How to prove that a particular set does not have measure zero Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Irrationals in $[0,1]$ does not have measure zeroProve that the a modified Cantor Set is not Jordan-MeasurableEvery Lebesgue measurable set contains a closed subset such that the set difference has small measureHelp with a proof that a countable union of measure zero sets is measure zero.Does every non empty open set has measure greater than zero?Let $A$ be a lebesgue null set in $mathbbR$. Prove that $B:=e^x:x in A$ is also a null setProve that a subset of a set of measure zero has measure zeroWhat is wrong in this proof: That $mathbbR$ has measure zeroConcerning the definition of zero measure setShow that the set $1/n_n=1^infty$ ⊆ ℝ has Jordan measure zero (content zero)
Why does tar appear to skip file contents when output file is /dev/null?
Cauchy Sequence Characterized only By Directly Neighbouring Sequence Members
Classification of bundles, Postnikov towers, obstruction theory, local coefficients
Why does this iterative way of solving of equation work?
Is there a documented rationale why the House Ways and Means chairman can demand tax info?
Complexity of many constant time steps with occasional logarithmic steps
Was credit for the black hole image misattributed?
Unexpected result with right shift after bitwise negation
Replacing HDD with SSD; what about non-APFS/APFS?
How can you insert a "times/divide" symbol similar to the "plus/minus" (±) one?
What loss function to use when labels are probabilities?
Active filter with series inductor and resistor - do these exist?
Cold is to Refrigerator as warm is to?
New Order #5: where Fibonacci and Beatty meet at Wythoff
Area of a 2D convex hull
The following signatures were invalid: EXPKEYSIG 1397BC53640DB551
Can a zero nonce be safely used with AES-GCM if the key is random and never used again?
Statistical model of ligand substitution
Is it possible to ask for a hotel room without minibar/extra services?
Is there a service that would inform me whenever a new direct route is scheduled from a given airport?
Single author papers against my advisor's will?
What can I do if my MacBook isn’t charging but already ran out?
Antler Helmet: Can it work?
Is 1 ppb equal to 1 μg/kg?
How to prove that a particular set does not have measure zero
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Irrationals in $[0,1]$ does not have measure zeroProve that the a modified Cantor Set is not Jordan-MeasurableEvery Lebesgue measurable set contains a closed subset such that the set difference has small measureHelp with a proof that a countable union of measure zero sets is measure zero.Does every non empty open set has measure greater than zero?Let $A$ be a lebesgue null set in $mathbbR$. Prove that $B:=e^x:x in A$ is also a null setProve that a subset of a set of measure zero has measure zeroWhat is wrong in this proof: That $mathbbR$ has measure zeroConcerning the definition of zero measure setShow that the set $1/n_n=1^infty$ ⊆ ℝ has Jordan measure zero (content zero)
$begingroup$
I would like to show that the set constructed from this algorithm:
- from $[0,1]$ I remove an open interval of length $frac14$ center in its middle point $ x=frac12 $;
- in both remaining closed intervals I remove an open interval of length $frac14^2$ centered in each of the middle points of the 2 closed intervals;
- in each of the $2^2$ remaining closed intervals I remove an open interval of length $frac14^3$ centered in...
...
is not a set of measure zero.
How could I show this fact, knowing only the definition of a measure zero set?
The definition is:
A set $X$ is of measure zero if $forall epsilon >0$ there exists a cover of $X$ made by an at most countable system of intervals $I_n$ such that:
$$sum_n|I_n|<epsilon.$$
I can't find the right way.
Could you give me just an hint, please?
Edit. We could consider as known the following properties:
- A single point is a set of measure zero;
- A countable union of measure zero set is a measure zero set;
- A subset of a measure zero set is of measure zero;
- Any closed interval $[a,b]$ with $a<b$ is not a set of measure zero.
real-analysis elementary-set-theory
edited Apr 9 at 12:08
Henno Brandsma
116k349127
asked Apr 8 at 20:50
NamelessNameless
8011
$endgroup$
|
show 5 more comments
$begingroup$
I would like to show that the set constructed from this algorithm:
- from $[0,1]$ I remove an open interval of length $frac14$ center in its middle point $ x=frac12 $;
- in both remaining closed intervals I remove an open interval of length $frac14^2$ centered in each of the middle points of the 2 closed intervals;
- in each of the $2^2$ remaining closed intervals I remove an open interval of length $frac14^3$ centered in...
...
is not a set of measure zero.
How could I show this fact, knowing only the definition of a measure zero set?
The definition is:
A set $X$ is of measure zero if $forall epsilon >0$ there exists a cover of $X$ made by an at most countable system of intervals $I_n$ such that:
$$sum_n|I_n|<epsilon.$$
I can't find the right way.
Could you give me just an hint, please?
Edit. We could consider as known the following properties:
- A single point is a set of measure zero;
- A countable union of measure zero set is a measure zero set;
- A subset of a measure zero set is of measure zero;
- Any closed interval $[a,b]$ with $a<b$ is not a set of measure zero.
real-analysis elementary-set-theory
edited Apr 9 at 12:08
Henno Brandsma
116k349127
asked Apr 8 at 20:50
NamelessNameless
8011
$endgroup$
2
$begingroup$
Think first about how much measure you are removing from the set at each step.
$endgroup$
– rubikscube09
Apr 8 at 20:55
$begingroup$
Yes, I know, but I, in principle, don't know what is the 'measure' of the intervals removed at each step. I could just say that every interval that I remove is not a measure zero set. But this is not useful for us.
$endgroup$
– Nameless
Apr 8 at 20:57
$begingroup$
Knowing only the measure-zero definition implies that I don't know what 'this set has measure x' means (if $xneq 0$).
$endgroup$
– Nameless
Apr 8 at 20:58
$begingroup$
@Nameless: the measure of the intervals removed at each step is the sum of their lengths. At step $n$ you remove $2^n-1$ intervals of length $frac 14^n$, so the total length is ???. Sum these up and it is less than $1$
$endgroup$
– Ross Millikan
Apr 8 at 20:59
$begingroup$
Thanks for your reply, but please read my last comment. I already know that proof, but I'm searching another kind of proof.
$endgroup$
– Nameless
Apr 8 at 21:01
|
show 5 more comments
$begingroup$
I would like to show that the set constructed from this algorithm:
- from $[0,1]$ I remove an open interval of length $frac14$ center in its middle point $ x=frac12 $;
- in both remaining closed intervals I remove an open interval of length $frac14^2$ centered in each of the middle points of the 2 closed intervals;
- in each of the $2^2$ remaining closed intervals I remove an open interval of length $frac14^3$ centered in...
...
is not a set of measure zero.
How could I show this fact, knowing only the definition of a measure zero set?
The definition is:
A set $X$ is of measure zero if $forall epsilon >0$ there exists a cover of $X$ made by an at most countable system of intervals $I_n$ such that:
$$sum_n|I_n|<epsilon.$$
I can't find the right way.
Could you give me just an hint, please?
Edit. We could consider as known the following properties:
- A single point is a set of measure zero;
- A countable union of measure zero set is a measure zero set;
- A subset of a measure zero set is of measure zero;
- Any closed interval $[a,b]$ with $a<b$ is not a set of measure zero.
real-analysis elementary-set-theory
edited Apr 9 at 12:08
Henno Brandsma
116k349127
asked Apr 8 at 20:50
NamelessNameless
8011
$endgroup$
I would like to show that the set constructed from this algorithm:
- from $[0,1]$ I remove an open interval of length $frac14$ center in its middle point $ x=frac12 $;
- in both remaining closed intervals I remove an open interval of length $frac14^2$ centered in each of the middle points of the 2 closed intervals;
- in each of the $2^2$ remaining closed intervals I remove an open interval of length $frac14^3$ centered in...
...
is not a set of measure zero.
How could I show this fact, knowing only the definition of a measure zero set?
The definition is:
A set $X$ is of measure zero if $forall epsilon >0$ there exists a cover of $X$ made by an at most countable system of intervals $I_n$ such that:
$$sum_n|I_n|<epsilon.$$
I can't find the right way.
Could you give me just an hint, please?
Edit. We could consider as known the following properties:
- A single point is a set of measure zero;
- A countable union of measure zero set is a measure zero set;
- A subset of a measure zero set is of measure zero;
- Any closed interval $[a,b]$ with $a<b$ is not a set of measure zero.
real-analysis elementary-set-theory
real-analysis elementary-set-theory
edited Apr 9 at 12:08
Henno Brandsma
116k349127
asked Apr 8 at 20:50
NamelessNameless
8011
edited Apr 9 at 12:08
Henno Brandsma
116k349127
asked Apr 8 at 20:50
NamelessNameless
8011
edited Apr 9 at 12:08
Henno Brandsma
116k349127
edited Apr 9 at 12:08
Henno Brandsma
116k349127
edited Apr 9 at 12:08
Henno Brandsma
116k349127
116k349127
asked Apr 8 at 20:50
NamelessNameless
8011
asked Apr 8 at 20:50
NamelessNameless
8011
asked Apr 8 at 20:50
NamelessNameless
8011
8011
2
$begingroup$
Think first about how much measure you are removing from the set at each step.
$endgroup$
– rubikscube09
Apr 8 at 20:55
$begingroup$
Yes, I know, but I, in principle, don't know what is the 'measure' of the intervals removed at each step. I could just say that every interval that I remove is not a measure zero set. But this is not useful for us.
$endgroup$
– Nameless
Apr 8 at 20:57
$begingroup$
Knowing only the measure-zero definition implies that I don't know what 'this set has measure x' means (if $xneq 0$).
$endgroup$
– Nameless
Apr 8 at 20:58
$begingroup$
@Nameless: the measure of the intervals removed at each step is the sum of their lengths. At step $n$ you remove $2^n-1$ intervals of length $frac 14^n$, so the total length is ???. Sum these up and it is less than $1$
$endgroup$
– Ross Millikan
Apr 8 at 20:59
$begingroup$
Thanks for your reply, but please read my last comment. I already know that proof, but I'm searching another kind of proof.
$endgroup$
– Nameless
Apr 8 at 21:01
|
show 5 more comments
2
$begingroup$
Think first about how much measure you are removing from the set at each step.
$endgroup$
– rubikscube09
Apr 8 at 20:55
$begingroup$
Yes, I know, but I, in principle, don't know what is the 'measure' of the intervals removed at each step. I could just say that every interval that I remove is not a measure zero set. But this is not useful for us.
$endgroup$
– Nameless
Apr 8 at 20:57
$begingroup$
Knowing only the measure-zero definition implies that I don't know what 'this set has measure x' means (if $xneq 0$).
$endgroup$
– Nameless
Apr 8 at 20:58
$begingroup$
@Nameless: the measure of the intervals removed at each step is the sum of their lengths. At step $n$ you remove $2^n-1$ intervals of length $frac 14^n$, so the total length is ???. Sum these up and it is less than $1$
$endgroup$
– Ross Millikan
Apr 8 at 20:59
$begingroup$
Thanks for your reply, but please read my last comment. I already know that proof, but I'm searching another kind of proof.
$endgroup$
– Nameless
Apr 8 at 21:01
2
2
$begingroup$
Think first about how much measure you are removing from the set at each step.
$endgroup$
– rubikscube09
Apr 8 at 20:55
$begingroup$
Think first about how much measure you are removing from the set at each step.
$endgroup$
– rubikscube09
Apr 8 at 20:55
$begingroup$
Yes, I know, but I, in principle, don't know what is the 'measure' of the intervals removed at each step. I could just say that every interval that I remove is not a measure zero set. But this is not useful for us.
$endgroup$
– Nameless
Apr 8 at 20:57
$begingroup$
Yes, I know, but I, in principle, don't know what is the 'measure' of the intervals removed at each step. I could just say that every interval that I remove is not a measure zero set. But this is not useful for us.
$endgroup$
– Nameless
Apr 8 at 20:57
$begingroup$
Knowing only the measure-zero definition implies that I don't know what 'this set has measure x' means (if $xneq 0$).
$endgroup$
– Nameless
Apr 8 at 20:58
$begingroup$
Knowing only the measure-zero definition implies that I don't know what 'this set has measure x' means (if $xneq 0$).
$endgroup$
– Nameless
Apr 8 at 20:58
$begingroup$
@Nameless: the measure of the intervals removed at each step is the sum of their lengths. At step $n$ you remove $2^n-1$ intervals of length $frac 14^n$, so the total length is ???. Sum these up and it is less than $1$
$endgroup$
– Ross Millikan
Apr 8 at 20:59
$begingroup$
@Nameless: the measure of the intervals removed at each step is the sum of their lengths. At step $n$ you remove $2^n-1$ intervals of length $frac 14^n$, so the total length is ???. Sum these up and it is less than $1$
$endgroup$
– Ross Millikan
Apr 8 at 20:59
$begingroup$
Thanks for your reply, but please read my last comment. I already know that proof, but I'm searching another kind of proof.
$endgroup$
– Nameless
Apr 8 at 21:01
$begingroup$
Thanks for your reply, but please read my last comment. I already know that proof, but I'm searching another kind of proof.
$endgroup$
– Nameless
Apr 8 at 21:01
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The set you presented is alike the Cantor set.
It is an uncountable union of measure zero sets, so it's not covered by your properties, in particular, it does not follow property 2.
It seems to me that, with only those properties, we can't state if it is of measure zero or not.
Edit 2:
Let $S$ be your set.
I will take a look at the complementary set $barS$ in the interval $[0; 1]$, i.e., $barS$ is the union of the removed intervals.
Let $I_j,k$ be the $k^textrmth$ interval removed in step $j$.
We have $2^j-1$ intervals $I_j,k$ at step $j$, hence, $k in 1, 2, ldots, 2^j-1 $.
Each interval $I_j,k$ has lenght $4^-j$.
From property 4 we only know that the measure $mu$ of the interval is not zero, so, $mu(I_j,k) > 0$.
It might be possible that measure could differ with the position of the interval. So, from what I understood, it might be possible, for instance, that $mu(I_3,1) neq mu(I_3,2)$.
So, I will add another property:
- The measure of intervals is proportional to their length.
With this property, the initial interval will have measure $c > 0$, i.e., $mu([0; 1]) = c $. Hence, the measure of the intervals $I_j,k$ will be $mu(I_j,k) = c frac14^j $.
So, the measure of $barS$ is the sum of the measures of the intervals, since they are disjoint. I.e., it is proportional to the total length of the removed intervals:
$$
mu(barS)
= sum_j=1^infty sum_k=1^2^j-1 c frac14^j
= sum_j=1^infty 2^j-1 c frac14^j
= sum_j=1^infty fracc2 times frac12^j
= fracc2 times 1
$$
From this we would get that summing measure of your set with the measure of the complementary set we will have the measure of the interval:
$$
mu(S) + mu(barS) = mu([0; 1])
$$
Hence,
$$
mu(S) = fracc2
$$
So, unlike the Cantor set that has measure $0$, the set $S$ has positive measure.
answered Apr 9 at 17:28
ErtxiemErtxiem
757212
$endgroup$
$begingroup$
Thanks for your reply. So, according to you, what are the key properties that will solve the problem? Can they be demonstrated in a relatively simple way?
$endgroup$
– Nameless
Apr 9 at 21:31
$begingroup$
@Nameless: I added another property and a computation of the measure. Did you intended something like this?
$endgroup$
– Ertxiem
Apr 10 at 0:03
$begingroup$
Unfortunately, this proof uses the concept of 'measure' as a function that maps sets in real numbers. I can know only the definition of 'measure zero set' without knowing what a measure is. Anyway I thank you for time you spent.
$endgroup$
– Nameless
Apr 10 at 20:27
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3180182%2fhow-to-prove-that-a-particular-set-does-not-have-measure-zero%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The set you presented is alike the Cantor set.
It is an uncountable union of measure zero sets, so it's not covered by your properties, in particular, it does not follow property 2.
It seems to me that, with only those properties, we can't state if it is of measure zero or not.
Edit 2:
Let $S$ be your set.
I will take a look at the complementary set $barS$ in the interval $[0; 1]$, i.e., $barS$ is the union of the removed intervals.
Let $I_j,k$ be the $k^textrmth$ interval removed in step $j$.
We have $2^j-1$ intervals $I_j,k$ at step $j$, hence, $k in 1, 2, ldots, 2^j-1 $.
Each interval $I_j,k$ has lenght $4^-j$.
From property 4 we only know that the measure $mu$ of the interval is not zero, so, $mu(I_j,k) > 0$.
It might be possible that measure could differ with the position of the interval. So, from what I understood, it might be possible, for instance, that $mu(I_3,1) neq mu(I_3,2)$.
So, I will add another property:
- The measure of intervals is proportional to their length.
With this property, the initial interval will have measure $c > 0$, i.e., $mu([0; 1]) = c $. Hence, the measure of the intervals $I_j,k$ will be $mu(I_j,k) = c frac14^j $.
So, the measure of $barS$ is the sum of the measures of the intervals, since they are disjoint. I.e., it is proportional to the total length of the removed intervals:
$$
mu(barS)
= sum_j=1^infty sum_k=1^2^j-1 c frac14^j
= sum_j=1^infty 2^j-1 c frac14^j
= sum_j=1^infty fracc2 times frac12^j
= fracc2 times 1
$$
From this we would get that summing measure of your set with the measure of the complementary set we will have the measure of the interval:
$$
mu(S) + mu(barS) = mu([0; 1])
$$
Hence,
$$
mu(S) = fracc2
$$
So, unlike the Cantor set that has measure $0$, the set $S$ has positive measure.
answered Apr 9 at 17:28
ErtxiemErtxiem
757212
$endgroup$
$begingroup$
Thanks for your reply. So, according to you, what are the key properties that will solve the problem? Can they be demonstrated in a relatively simple way?
$endgroup$
– Nameless
Apr 9 at 21:31
$begingroup$
@Nameless: I added another property and a computation of the measure. Did you intended something like this?
$endgroup$
– Ertxiem
Apr 10 at 0:03
$begingroup$
Unfortunately, this proof uses the concept of 'measure' as a function that maps sets in real numbers. I can know only the definition of 'measure zero set' without knowing what a measure is. Anyway I thank you for time you spent.
$endgroup$
– Nameless
Apr 10 at 20:27
add a comment |
$begingroup$
The set you presented is alike the Cantor set.
It is an uncountable union of measure zero sets, so it's not covered by your properties, in particular, it does not follow property 2.
It seems to me that, with only those properties, we can't state if it is of measure zero or not.
Edit 2:
Let $S$ be your set.
I will take a look at the complementary set $barS$ in the interval $[0; 1]$, i.e., $barS$ is the union of the removed intervals.
Let $I_j,k$ be the $k^textrmth$ interval removed in step $j$.
We have $2^j-1$ intervals $I_j,k$ at step $j$, hence, $k in 1, 2, ldots, 2^j-1 $.
Each interval $I_j,k$ has lenght $4^-j$.
From property 4 we only know that the measure $mu$ of the interval is not zero, so, $mu(I_j,k) > 0$.
It might be possible that measure could differ with the position of the interval. So, from what I understood, it might be possible, for instance, that $mu(I_3,1) neq mu(I_3,2)$.
So, I will add another property:
- The measure of intervals is proportional to their length.
With this property, the initial interval will have measure $c > 0$, i.e., $mu([0; 1]) = c $. Hence, the measure of the intervals $I_j,k$ will be $mu(I_j,k) = c frac14^j $.
So, the measure of $barS$ is the sum of the measures of the intervals, since they are disjoint. I.e., it is proportional to the total length of the removed intervals:
$$
mu(barS)
= sum_j=1^infty sum_k=1^2^j-1 c frac14^j
= sum_j=1^infty 2^j-1 c frac14^j
= sum_j=1^infty fracc2 times frac12^j
= fracc2 times 1
$$
From this we would get that summing measure of your set with the measure of the complementary set we will have the measure of the interval:
$$
mu(S) + mu(barS) = mu([0; 1])
$$
Hence,
$$
mu(S) = fracc2
$$
So, unlike the Cantor set that has measure $0$, the set $S$ has positive measure.
answered Apr 9 at 17:28
ErtxiemErtxiem
757212
$endgroup$
$begingroup$
Thanks for your reply. So, according to you, what are the key properties that will solve the problem? Can they be demonstrated in a relatively simple way?
$endgroup$
– Nameless
Apr 9 at 21:31
$begingroup$
@Nameless: I added another property and a computation of the measure. Did you intended something like this?
$endgroup$
– Ertxiem
Apr 10 at 0:03
$begingroup$
Unfortunately, this proof uses the concept of 'measure' as a function that maps sets in real numbers. I can know only the definition of 'measure zero set' without knowing what a measure is. Anyway I thank you for time you spent.
$endgroup$
– Nameless
Apr 10 at 20:27
add a comment |
$begingroup$
The set you presented is alike the Cantor set.
It is an uncountable union of measure zero sets, so it's not covered by your properties, in particular, it does not follow property 2.
It seems to me that, with only those properties, we can't state if it is of measure zero or not.
Edit 2:
Let $S$ be your set.
I will take a look at the complementary set $barS$ in the interval $[0; 1]$, i.e., $barS$ is the union of the removed intervals.
Let $I_j,k$ be the $k^textrmth$ interval removed in step $j$.
We have $2^j-1$ intervals $I_j,k$ at step $j$, hence, $k in 1, 2, ldots, 2^j-1 $.
Each interval $I_j,k$ has lenght $4^-j$.
From property 4 we only know that the measure $mu$ of the interval is not zero, so, $mu(I_j,k) > 0$.
It might be possible that measure could differ with the position of the interval. So, from what I understood, it might be possible, for instance, that $mu(I_3,1) neq mu(I_3,2)$.
So, I will add another property:
- The measure of intervals is proportional to their length.
With this property, the initial interval will have measure $c > 0$, i.e., $mu([0; 1]) = c $. Hence, the measure of the intervals $I_j,k$ will be $mu(I_j,k) = c frac14^j $.
So, the measure of $barS$ is the sum of the measures of the intervals, since they are disjoint. I.e., it is proportional to the total length of the removed intervals:
$$
mu(barS)
= sum_j=1^infty sum_k=1^2^j-1 c frac14^j
= sum_j=1^infty 2^j-1 c frac14^j
= sum_j=1^infty fracc2 times frac12^j
= fracc2 times 1
$$
From this we would get that summing measure of your set with the measure of the complementary set we will have the measure of the interval:
$$
mu(S) + mu(barS) = mu([0; 1])
$$
Hence,
$$
mu(S) = fracc2
$$
So, unlike the Cantor set that has measure $0$, the set $S$ has positive measure.
answered Apr 9 at 17:28
ErtxiemErtxiem
757212
$endgroup$
The set you presented is alike the Cantor set.
It is an uncountable union of measure zero sets, so it's not covered by your properties, in particular, it does not follow property 2.
It seems to me that, with only those properties, we can't state if it is of measure zero or not.
Edit 2:
Let $S$ be your set.
I will take a look at the complementary set $barS$ in the interval $[0; 1]$, i.e., $barS$ is the union of the removed intervals.
Let $I_j,k$ be the $k^textrmth$ interval removed in step $j$.
We have $2^j-1$ intervals $I_j,k$ at step $j$, hence, $k in 1, 2, ldots, 2^j-1 $.
Each interval $I_j,k$ has lenght $4^-j$.
From property 4 we only know that the measure $mu$ of the interval is not zero, so, $mu(I_j,k) > 0$.
It might be possible that measure could differ with the position of the interval. So, from what I understood, it might be possible, for instance, that $mu(I_3,1) neq mu(I_3,2)$.
So, I will add another property:
- The measure of intervals is proportional to their length.
With this property, the initial interval will have measure $c > 0$, i.e., $mu([0; 1]) = c $. Hence, the measure of the intervals $I_j,k$ will be $mu(I_j,k) = c frac14^j $.
So, the measure of $barS$ is the sum of the measures of the intervals, since they are disjoint. I.e., it is proportional to the total length of the removed intervals:
$$
mu(barS)
= sum_j=1^infty sum_k=1^2^j-1 c frac14^j
= sum_j=1^infty 2^j-1 c frac14^j
= sum_j=1^infty fracc2 times frac12^j
= fracc2 times 1
$$
From this we would get that summing measure of your set with the measure of the complementary set we will have the measure of the interval:
$$
mu(S) + mu(barS) = mu([0; 1])
$$
Hence,
$$
mu(S) = fracc2
$$
So, unlike the Cantor set that has measure $0$, the set $S$ has positive measure.
answered Apr 9 at 17:28
ErtxiemErtxiem
757212
edited Apr 10 at 0:02
answered Apr 9 at 17:28
ErtxiemErtxiem
757212
answered Apr 9 at 17:28
ErtxiemErtxiem
757212
answered Apr 9 at 17:28
ErtxiemErtxiem
757212
757212
$begingroup$
Thanks for your reply. So, according to you, what are the key properties that will solve the problem? Can they be demonstrated in a relatively simple way?
$endgroup$
– Nameless
Apr 9 at 21:31
$begingroup$
@Nameless: I added another property and a computation of the measure. Did you intended something like this?
$endgroup$
– Ertxiem
Apr 10 at 0:03
$begingroup$
Unfortunately, this proof uses the concept of 'measure' as a function that maps sets in real numbers. I can know only the definition of 'measure zero set' without knowing what a measure is. Anyway I thank you for time you spent.
$endgroup$
– Nameless
Apr 10 at 20:27
add a comment |
$begingroup$
Thanks for your reply. So, according to you, what are the key properties that will solve the problem? Can they be demonstrated in a relatively simple way?
$endgroup$
– Nameless
Apr 9 at 21:31
$begingroup$
@Nameless: I added another property and a computation of the measure. Did you intended something like this?
$endgroup$
– Ertxiem
Apr 10 at 0:03
$begingroup$
Unfortunately, this proof uses the concept of 'measure' as a function that maps sets in real numbers. I can know only the definition of 'measure zero set' without knowing what a measure is. Anyway I thank you for time you spent.
$endgroup$
– Nameless
Apr 10 at 20:27
$begingroup$
Thanks for your reply. So, according to you, what are the key properties that will solve the problem? Can they be demonstrated in a relatively simple way?
$endgroup$
– Nameless
Apr 9 at 21:31
$begingroup$
Thanks for your reply. So, according to you, what are the key properties that will solve the problem? Can they be demonstrated in a relatively simple way?
$endgroup$
– Nameless
Apr 9 at 21:31
$begingroup$
@Nameless: I added another property and a computation of the measure. Did you intended something like this?
$endgroup$
– Ertxiem
Apr 10 at 0:03
$begingroup$
@Nameless: I added another property and a computation of the measure. Did you intended something like this?
$endgroup$
– Ertxiem
Apr 10 at 0:03
$begingroup$
Unfortunately, this proof uses the concept of 'measure' as a function that maps sets in real numbers. I can know only the definition of 'measure zero set' without knowing what a measure is. Anyway I thank you for time you spent.
$endgroup$
– Nameless
Apr 10 at 20:27
$begingroup$
Unfortunately, this proof uses the concept of 'measure' as a function that maps sets in real numbers. I can know only the definition of 'measure zero set' without knowing what a measure is. Anyway I thank you for time you spent.
$endgroup$
– Nameless
Apr 10 at 20:27
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3180182%2fhow-to-prove-that-a-particular-set-does-not-have-measure-zero%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
$begingroup$
Think first about how much measure you are removing from the set at each step.
$endgroup$
– rubikscube09
Apr 8 at 20:55
$begingroup$
Yes, I know, but I, in principle, don't know what is the 'measure' of the intervals removed at each step. I could just say that every interval that I remove is not a measure zero set. But this is not useful for us.
$endgroup$
– Nameless
Apr 8 at 20:57
$begingroup$
Knowing only the measure-zero definition implies that I don't know what 'this set has measure x' means (if $xneq 0$).
$endgroup$
– Nameless
Apr 8 at 20:58
$begingroup$
@Nameless: the measure of the intervals removed at each step is the sum of their lengths. At step $n$ you remove $2^n-1$ intervals of length $frac 14^n$, so the total length is ???. Sum these up and it is less than $1$
$endgroup$
– Ross Millikan
Apr 8 at 20:59
$begingroup$
Thanks for your reply, but please read my last comment. I already know that proof, but I'm searching another kind of proof.
$endgroup$
– Nameless
Apr 8 at 21:01