Set and Set Complement of Uniform Distribution on [0,1] The 2019 Stack Overflow Developer...
Can withdrawing asylum be illegal?
Old scifi movie from the 50s or 60s with men in solid red uniforms who interrogate a spy from the past
Button changing its text & action. Good or terrible?
If I score a critical hit on an 18 or higher, what are my chances of getting a critical hit if I roll 3d20?
What is preventing me from simply constructing a hash that's lower than the current target?
Why can't devices on different VLANs, but on the same subnet, communicate?
The phrase "to the numbers born"?
Why can I use a list index as an indexing variable in a for loop?
What can I do if neighbor is blocking my solar panels intentionally
Why not take a picture of a closer black hole?
Kerning for subscripts of sigma?
If climate change impact can be observed in nature, has that had any effect on rural, i.e. farming community, perception of the scientific consensus?
Why does the nucleus not repel itself?
Is it ethical to upload a automatically generated paper to a non peer-reviewed site as part of a larger research?
For what reasons would an animal species NOT cross a *horizontal* land bridge?
Why couldn't they take pictures of a closer black hole?
Pokemon Turn Based battle (Python)
Is bread bad for ducks?
Why isn't the circumferential light around the M87 black hole's event horizon symmetric?
How can I have a shield and a way of attacking with a ranged weapon at the same time?
Dropping list elements from nested list after evaluation
Output the Arecibo Message
Is it safe to harvest rainwater that fell on solar panels?
Is Cinnamon a desktop environment or a window manager? (Or both?)
Set and Set Complement of Uniform Distribution on [0,1]
The 2019 Stack Overflow Developer Survey Results Are InA question regarding Borel sets, the Lebesgue measure, and Cantor-type setsUnion of Sets Dense in $[0,1]$For set of positive measure $E$, $alpha in (0, 1)$, there is interval $I$ such that $m(E cap I) > alpha , m(I)$Measure Theory - working with unusual measures and set functionsShowing $chi_Gnotinmathcal{R}[0,1]$ where: $G$ open, $mathbb{Q}cap[0,1] subset G,$ and $m(G)<frac{1}{2}$How can I construct such a Borel subset?How to prove complement of generalized Cantor set is dense in $[0,1]$A discontinuous function at every point in $[0,1]$Asking about a hint: constructing a cantor-like setCantor-Like Sets
$begingroup$
This questions first came to mind a few years ago when I was taking a course on real analysis as an undergraduate. I posed it to my instructor but he did not know a means of solving my inquiry. But to the point, does there exist a set (or is there a means of constructing a set) such that for a set of points $S in [0,1]$ and $S^{C} in [0,1]$ $S$ and $S^{C}$ both have measure $frac{1}{2}$ and for any arbitrary sub-interval $(a,b)$ of $[0,1]$ the value $mu_{S} = mu (S : cap : (a,b))=mu (S^{C} : cap : (a,b)) = mu_{S^{C}}$ ? Just to clarify $S^{C}$ is the complement of $S$ on the unit interval, or put differently $S^{C} = S^{complement} : cap : [0,1]$
real-analysis measure-theory
edited Mar 22 at 3:05
Andrés E. Caicedo
65.9k8160252
asked Mar 22 at 1:16
David G.David G.
156
$endgroup$
add a comment |
$begingroup$
This questions first came to mind a few years ago when I was taking a course on real analysis as an undergraduate. I posed it to my instructor but he did not know a means of solving my inquiry. But to the point, does there exist a set (or is there a means of constructing a set) such that for a set of points $S in [0,1]$ and $S^{C} in [0,1]$ $S$ and $S^{C}$ both have measure $frac{1}{2}$ and for any arbitrary sub-interval $(a,b)$ of $[0,1]$ the value $mu_{S} = mu (S : cap : (a,b))=mu (S^{C} : cap : (a,b)) = mu_{S^{C}}$ ? Just to clarify $S^{C}$ is the complement of $S$ on the unit interval, or put differently $S^{C} = S^{complement} : cap : [0,1]$
real-analysis measure-theory
edited Mar 22 at 3:05
Andrés E. Caicedo
65.9k8160252
asked Mar 22 at 1:16
David G.David G.
156
$endgroup$
add a comment |
$begingroup$
This questions first came to mind a few years ago when I was taking a course on real analysis as an undergraduate. I posed it to my instructor but he did not know a means of solving my inquiry. But to the point, does there exist a set (or is there a means of constructing a set) such that for a set of points $S in [0,1]$ and $S^{C} in [0,1]$ $S$ and $S^{C}$ both have measure $frac{1}{2}$ and for any arbitrary sub-interval $(a,b)$ of $[0,1]$ the value $mu_{S} = mu (S : cap : (a,b))=mu (S^{C} : cap : (a,b)) = mu_{S^{C}}$ ? Just to clarify $S^{C}$ is the complement of $S$ on the unit interval, or put differently $S^{C} = S^{complement} : cap : [0,1]$
real-analysis measure-theory
edited Mar 22 at 3:05
Andrés E. Caicedo
65.9k8160252
asked Mar 22 at 1:16
David G.David G.
156
$endgroup$
This questions first came to mind a few years ago when I was taking a course on real analysis as an undergraduate. I posed it to my instructor but he did not know a means of solving my inquiry. But to the point, does there exist a set (or is there a means of constructing a set) such that for a set of points $S in [0,1]$ and $S^{C} in [0,1]$ $S$ and $S^{C}$ both have measure $frac{1}{2}$ and for any arbitrary sub-interval $(a,b)$ of $[0,1]$ the value $mu_{S} = mu (S : cap : (a,b))=mu (S^{C} : cap : (a,b)) = mu_{S^{C}}$ ? Just to clarify $S^{C}$ is the complement of $S$ on the unit interval, or put differently $S^{C} = S^{complement} : cap : [0,1]$
real-analysis measure-theory
real-analysis measure-theory
edited Mar 22 at 3:05
Andrés E. Caicedo
65.9k8160252
asked Mar 22 at 1:16
David G.David G.
156
edited Mar 22 at 3:05
Andrés E. Caicedo
65.9k8160252
asked Mar 22 at 1:16
David G.David G.
156
edited Mar 22 at 3:05
Andrés E. Caicedo
65.9k8160252
edited Mar 22 at 3:05
Andrés E. Caicedo
65.9k8160252
edited Mar 22 at 3:05
Andrés E. Caicedo
65.9k8160252
65.9k8160252
asked Mar 22 at 1:16
David G.David G.
156
asked Mar 22 at 1:16
David G.David G.
156
asked Mar 22 at 1:16
David G.David G.
156
156
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No, there is no such measurable set. If there were you can get a contradiction as follows.
Note that $mu_Sllmu$ because $mu(A)=0$ implies $mu_S(A)=mu(Acap S)=0$, so the Radon Nikodym theorem applies: there exists a function $f$ such that $mu_S(A)=int_A f dx$, unique up to measure $0$ modifications. Calculate this Radon Nikodym derivative two different ways. On the one hand, $f=chi_S$, the indicator function of $S$, since $mu_S(A)=int_A chi_S(x)dx$. On the other, it is the constant function $1/2$. (Evaluate $F(x)=mu_S([0,x]) = x/2$, and so on.)
These two formulas do not agree almost everywhere.
answered Mar 22 at 1:46
kimchi loverkimchi lover
11.8k31229
$endgroup$
$begingroup$
So unfortunately, I don't quite understand why the RN derivative is equal to the indicator function $chi_{S}$. Could you please expand?
$endgroup$
– David G.
Mar 22 at 2:27
$begingroup$
@DavidG. I have edited my answer. I hope it makes it clearer.
$endgroup$
– kimchi lover
Mar 22 at 11:10
$begingroup$
I don't see why $mu (A) = 0$ since $A = (a,b)$ wouldn't it necessarily be the case that $mu (A) = b-a$?
$endgroup$
– David G.
Mar 22 at 15:53
$begingroup$
The condition for $mu_Sllmu$ is that whenever $mu(A)=0$ we also have $mu_S(A)=0$, too. Which is satisfied by your $mu_S$.
$endgroup$
– kimchi lover
Mar 22 at 15:58
$begingroup$
Could you clarify why $mu_{S} ll mu$? It seems to me that the relation should be $mu_{s} = frac{1}{2} mu$ which still holds when $mu (A) = 0$
$endgroup$
– David G.
Mar 22 at 16:11
|
show 1 more comment
$begingroup$
By a standard argument $mu(Scap (a,b))=mu(S^{c}cap (a,b))$ for every subinterval $(a,b)$ implies $mu(Scap A)=mu(S^{c}cap A)$ for every Bore set $A$. Taking $A=S$ we get a contradiction.
answered Mar 22 at 6:06
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
$endgroup$
$begingroup$
I understand your point, however, I'm inclined to believe that $S$ cannot equal $A$ as I don't see how $S$ is an interval.
$endgroup$
– David G.
Mar 22 at 6:38
$begingroup$
If you extend the equality from intervals to measurable sets you can then take $A=S$.
$endgroup$
– Kavi Rama Murthy
Mar 22 at 6:40
$begingroup$
Though this is true, that breaks the statement of the question and circumvents the basis of the question. The subintervals are there to illustrated uniform distribution of points across the unit interval
$endgroup$
– David G.
Mar 22 at 7:24
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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%2f3157611%2fset-and-set-complement-of-uniform-distribution-on-0-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No, there is no such measurable set. If there were you can get a contradiction as follows.
Note that $mu_Sllmu$ because $mu(A)=0$ implies $mu_S(A)=mu(Acap S)=0$, so the Radon Nikodym theorem applies: there exists a function $f$ such that $mu_S(A)=int_A f dx$, unique up to measure $0$ modifications. Calculate this Radon Nikodym derivative two different ways. On the one hand, $f=chi_S$, the indicator function of $S$, since $mu_S(A)=int_A chi_S(x)dx$. On the other, it is the constant function $1/2$. (Evaluate $F(x)=mu_S([0,x]) = x/2$, and so on.)
These two formulas do not agree almost everywhere.
answered Mar 22 at 1:46
kimchi loverkimchi lover
11.8k31229
$endgroup$
$begingroup$
So unfortunately, I don't quite understand why the RN derivative is equal to the indicator function $chi_{S}$. Could you please expand?
$endgroup$
– David G.
Mar 22 at 2:27
$begingroup$
@DavidG. I have edited my answer. I hope it makes it clearer.
$endgroup$
– kimchi lover
Mar 22 at 11:10
$begingroup$
I don't see why $mu (A) = 0$ since $A = (a,b)$ wouldn't it necessarily be the case that $mu (A) = b-a$?
$endgroup$
– David G.
Mar 22 at 15:53
$begingroup$
The condition for $mu_Sllmu$ is that whenever $mu(A)=0$ we also have $mu_S(A)=0$, too. Which is satisfied by your $mu_S$.
$endgroup$
– kimchi lover
Mar 22 at 15:58
$begingroup$
Could you clarify why $mu_{S} ll mu$? It seems to me that the relation should be $mu_{s} = frac{1}{2} mu$ which still holds when $mu (A) = 0$
$endgroup$
– David G.
Mar 22 at 16:11
|
show 1 more comment
$begingroup$
No, there is no such measurable set. If there were you can get a contradiction as follows.
Note that $mu_Sllmu$ because $mu(A)=0$ implies $mu_S(A)=mu(Acap S)=0$, so the Radon Nikodym theorem applies: there exists a function $f$ such that $mu_S(A)=int_A f dx$, unique up to measure $0$ modifications. Calculate this Radon Nikodym derivative two different ways. On the one hand, $f=chi_S$, the indicator function of $S$, since $mu_S(A)=int_A chi_S(x)dx$. On the other, it is the constant function $1/2$. (Evaluate $F(x)=mu_S([0,x]) = x/2$, and so on.)
These two formulas do not agree almost everywhere.
answered Mar 22 at 1:46
kimchi loverkimchi lover
11.8k31229
$endgroup$
$begingroup$
So unfortunately, I don't quite understand why the RN derivative is equal to the indicator function $chi_{S}$. Could you please expand?
$endgroup$
– David G.
Mar 22 at 2:27
$begingroup$
@DavidG. I have edited my answer. I hope it makes it clearer.
$endgroup$
– kimchi lover
Mar 22 at 11:10
$begingroup$
I don't see why $mu (A) = 0$ since $A = (a,b)$ wouldn't it necessarily be the case that $mu (A) = b-a$?
$endgroup$
– David G.
Mar 22 at 15:53
$begingroup$
The condition for $mu_Sllmu$ is that whenever $mu(A)=0$ we also have $mu_S(A)=0$, too. Which is satisfied by your $mu_S$.
$endgroup$
– kimchi lover
Mar 22 at 15:58
$begingroup$
Could you clarify why $mu_{S} ll mu$? It seems to me that the relation should be $mu_{s} = frac{1}{2} mu$ which still holds when $mu (A) = 0$
$endgroup$
– David G.
Mar 22 at 16:11
|
show 1 more comment
$begingroup$
No, there is no such measurable set. If there were you can get a contradiction as follows.
Note that $mu_Sllmu$ because $mu(A)=0$ implies $mu_S(A)=mu(Acap S)=0$, so the Radon Nikodym theorem applies: there exists a function $f$ such that $mu_S(A)=int_A f dx$, unique up to measure $0$ modifications. Calculate this Radon Nikodym derivative two different ways. On the one hand, $f=chi_S$, the indicator function of $S$, since $mu_S(A)=int_A chi_S(x)dx$. On the other, it is the constant function $1/2$. (Evaluate $F(x)=mu_S([0,x]) = x/2$, and so on.)
These two formulas do not agree almost everywhere.
answered Mar 22 at 1:46
kimchi loverkimchi lover
11.8k31229
$endgroup$
No, there is no such measurable set. If there were you can get a contradiction as follows.
Note that $mu_Sllmu$ because $mu(A)=0$ implies $mu_S(A)=mu(Acap S)=0$, so the Radon Nikodym theorem applies: there exists a function $f$ such that $mu_S(A)=int_A f dx$, unique up to measure $0$ modifications. Calculate this Radon Nikodym derivative two different ways. On the one hand, $f=chi_S$, the indicator function of $S$, since $mu_S(A)=int_A chi_S(x)dx$. On the other, it is the constant function $1/2$. (Evaluate $F(x)=mu_S([0,x]) = x/2$, and so on.)
These two formulas do not agree almost everywhere.
answered Mar 22 at 1:46
kimchi loverkimchi lover
11.8k31229
edited Mar 22 at 11:38
answered Mar 22 at 1:46
kimchi loverkimchi lover
11.8k31229
answered Mar 22 at 1:46
kimchi loverkimchi lover
11.8k31229
answered Mar 22 at 1:46
kimchi loverkimchi lover
11.8k31229
11.8k31229
$begingroup$
So unfortunately, I don't quite understand why the RN derivative is equal to the indicator function $chi_{S}$. Could you please expand?
$endgroup$
– David G.
Mar 22 at 2:27
$begingroup$
@DavidG. I have edited my answer. I hope it makes it clearer.
$endgroup$
– kimchi lover
Mar 22 at 11:10
$begingroup$
I don't see why $mu (A) = 0$ since $A = (a,b)$ wouldn't it necessarily be the case that $mu (A) = b-a$?
$endgroup$
– David G.
Mar 22 at 15:53
$begingroup$
The condition for $mu_Sllmu$ is that whenever $mu(A)=0$ we also have $mu_S(A)=0$, too. Which is satisfied by your $mu_S$.
$endgroup$
– kimchi lover
Mar 22 at 15:58
$begingroup$
Could you clarify why $mu_{S} ll mu$? It seems to me that the relation should be $mu_{s} = frac{1}{2} mu$ which still holds when $mu (A) = 0$
$endgroup$
– David G.
Mar 22 at 16:11
|
show 1 more comment
$begingroup$
So unfortunately, I don't quite understand why the RN derivative is equal to the indicator function $chi_{S}$. Could you please expand?
$endgroup$
– David G.
Mar 22 at 2:27
$begingroup$
@DavidG. I have edited my answer. I hope it makes it clearer.
$endgroup$
– kimchi lover
Mar 22 at 11:10
$begingroup$
I don't see why $mu (A) = 0$ since $A = (a,b)$ wouldn't it necessarily be the case that $mu (A) = b-a$?
$endgroup$
– David G.
Mar 22 at 15:53
$begingroup$
The condition for $mu_Sllmu$ is that whenever $mu(A)=0$ we also have $mu_S(A)=0$, too. Which is satisfied by your $mu_S$.
$endgroup$
– kimchi lover
Mar 22 at 15:58
$begingroup$
Could you clarify why $mu_{S} ll mu$? It seems to me that the relation should be $mu_{s} = frac{1}{2} mu$ which still holds when $mu (A) = 0$
$endgroup$
– David G.
Mar 22 at 16:11
$begingroup$
So unfortunately, I don't quite understand why the RN derivative is equal to the indicator function $chi_{S}$. Could you please expand?
$endgroup$
– David G.
Mar 22 at 2:27
$begingroup$
So unfortunately, I don't quite understand why the RN derivative is equal to the indicator function $chi_{S}$. Could you please expand?
$endgroup$
– David G.
Mar 22 at 2:27
$begingroup$
@DavidG. I have edited my answer. I hope it makes it clearer.
$endgroup$
– kimchi lover
Mar 22 at 11:10
$begingroup$
@DavidG. I have edited my answer. I hope it makes it clearer.
$endgroup$
– kimchi lover
Mar 22 at 11:10
$begingroup$
I don't see why $mu (A) = 0$ since $A = (a,b)$ wouldn't it necessarily be the case that $mu (A) = b-a$?
$endgroup$
– David G.
Mar 22 at 15:53
$begingroup$
I don't see why $mu (A) = 0$ since $A = (a,b)$ wouldn't it necessarily be the case that $mu (A) = b-a$?
$endgroup$
– David G.
Mar 22 at 15:53
$begingroup$
The condition for $mu_Sllmu$ is that whenever $mu(A)=0$ we also have $mu_S(A)=0$, too. Which is satisfied by your $mu_S$.
$endgroup$
– kimchi lover
Mar 22 at 15:58
$begingroup$
The condition for $mu_Sllmu$ is that whenever $mu(A)=0$ we also have $mu_S(A)=0$, too. Which is satisfied by your $mu_S$.
$endgroup$
– kimchi lover
Mar 22 at 15:58
$begingroup$
Could you clarify why $mu_{S} ll mu$? It seems to me that the relation should be $mu_{s} = frac{1}{2} mu$ which still holds when $mu (A) = 0$
$endgroup$
– David G.
Mar 22 at 16:11
$begingroup$
Could you clarify why $mu_{S} ll mu$? It seems to me that the relation should be $mu_{s} = frac{1}{2} mu$ which still holds when $mu (A) = 0$
$endgroup$
– David G.
Mar 22 at 16:11
|
show 1 more comment
$begingroup$
By a standard argument $mu(Scap (a,b))=mu(S^{c}cap (a,b))$ for every subinterval $(a,b)$ implies $mu(Scap A)=mu(S^{c}cap A)$ for every Bore set $A$. Taking $A=S$ we get a contradiction.
answered Mar 22 at 6:06
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
$endgroup$
$begingroup$
I understand your point, however, I'm inclined to believe that $S$ cannot equal $A$ as I don't see how $S$ is an interval.
$endgroup$
– David G.
Mar 22 at 6:38
$begingroup$
If you extend the equality from intervals to measurable sets you can then take $A=S$.
$endgroup$
– Kavi Rama Murthy
Mar 22 at 6:40
$begingroup$
Though this is true, that breaks the statement of the question and circumvents the basis of the question. The subintervals are there to illustrated uniform distribution of points across the unit interval
$endgroup$
– David G.
Mar 22 at 7:24
add a comment |
$begingroup$
By a standard argument $mu(Scap (a,b))=mu(S^{c}cap (a,b))$ for every subinterval $(a,b)$ implies $mu(Scap A)=mu(S^{c}cap A)$ for every Bore set $A$. Taking $A=S$ we get a contradiction.
answered Mar 22 at 6:06
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
$endgroup$
$begingroup$
I understand your point, however, I'm inclined to believe that $S$ cannot equal $A$ as I don't see how $S$ is an interval.
$endgroup$
– David G.
Mar 22 at 6:38
$begingroup$
If you extend the equality from intervals to measurable sets you can then take $A=S$.
$endgroup$
– Kavi Rama Murthy
Mar 22 at 6:40
$begingroup$
Though this is true, that breaks the statement of the question and circumvents the basis of the question. The subintervals are there to illustrated uniform distribution of points across the unit interval
$endgroup$
– David G.
Mar 22 at 7:24
add a comment |
$begingroup$
By a standard argument $mu(Scap (a,b))=mu(S^{c}cap (a,b))$ for every subinterval $(a,b)$ implies $mu(Scap A)=mu(S^{c}cap A)$ for every Bore set $A$. Taking $A=S$ we get a contradiction.
answered Mar 22 at 6:06
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
$endgroup$
By a standard argument $mu(Scap (a,b))=mu(S^{c}cap (a,b))$ for every subinterval $(a,b)$ implies $mu(Scap A)=mu(S^{c}cap A)$ for every Bore set $A$. Taking $A=S$ we get a contradiction.
answered Mar 22 at 6:06
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
answered Mar 22 at 6:06
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
answered Mar 22 at 6:06
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
answered Mar 22 at 6:06
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
74.1k53270
$begingroup$
I understand your point, however, I'm inclined to believe that $S$ cannot equal $A$ as I don't see how $S$ is an interval.
$endgroup$
– David G.
Mar 22 at 6:38
$begingroup$
If you extend the equality from intervals to measurable sets you can then take $A=S$.
$endgroup$
– Kavi Rama Murthy
Mar 22 at 6:40
$begingroup$
Though this is true, that breaks the statement of the question and circumvents the basis of the question. The subintervals are there to illustrated uniform distribution of points across the unit interval
$endgroup$
– David G.
Mar 22 at 7:24
add a comment |
$begingroup$
I understand your point, however, I'm inclined to believe that $S$ cannot equal $A$ as I don't see how $S$ is an interval.
$endgroup$
– David G.
Mar 22 at 6:38
$begingroup$
If you extend the equality from intervals to measurable sets you can then take $A=S$.
$endgroup$
– Kavi Rama Murthy
Mar 22 at 6:40
$begingroup$
Though this is true, that breaks the statement of the question and circumvents the basis of the question. The subintervals are there to illustrated uniform distribution of points across the unit interval
$endgroup$
– David G.
Mar 22 at 7:24
$begingroup$
I understand your point, however, I'm inclined to believe that $S$ cannot equal $A$ as I don't see how $S$ is an interval.
$endgroup$
– David G.
Mar 22 at 6:38
$begingroup$
I understand your point, however, I'm inclined to believe that $S$ cannot equal $A$ as I don't see how $S$ is an interval.
$endgroup$
– David G.
Mar 22 at 6:38
$begingroup$
If you extend the equality from intervals to measurable sets you can then take $A=S$.
$endgroup$
– Kavi Rama Murthy
Mar 22 at 6:40
$begingroup$
If you extend the equality from intervals to measurable sets you can then take $A=S$.
$endgroup$
– Kavi Rama Murthy
Mar 22 at 6:40
$begingroup$
Though this is true, that breaks the statement of the question and circumvents the basis of the question. The subintervals are there to illustrated uniform distribution of points across the unit interval
$endgroup$
– David G.
Mar 22 at 7:24
$begingroup$
Though this is true, that breaks the statement of the question and circumvents the basis of the question. The subintervals are there to illustrated uniform distribution of points across the unit interval
$endgroup$
– David G.
Mar 22 at 7:24
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%2f3157611%2fset-and-set-complement-of-uniform-distribution-on-0-1%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
