Any countable set has measurable zero
1
$begingroup$
To demonstrate that ANY countable set has measure zero, is it sufficient to show that the natural numbers have a measure zero? If so, why; and, if not, why not?
Thank you :)
real-analysis general-topology elementary-set-theory
asked Dec 10 '18 at 1:49
Rafael VergnaudRafael Vergnaud
348217
$endgroup$
|
show 1 more comment
1
$begingroup$
To demonstrate that ANY countable set has measure zero, is it sufficient to show that the natural numbers have a measure zero? If so, why; and, if not, why not?
Thank you :)
real-analysis general-topology elementary-set-theory
asked Dec 10 '18 at 1:49
Rafael VergnaudRafael Vergnaud
348217
$endgroup$
1
$begingroup$
Are you referring specifically to Lebesgue measure? Because this isn't true for an arbitrary measure.
$endgroup$
– carmichael561
Dec 10 '18 at 2:04
1
$begingroup$
No, it is not sufficient. But, if you have a countable set, you can write it as something similar to the natural numbers. Specifically, you can sequence the set, so that each element has a unique corresponding natural number, i.e $S = {a_1, a_2 ,a_3, cdots}$
$endgroup$
– rubikscube09
Dec 10 '18 at 2:05
$begingroup$
Hey Carmichael. I'm not sure. My book has only mentioned the term "measure zero," without additional qualifying adjectives. To quote, "A set $A subseteq mathbb{R}$ has measure zero if, for all $epsilon > 0,$ there exists a countable collection of open intervals $O_n$ with the property that $A$ is contained in the union of all of the intervals $O_n$ and the sum of the lengths of all of the intervals is less than or equal to $epsilon.$"
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:06
1
$begingroup$
Yes it is identical. To answer the other question, your book is referring to the Lebesgue measure on $mathbb{R}$. There are other measures of course.
$endgroup$
– rubikscube09
Dec 10 '18 at 2:14
1
$begingroup$
Thanks, Rubik! :)
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:14
|
show 1 more comment
1
1
1
1
$begingroup$
To demonstrate that ANY countable set has measure zero, is it sufficient to show that the natural numbers have a measure zero? If so, why; and, if not, why not?
Thank you :)
real-analysis general-topology elementary-set-theory
asked Dec 10 '18 at 1:49
Rafael VergnaudRafael Vergnaud
348217
$endgroup$
To demonstrate that ANY countable set has measure zero, is it sufficient to show that the natural numbers have a measure zero? If so, why; and, if not, why not?
Thank you :)
real-analysis general-topology elementary-set-theory
real-analysis general-topology elementary-set-theory
asked Dec 10 '18 at 1:49
Rafael VergnaudRafael Vergnaud
348217
asked Dec 10 '18 at 1:49
Rafael VergnaudRafael Vergnaud
348217
asked Dec 10 '18 at 1:49
Rafael VergnaudRafael Vergnaud
348217
asked Dec 10 '18 at 1:49
Rafael VergnaudRafael Vergnaud
348217
asked Dec 10 '18 at 1:49
Rafael VergnaudRafael Vergnaud
348217
348217
1
$begingroup$
Are you referring specifically to Lebesgue measure? Because this isn't true for an arbitrary measure.
$endgroup$
– carmichael561
Dec 10 '18 at 2:04
1
$begingroup$
No, it is not sufficient. But, if you have a countable set, you can write it as something similar to the natural numbers. Specifically, you can sequence the set, so that each element has a unique corresponding natural number, i.e $S = {a_1, a_2 ,a_3, cdots}$
$endgroup$
– rubikscube09
Dec 10 '18 at 2:05
$begingroup$
Hey Carmichael. I'm not sure. My book has only mentioned the term "measure zero," without additional qualifying adjectives. To quote, "A set $A subseteq mathbb{R}$ has measure zero if, for all $epsilon > 0,$ there exists a countable collection of open intervals $O_n$ with the property that $A$ is contained in the union of all of the intervals $O_n$ and the sum of the lengths of all of the intervals is less than or equal to $epsilon.$"
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:06
1
$begingroup$
Yes it is identical. To answer the other question, your book is referring to the Lebesgue measure on $mathbb{R}$. There are other measures of course.
$endgroup$
– rubikscube09
Dec 10 '18 at 2:14
1
$begingroup$
Thanks, Rubik! :)
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:14
|
show 1 more comment
1
$begingroup$
Are you referring specifically to Lebesgue measure? Because this isn't true for an arbitrary measure.
$endgroup$
– carmichael561
Dec 10 '18 at 2:04
1
$begingroup$
No, it is not sufficient. But, if you have a countable set, you can write it as something similar to the natural numbers. Specifically, you can sequence the set, so that each element has a unique corresponding natural number, i.e $S = {a_1, a_2 ,a_3, cdots}$
$endgroup$
– rubikscube09
Dec 10 '18 at 2:05
$begingroup$
Hey Carmichael. I'm not sure. My book has only mentioned the term "measure zero," without additional qualifying adjectives. To quote, "A set $A subseteq mathbb{R}$ has measure zero if, for all $epsilon > 0,$ there exists a countable collection of open intervals $O_n$ with the property that $A$ is contained in the union of all of the intervals $O_n$ and the sum of the lengths of all of the intervals is less than or equal to $epsilon.$"
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:06
1
$begingroup$
Yes it is identical. To answer the other question, your book is referring to the Lebesgue measure on $mathbb{R}$. There are other measures of course.
$endgroup$
– rubikscube09
Dec 10 '18 at 2:14
1
$begingroup$
Thanks, Rubik! :)
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:14
1
1
$begingroup$
Are you referring specifically to Lebesgue measure? Because this isn't true for an arbitrary measure.
$endgroup$
– carmichael561
Dec 10 '18 at 2:04
$begingroup$
Are you referring specifically to Lebesgue measure? Because this isn't true for an arbitrary measure.
$endgroup$
– carmichael561
Dec 10 '18 at 2:04
1
1
$begingroup$
No, it is not sufficient. But, if you have a countable set, you can write it as something similar to the natural numbers. Specifically, you can sequence the set, so that each element has a unique corresponding natural number, i.e $S = {a_1, a_2 ,a_3, cdots}$
$endgroup$
– rubikscube09
Dec 10 '18 at 2:05
$begingroup$
No, it is not sufficient. But, if you have a countable set, you can write it as something similar to the natural numbers. Specifically, you can sequence the set, so that each element has a unique corresponding natural number, i.e $S = {a_1, a_2 ,a_3, cdots}$
$endgroup$
– rubikscube09
Dec 10 '18 at 2:05
$begingroup$
Hey Carmichael. I'm not sure. My book has only mentioned the term "measure zero," without additional qualifying adjectives. To quote, "A set $A subseteq mathbb{R}$ has measure zero if, for all $epsilon > 0,$ there exists a countable collection of open intervals $O_n$ with the property that $A$ is contained in the union of all of the intervals $O_n$ and the sum of the lengths of all of the intervals is less than or equal to $epsilon.$"
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:06
$begingroup$
Hey Carmichael. I'm not sure. My book has only mentioned the term "measure zero," without additional qualifying adjectives. To quote, "A set $A subseteq mathbb{R}$ has measure zero if, for all $epsilon > 0,$ there exists a countable collection of open intervals $O_n$ with the property that $A$ is contained in the union of all of the intervals $O_n$ and the sum of the lengths of all of the intervals is less than or equal to $epsilon.$"
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:06
1
1
$begingroup$
Yes it is identical. To answer the other question, your book is referring to the Lebesgue measure on $mathbb{R}$. There are other measures of course.
$endgroup$
– rubikscube09
Dec 10 '18 at 2:14
$begingroup$
Yes it is identical. To answer the other question, your book is referring to the Lebesgue measure on $mathbb{R}$. There are other measures of course.
$endgroup$
– rubikscube09
Dec 10 '18 at 2:14
1
1
$begingroup$
Thanks, Rubik! :)
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:14
$begingroup$
Thanks, Rubik! :)
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:14
|
show 1 more comment
1 Answer
1
active
oldest
votes
2
$begingroup$
(Asuuming we are talking about Lebesgue measur)
Let $a_i$ be a sequence of a countable set $A$, then $A_i=(a_i-2^{-i-1}varepsilon,a_i+2^{-i-1}varepsilon)$.
Now $Asubseteq bigcup A_i$, and $muleft(bigcup A_iright)lesum2^{-i}varepsilon=varepsilon$
Showing that $Bbb N$ is not enough because measure does not preserve by bijection, a simple example is the cantor set and the interval [0,1], where they have the same cardinality(hence there exists bijction between them) but one is measure 0 and the other 1.
answered Dec 10 '18 at 2:10
HoloHolo
5,75421131
$endgroup$
$begingroup$
Hey, Holo. Thanks for the response. My proof was the following: take the union of the intervals $(a_i - frac{epsilon}{2^{i+1}}, a_i + frac{epsilon}{2^{i+1}}).$ The sum would be $sum_{i = 1}^{infty} frac{epsilon}{2} (frac{1}{2})^{i} = epsilon.$ Is this correct? My summation looks different than your's!
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:11
$begingroup$
@RafaelVergnaud you did the exact same things as I did, but you have a little(algebra) mistake, the summation of yours should be with $[...])^{i-1}$, then the $epsilon/2$ will make it $[...])^i$
$endgroup$
– Holo
Dec 10 '18 at 2:14
1
$begingroup$
Right. Got it! Thanks a lot Holo!
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:15
1
$begingroup$
@RafaelVergnaud I added an example where 2 sets has the same cardinality but not the same measure to explain why to show that $mu(Bbb N)=0$ is not enough
$endgroup$
– Holo
Dec 10 '18 at 2:29
add a comment |
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$begingroup$
(Asuuming we are talking about Lebesgue measur)
Let $a_i$ be a sequence of a countable set $A$, then $A_i=(a_i-2^{-i-1}varepsilon,a_i+2^{-i-1}varepsilon)$.
Now $Asubseteq bigcup A_i$, and $muleft(bigcup A_iright)lesum2^{-i}varepsilon=varepsilon$
Showing that $Bbb N$ is not enough because measure does not preserve by bijection, a simple example is the cantor set and the interval [0,1], where they have the same cardinality(hence there exists bijction between them) but one is measure 0 and the other 1.
answered Dec 10 '18 at 2:10
HoloHolo
5,75421131
$endgroup$
$begingroup$
Hey, Holo. Thanks for the response. My proof was the following: take the union of the intervals $(a_i - frac{epsilon}{2^{i+1}}, a_i + frac{epsilon}{2^{i+1}}).$ The sum would be $sum_{i = 1}^{infty} frac{epsilon}{2} (frac{1}{2})^{i} = epsilon.$ Is this correct? My summation looks different than your's!
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:11
$begingroup$
@RafaelVergnaud you did the exact same things as I did, but you have a little(algebra) mistake, the summation of yours should be with $[...])^{i-1}$, then the $epsilon/2$ will make it $[...])^i$
$endgroup$
– Holo
Dec 10 '18 at 2:14
1
$begingroup$
Right. Got it! Thanks a lot Holo!
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:15
1
$begingroup$
@RafaelVergnaud I added an example where 2 sets has the same cardinality but not the same measure to explain why to show that $mu(Bbb N)=0$ is not enough
$endgroup$
– Holo
Dec 10 '18 at 2:29
add a comment |
2
$begingroup$
(Asuuming we are talking about Lebesgue measur)
Let $a_i$ be a sequence of a countable set $A$, then $A_i=(a_i-2^{-i-1}varepsilon,a_i+2^{-i-1}varepsilon)$.
Now $Asubseteq bigcup A_i$, and $muleft(bigcup A_iright)lesum2^{-i}varepsilon=varepsilon$
Showing that $Bbb N$ is not enough because measure does not preserve by bijection, a simple example is the cantor set and the interval [0,1], where they have the same cardinality(hence there exists bijction between them) but one is measure 0 and the other 1.
answered Dec 10 '18 at 2:10
HoloHolo
5,75421131
$endgroup$
$begingroup$
Hey, Holo. Thanks for the response. My proof was the following: take the union of the intervals $(a_i - frac{epsilon}{2^{i+1}}, a_i + frac{epsilon}{2^{i+1}}).$ The sum would be $sum_{i = 1}^{infty} frac{epsilon}{2} (frac{1}{2})^{i} = epsilon.$ Is this correct? My summation looks different than your's!
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:11
$begingroup$
@RafaelVergnaud you did the exact same things as I did, but you have a little(algebra) mistake, the summation of yours should be with $[...])^{i-1}$, then the $epsilon/2$ will make it $[...])^i$
$endgroup$
– Holo
Dec 10 '18 at 2:14
1
$begingroup$
Right. Got it! Thanks a lot Holo!
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:15
1
$begingroup$
@RafaelVergnaud I added an example where 2 sets has the same cardinality but not the same measure to explain why to show that $mu(Bbb N)=0$ is not enough
$endgroup$
– Holo
Dec 10 '18 at 2:29
add a comment |
2
2
2
$begingroup$
(Asuuming we are talking about Lebesgue measur)
Let $a_i$ be a sequence of a countable set $A$, then $A_i=(a_i-2^{-i-1}varepsilon,a_i+2^{-i-1}varepsilon)$.
Now $Asubseteq bigcup A_i$, and $muleft(bigcup A_iright)lesum2^{-i}varepsilon=varepsilon$
Showing that $Bbb N$ is not enough because measure does not preserve by bijection, a simple example is the cantor set and the interval [0,1], where they have the same cardinality(hence there exists bijction between them) but one is measure 0 and the other 1.
answered Dec 10 '18 at 2:10
HoloHolo
5,75421131
$endgroup$
(Asuuming we are talking about Lebesgue measur)
Let $a_i$ be a sequence of a countable set $A$, then $A_i=(a_i-2^{-i-1}varepsilon,a_i+2^{-i-1}varepsilon)$.
Now $Asubseteq bigcup A_i$, and $muleft(bigcup A_iright)lesum2^{-i}varepsilon=varepsilon$
Showing that $Bbb N$ is not enough because measure does not preserve by bijection, a simple example is the cantor set and the interval [0,1], where they have the same cardinality(hence there exists bijction between them) but one is measure 0 and the other 1.
answered Dec 10 '18 at 2:10
HoloHolo
5,75421131
edited Dec 10 '18 at 2:29
answered Dec 10 '18 at 2:10
HoloHolo
5,75421131
answered Dec 10 '18 at 2:10
HoloHolo
5,75421131
answered Dec 10 '18 at 2:10
HoloHolo
5,75421131
5,75421131
$begingroup$
Hey, Holo. Thanks for the response. My proof was the following: take the union of the intervals $(a_i - frac{epsilon}{2^{i+1}}, a_i + frac{epsilon}{2^{i+1}}).$ The sum would be $sum_{i = 1}^{infty} frac{epsilon}{2} (frac{1}{2})^{i} = epsilon.$ Is this correct? My summation looks different than your's!
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:11
$begingroup$
@RafaelVergnaud you did the exact same things as I did, but you have a little(algebra) mistake, the summation of yours should be with $[...])^{i-1}$, then the $epsilon/2$ will make it $[...])^i$
$endgroup$
– Holo
Dec 10 '18 at 2:14
1
$begingroup$
Right. Got it! Thanks a lot Holo!
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:15
1
$begingroup$
@RafaelVergnaud I added an example where 2 sets has the same cardinality but not the same measure to explain why to show that $mu(Bbb N)=0$ is not enough
$endgroup$
– Holo
Dec 10 '18 at 2:29
add a comment |
$begingroup$
Hey, Holo. Thanks for the response. My proof was the following: take the union of the intervals $(a_i - frac{epsilon}{2^{i+1}}, a_i + frac{epsilon}{2^{i+1}}).$ The sum would be $sum_{i = 1}^{infty} frac{epsilon}{2} (frac{1}{2})^{i} = epsilon.$ Is this correct? My summation looks different than your's!
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:11
$begingroup$
@RafaelVergnaud you did the exact same things as I did, but you have a little(algebra) mistake, the summation of yours should be with $[...])^{i-1}$, then the $epsilon/2$ will make it $[...])^i$
$endgroup$
– Holo
Dec 10 '18 at 2:14
1
$begingroup$
Right. Got it! Thanks a lot Holo!
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:15
1
$begingroup$
@RafaelVergnaud I added an example where 2 sets has the same cardinality but not the same measure to explain why to show that $mu(Bbb N)=0$ is not enough
$endgroup$
– Holo
Dec 10 '18 at 2:29
$begingroup$
Hey, Holo. Thanks for the response. My proof was the following: take the union of the intervals $(a_i - frac{epsilon}{2^{i+1}}, a_i + frac{epsilon}{2^{i+1}}).$ The sum would be $sum_{i = 1}^{infty} frac{epsilon}{2} (frac{1}{2})^{i} = epsilon.$ Is this correct? My summation looks different than your's!
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:11
$begingroup$
Hey, Holo. Thanks for the response. My proof was the following: take the union of the intervals $(a_i - frac{epsilon}{2^{i+1}}, a_i + frac{epsilon}{2^{i+1}}).$ The sum would be $sum_{i = 1}^{infty} frac{epsilon}{2} (frac{1}{2})^{i} = epsilon.$ Is this correct? My summation looks different than your's!
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:11
$begingroup$
@RafaelVergnaud you did the exact same things as I did, but you have a little(algebra) mistake, the summation of yours should be with $[...])^{i-1}$, then the $epsilon/2$ will make it $[...])^i$
$endgroup$
– Holo
Dec 10 '18 at 2:14
$begingroup$
@RafaelVergnaud you did the exact same things as I did, but you have a little(algebra) mistake, the summation of yours should be with $[...])^{i-1}$, then the $epsilon/2$ will make it $[...])^i$
$endgroup$
– Holo
Dec 10 '18 at 2:14
1
1
$begingroup$
Right. Got it! Thanks a lot Holo!
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:15
$begingroup$
Right. Got it! Thanks a lot Holo!
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:15
1
1
$begingroup$
@RafaelVergnaud I added an example where 2 sets has the same cardinality but not the same measure to explain why to show that $mu(Bbb N)=0$ is not enough
$endgroup$
– Holo
Dec 10 '18 at 2:29
$begingroup$
@RafaelVergnaud I added an example where 2 sets has the same cardinality but not the same measure to explain why to show that $mu(Bbb N)=0$ is not enough
$endgroup$
– Holo
Dec 10 '18 at 2:29
add a comment |
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1
$begingroup$
Are you referring specifically to Lebesgue measure? Because this isn't true for an arbitrary measure.
$endgroup$
– carmichael561
Dec 10 '18 at 2:04
1
$begingroup$
No, it is not sufficient. But, if you have a countable set, you can write it as something similar to the natural numbers. Specifically, you can sequence the set, so that each element has a unique corresponding natural number, i.e $S = {a_1, a_2 ,a_3, cdots}$
$endgroup$
– rubikscube09
Dec 10 '18 at 2:05
$begingroup$
Hey Carmichael. I'm not sure. My book has only mentioned the term "measure zero," without additional qualifying adjectives. To quote, "A set $A subseteq mathbb{R}$ has measure zero if, for all $epsilon > 0,$ there exists a countable collection of open intervals $O_n$ with the property that $A$ is contained in the union of all of the intervals $O_n$ and the sum of the lengths of all of the intervals is less than or equal to $epsilon.$"
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:06
1
$begingroup$
Yes it is identical. To answer the other question, your book is referring to the Lebesgue measure on $mathbb{R}$. There are other measures of course.
$endgroup$
– rubikscube09
Dec 10 '18 at 2:14
1
$begingroup$
Thanks, Rubik! :)
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:14