Covering measure zero set by finitely many intervals which sum to desired length
up vote
1
down vote
favorite
I know that by definition of Lebesgue measure, if the set N has Lebesgue measure zero i.e. m(N) = 0 then we can cover the set by countably many open intervals {a$_{j}$, b$_{j}$}$_{jinmathbb{N}}$ such that $sum_{j}$ b$_{j}$ - a$_{j}$ $<$ $epsilon$.
I was wondering if the set N is contained in some bounded interval [a,b], would I be able to replace the countable open cover by finitely many open (left open or right open) intervals.
In other words, is it possible to say that I can cover N by finitely many intervals {a$_{j}$, b$_{j}$}$_{j=1}^{n}$ such that $sum_{j=1}^{n}$ b$_{j}$ - a$_{j}$ $<$ $epsilon$.
measure-theory lebesgue-measure
asked Nov 27 at 3:10
HumbleStudent
660311
add a comment |
up vote
1
down vote
favorite
I know that by definition of Lebesgue measure, if the set N has Lebesgue measure zero i.e. m(N) = 0 then we can cover the set by countably many open intervals {a$_{j}$, b$_{j}$}$_{jinmathbb{N}}$ such that $sum_{j}$ b$_{j}$ - a$_{j}$ $<$ $epsilon$.
I was wondering if the set N is contained in some bounded interval [a,b], would I be able to replace the countable open cover by finitely many open (left open or right open) intervals.
In other words, is it possible to say that I can cover N by finitely many intervals {a$_{j}$, b$_{j}$}$_{j=1}^{n}$ such that $sum_{j=1}^{n}$ b$_{j}$ - a$_{j}$ $<$ $epsilon$.
measure-theory lebesgue-measure
asked Nov 27 at 3:10
HumbleStudent
660311
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I know that by definition of Lebesgue measure, if the set N has Lebesgue measure zero i.e. m(N) = 0 then we can cover the set by countably many open intervals {a$_{j}$, b$_{j}$}$_{jinmathbb{N}}$ such that $sum_{j}$ b$_{j}$ - a$_{j}$ $<$ $epsilon$.
I was wondering if the set N is contained in some bounded interval [a,b], would I be able to replace the countable open cover by finitely many open (left open or right open) intervals.
In other words, is it possible to say that I can cover N by finitely many intervals {a$_{j}$, b$_{j}$}$_{j=1}^{n}$ such that $sum_{j=1}^{n}$ b$_{j}$ - a$_{j}$ $<$ $epsilon$.
measure-theory lebesgue-measure
asked Nov 27 at 3:10
HumbleStudent
660311
I know that by definition of Lebesgue measure, if the set N has Lebesgue measure zero i.e. m(N) = 0 then we can cover the set by countably many open intervals {a$_{j}$, b$_{j}$}$_{jinmathbb{N}}$ such that $sum_{j}$ b$_{j}$ - a$_{j}$ $<$ $epsilon$.
I was wondering if the set N is contained in some bounded interval [a,b], would I be able to replace the countable open cover by finitely many open (left open or right open) intervals.
In other words, is it possible to say that I can cover N by finitely many intervals {a$_{j}$, b$_{j}$}$_{j=1}^{n}$ such that $sum_{j=1}^{n}$ b$_{j}$ - a$_{j}$ $<$ $epsilon$.
measure-theory lebesgue-measure
measure-theory lebesgue-measure
asked Nov 27 at 3:10
HumbleStudent
660311
asked Nov 27 at 3:10
HumbleStudent
660311
asked Nov 27 at 3:10
HumbleStudent
660311
asked Nov 27 at 3:10
HumbleStudent
660311
asked Nov 27 at 3:10
HumbleStudent
660311
660311
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
The quick answer is no.
Here is an interesting exercise: see if you can prove that if $N$ is the set of rational numbers in between $0$ and $1$ and $N subset bigcup_{k=1}^n (a_k,b_k)$, then $sum_{k=1}^n (b_k - a_k) ge 1$.
answered Nov 27 at 3:27
Umberto P.
38.3k13063
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The quick answer is no.
Here is an interesting exercise: see if you can prove that if $N$ is the set of rational numbers in between $0$ and $1$ and $N subset bigcup_{k=1}^n (a_k,b_k)$, then $sum_{k=1}^n (b_k - a_k) ge 1$.
answered Nov 27 at 3:27
Umberto P.
38.3k13063
add a comment |
up vote
2
down vote
accepted
The quick answer is no.
Here is an interesting exercise: see if you can prove that if $N$ is the set of rational numbers in between $0$ and $1$ and $N subset bigcup_{k=1}^n (a_k,b_k)$, then $sum_{k=1}^n (b_k - a_k) ge 1$.
answered Nov 27 at 3:27
Umberto P.
38.3k13063
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The quick answer is no.
Here is an interesting exercise: see if you can prove that if $N$ is the set of rational numbers in between $0$ and $1$ and $N subset bigcup_{k=1}^n (a_k,b_k)$, then $sum_{k=1}^n (b_k - a_k) ge 1$.
answered Nov 27 at 3:27
Umberto P.
38.3k13063
The quick answer is no.
Here is an interesting exercise: see if you can prove that if $N$ is the set of rational numbers in between $0$ and $1$ and $N subset bigcup_{k=1}^n (a_k,b_k)$, then $sum_{k=1}^n (b_k - a_k) ge 1$.
answered Nov 27 at 3:27
Umberto P.
38.3k13063
answered Nov 27 at 3:27
Umberto P.
38.3k13063
answered Nov 27 at 3:27
Umberto P.
38.3k13063
answered Nov 27 at 3:27
Umberto P.
38.3k13063
38.3k13063
add a comment |
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3015278%2fcovering-measure-zero-set-by-finitely-many-intervals-which-sum-to-desired-length%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
