Covering measure zero set by finitely many intervals which sum to desired length
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
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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 |
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
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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.
38.5k13064
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
38.5k13064
add a comment |
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.
38.5k13064
add a comment |
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.
38.5k13064
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.
38.5k13064
answered Nov 27 at 3:27
Umberto P.
38.5k13064
answered Nov 27 at 3:27
Umberto P.
38.5k13064
answered Nov 27 at 3:27
Umberto P.
38.5k13064
38.5k13064
add a comment |
add a comment |
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