I'm not able to figure out where went wrong: Let $E = A cup B$. Then $AB subseteq E^2$. We know $E$ has measure zero. Since the function $f(x) = x^2$ is Lipschitz, it maps sets of measure zero to sets of measure zero, so $E^2$ has measure zero, which means its subset $AB$ is also measure zero.
– Ralph Xu
Nov 23 at 8:58
$AB$ is the image of $C+C=[0,1]$ under the absolutely continuous function $e^x$. We know that every absolutely continuous function sends a set of measure zero to a set of measure zero. But can we say that it sends a set of positive measure to a set of positive measure?
– Anupam
Nov 23 at 9:03
@Anupam No you cannot say that any absolutely continuous function maps a set of positive measure to a set of positive measure. (An obvious counterexample is given by a constant function). But in my argument $AB$ contains all the numbers from $e^{0}$ to $e^{2}$ so it has positive measure.
– Kavi Rama Murthy
Nov 23 at 9:12
Ok. Sorry, I got your argument. May be $C+C=[0,2]$ and $AB$ contains all elements from $e^0$ to $e^4$.
– Anupam
Nov 23 at 9:16
1
@RalphXu You are getting confused with notations. It is not true that $AB$ is contained in the image of $E$ under $f(x)=x^{2}$. The image consists of the points $t^{2}$, with $t in E$ not products of any two elements of $E$.
– Kavi Rama Murthy
Nov 23 at 9:16
|
show 2 more comments
I followed your argument, but I was thinking about the following, but I'm not able to figure out where went wrong: Let $E = A cup B$. Then $AB subseteq E^2$. We know $E$ has measure zero. Since the function $f(x) = x^2$ is Lipschitz, it maps sets of measure zero to sets of measure zero, so $E^2$ has measure zero, which means its subset $AB$ is also measure zero.
– Ralph Xu
Nov 23 at 8:58
$AB$ is the image of $C+C=[0,1]$ under the absolutely continuous function $e^x$. We know that every absolutely continuous function sends a set of measure zero to a set of measure zero. But can we say that it sends a set of positive measure to a set of positive measure?
– Anupam
Nov 23 at 9:03
@Anupam No you cannot say that any absolutely continuous function maps a set of positive measure to a set of positive measure. (An obvious counterexample is given by a constant function). But in my argument $AB$ contains all the numbers from $e^{0}$ to $e^{2}$ so it has positive measure.
– Kavi Rama Murthy
Nov 23 at 9:12
Ok. Sorry, I got your argument. May be $C+C=[0,2]$ and $AB$ contains all elements from $e^0$ to $e^4$.
– Anupam
Nov 23 at 9:16
1
@RalphXu You are getting confused with notations. It is not true that $AB$ is contained in the image of $E$ under $f(x)=x^{2}$. The image consists of the points $t^{2}$, with $t in E$ not products of any two elements of $E$.
– Kavi Rama Murthy
Nov 23 at 9:16
I followed your argument, but I was thinking about the following, but I'm not able to figure out where went wrong: Let $E = A cup B$. Then $AB subseteq E^2$. We know $E$ has measure zero. Since the function $f(x) = x^2$ is Lipschitz, it maps sets of measure zero to sets of measure zero, so $E^2$ has measure zero, which means its subset $AB$ is also measure zero.
– Ralph Xu
Nov 23 at 8:58
I followed your argument, but I was thinking about the following, but I'm not able to figure out where went wrong: Let $E = A cup B$. Then $AB subseteq E^2$. We know $E$ has measure zero. Since the function $f(x) = x^2$ is Lipschitz, it maps sets of measure zero to sets of measure zero, so $E^2$ has measure zero, which means its subset $AB$ is also measure zero.
– Ralph Xu
Nov 23 at 8:58
$AB$ is the image of $C+C=[0,1]$ under the absolutely continuous function $e^x$. We know that every absolutely continuous function sends a set of measure zero to a set of measure zero. But can we say that it sends a set of positive measure to a set of positive measure?
– Anupam
Nov 23 at 9:03
$AB$ is the image of $C+C=[0,1]$ under the absolutely continuous function $e^x$. We know that every absolutely continuous function sends a set of measure zero to a set of measure zero. But can we say that it sends a set of positive measure to a set of positive measure?
– Anupam
Nov 23 at 9:03
@Anupam No you cannot say that any absolutely continuous function maps a set of positive measure to a set of positive measure. (An obvious counterexample is given by a constant function). But in my argument $AB$ contains all the numbers from $e^{0}$ to $e^{2}$ so it has positive measure.
– Kavi Rama Murthy
Nov 23 at 9:12
@Anupam No you cannot say that any absolutely continuous function maps a set of positive measure to a set of positive measure. (An obvious counterexample is given by a constant function). But in my argument $AB$ contains all the numbers from $e^{0}$ to $e^{2}$ so it has positive measure.
– Kavi Rama Murthy
Nov 23 at 9:12
Ok. Sorry, I got your argument. May be $C+C=[0,2]$ and $AB$ contains all elements from $e^0$ to $e^4$.
– Anupam
Nov 23 at 9:16
Ok. Sorry, I got your argument. May be $C+C=[0,2]$ and $AB$ contains all elements from $e^0$ to $e^4$.
– Anupam
Nov 23 at 9:16
1
1
@RalphXu You are getting confused with notations. It is not true that $AB$ is contained in the image of $E$ under $f(x)=x^{2}$. The image consists of the points $t^{2}$, with $t in E$ not products of any two elements of $E$.
– Kavi Rama Murthy
Nov 23 at 9:16
@RalphXu You are getting confused with notations. It is not true that $AB$ is contained in the image of $E$ under $f(x)=x^{2}$. The image consists of the points $t^{2}$, with $t in E$ not products of any two elements of $E$.
– Kavi Rama Murthy
Nov 23 at 9:16
|
show 2 more comments
