Mixing
Borel Sigma Algebra generated by Open Intervals
So I know that the Borel $sigma$-algebra of $mathbb{R}$ is the $sigma$-algebra generated by open sets. I have been able to prove that this Borel $sigma$-algebra is also generated by the family of open intervals of the form $(a,b), a,b in mathbb{R}$ Now I want to show that the family of open intervals $(a,infty)$ also generate the Borel $sigma$-algebra. So what I have done is first show that
$(a,b) = bigcup_{q in mathbb{Q}, q<b} ((a, infty)-(q,infty))$. This shows that every open interval is a countable union of intervals of the form $(a,infty)$ and thus $sigma$-algebra generated by $(a,b)$ is contained in the $sigma$-algebra generated by $(a, infty)$ But how would I show the other way around, i.e. that every interval $(a, infty)$ is in the Borel $sigma$- algebra?
(Can we just say that every interval $(a, infty)$ is infact an open interval in itself and thus belongs to the $sigma$-algebra generated by $(a,b)$?)
measure-theory
asked Apr 18 '15 at 7:24
user1314
628719
add a comment |
So I know that the Borel $sigma$-algebra of $mathbb{R}$ is the $sigma$-algebra generated by open sets. I have been able to prove that this Borel $sigma$-algebra is also generated by the family of open intervals of the form $(a,b), a,b in mathbb{R}$ Now I want to show that the family of open intervals $(a,infty)$ also generate the Borel $sigma$-algebra. So what I have done is first show that
$(a,b) = bigcup_{q in mathbb{Q}, q<b} ((a, infty)-(q,infty))$. This shows that every open interval is a countable union of intervals of the form $(a,infty)$ and thus $sigma$-algebra generated by $(a,b)$ is contained in the $sigma$-algebra generated by $(a, infty)$ But how would I show the other way around, i.e. that every interval $(a, infty)$ is in the Borel $sigma$- algebra?
(Can we just say that every interval $(a, infty)$ is infact an open interval in itself and thus belongs to the $sigma$-algebra generated by $(a,b)$?)
measure-theory
asked Apr 18 '15 at 7:24
user1314
628719
add a comment |
So I know that the Borel $sigma$-algebra of $mathbb{R}$ is the $sigma$-algebra generated by open sets. I have been able to prove that this Borel $sigma$-algebra is also generated by the family of open intervals of the form $(a,b), a,b in mathbb{R}$ Now I want to show that the family of open intervals $(a,infty)$ also generate the Borel $sigma$-algebra. So what I have done is first show that
$(a,b) = bigcup_{q in mathbb{Q}, q<b} ((a, infty)-(q,infty))$. This shows that every open interval is a countable union of intervals of the form $(a,infty)$ and thus $sigma$-algebra generated by $(a,b)$ is contained in the $sigma$-algebra generated by $(a, infty)$ But how would I show the other way around, i.e. that every interval $(a, infty)$ is in the Borel $sigma$- algebra?
(Can we just say that every interval $(a, infty)$ is infact an open interval in itself and thus belongs to the $sigma$-algebra generated by $(a,b)$?)
measure-theory
asked Apr 18 '15 at 7:24
user1314
628719
So I know that the Borel $sigma$-algebra of $mathbb{R}$ is the $sigma$-algebra generated by open sets. I have been able to prove that this Borel $sigma$-algebra is also generated by the family of open intervals of the form $(a,b), a,b in mathbb{R}$ Now I want to show that the family of open intervals $(a,infty)$ also generate the Borel $sigma$-algebra. So what I have done is first show that
$(a,b) = bigcup_{q in mathbb{Q}, q<b} ((a, infty)-(q,infty))$. This shows that every open interval is a countable union of intervals of the form $(a,infty)$ and thus $sigma$-algebra generated by $(a,b)$ is contained in the $sigma$-algebra generated by $(a, infty)$ But how would I show the other way around, i.e. that every interval $(a, infty)$ is in the Borel $sigma$- algebra?
(Can we just say that every interval $(a, infty)$ is infact an open interval in itself and thus belongs to the $sigma$-algebra generated by $(a,b)$?)
measure-theory
measure-theory
asked Apr 18 '15 at 7:24
user1314
628719
asked Apr 18 '15 at 7:24
user1314
628719
asked Apr 18 '15 at 7:24
user1314
628719
asked Apr 18 '15 at 7:24
user1314
628719
asked Apr 18 '15 at 7:24
user1314
628719
628719
add a comment |
add a comment |
2 Answers
2
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Hint: $$(a,infty) = bigcup_{n in mathbb{N}_0} (a+n,a+n+1).$$
And yes: If you know that the Borel-$sigma$ algebra contains all open sets, then $(a,infty) in mathcal{B}(mathbb{R})$ follows also from the fact that $(a,infty)$ is an open set.
answered Apr 18 '15 at 7:31
saz
78.2k758123
Ok, so even though the identity you gave is correct I think I will just use the fact $(a, infty)$ is an open interval, hence an open set. Just one last question, how would I show the above for $[a, infty)$. I can express $(a,b)$ as a union of intervals of the form $[a, infty)$ but can't show the other way around
– user1314
Apr 18 '15 at 7:36
@user1314 Hint: Use $$[a,infty) = bigcap_{n in mathbb{N}} (a- frac{1}{n},infty)$$
– saz
Apr 18 '15 at 7:43
By the way a small correction, a bit pedantic maybe, but in the above formua, don't you mean $nin mathbb{N}_{0}$, since you used that in the earlier one.
– user1314
Apr 18 '15 at 7:50
@user1314 You mean the formula in my previous comment? No, I don't mean $n in mathbb{N}_0$ - $frac{1}{n}$ is not well-defined for $n=0$.
– saz
Apr 18 '15 at 7:52
oh of course, sorry, I kept thinking $mathbb{N}_{0}$ means $mathbb{N}$ without the 0, yes you are absolutely correct. Thanks for the help
– user1314
Apr 18 '15 at 7:54
|
show 1 more comment
Denote $mathcal{B}=sigmaleft(tauright)=sigmaleft(left{ left(a,bright)mid a,binmathbb{R}right} right)$ (this equality was found out by you allready)
where $tau$ denotes the topology and $mathcal{B}_{1}=sigmaleft(left{ left(a,inftyright)mid ainmathbb{R}right} right)$.
Then $left{ left(a,inftyright)mid ainmathbb{R}right} subsettau$
so that $mathcal{B}_{1}subseteqmathcal{B}$.
To prove the converse
inclusion it is enough to show that $left{ left(a,bright)mid a,binmathbb{R}right} subsetmathcal{B}_{1}$.
We have $left(a,cright]=left(a,inftyright)-left(c,inftyright)inmathcal{B}_{1}$
for each $c$, and consequently $left(a,bright)=bigcup_{ninmathbb{N}}left(a,b-frac{1}{n}right]inmathcal{B}_{1}$.
answered Apr 18 '15 at 7:54
drhab
97.9k544129
I don't see any contradiction: $left(a,bright]=cap_{ninmathbb{N}}left(a,b+frac{1}{n}right]inmathcal{B}_{1}$. Maybe, $left(a,bright)=cup_{ninmathbb{N}}left(a,b-frac{1}{n}right]inmathcal{B}_{1}$.
– Aidas
Nov 20 '18 at 15:38
@Aidas You are right, and thank you for attending me. I repaired. I really don't understand (and remember) how I could write that.
– drhab
Nov 20 '18 at 16:12
add a comment |
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2 Answers
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Hint: $$(a,infty) = bigcup_{n in mathbb{N}_0} (a+n,a+n+1).$$
And yes: If you know that the Borel-$sigma$ algebra contains all open sets, then $(a,infty) in mathcal{B}(mathbb{R})$ follows also from the fact that $(a,infty)$ is an open set.
answered Apr 18 '15 at 7:31
saz
78.2k758123
Ok, so even though the identity you gave is correct I think I will just use the fact $(a, infty)$ is an open interval, hence an open set. Just one last question, how would I show the above for $[a, infty)$. I can express $(a,b)$ as a union of intervals of the form $[a, infty)$ but can't show the other way around
– user1314
Apr 18 '15 at 7:36
@user1314 Hint: Use $$[a,infty) = bigcap_{n in mathbb{N}} (a- frac{1}{n},infty)$$
– saz
Apr 18 '15 at 7:43
By the way a small correction, a bit pedantic maybe, but in the above formua, don't you mean $nin mathbb{N}_{0}$, since you used that in the earlier one.
– user1314
Apr 18 '15 at 7:50
@user1314 You mean the formula in my previous comment? No, I don't mean $n in mathbb{N}_0$ - $frac{1}{n}$ is not well-defined for $n=0$.
– saz
Apr 18 '15 at 7:52
oh of course, sorry, I kept thinking $mathbb{N}_{0}$ means $mathbb{N}$ without the 0, yes you are absolutely correct. Thanks for the help
– user1314
Apr 18 '15 at 7:54
|
show 1 more comment
Hint: $$(a,infty) = bigcup_{n in mathbb{N}_0} (a+n,a+n+1).$$
And yes: If you know that the Borel-$sigma$ algebra contains all open sets, then $(a,infty) in mathcal{B}(mathbb{R})$ follows also from the fact that $(a,infty)$ is an open set.
answered Apr 18 '15 at 7:31
saz
78.2k758123
Ok, so even though the identity you gave is correct I think I will just use the fact $(a, infty)$ is an open interval, hence an open set. Just one last question, how would I show the above for $[a, infty)$. I can express $(a,b)$ as a union of intervals of the form $[a, infty)$ but can't show the other way around
– user1314
Apr 18 '15 at 7:36
@user1314 Hint: Use $$[a,infty) = bigcap_{n in mathbb{N}} (a- frac{1}{n},infty)$$
– saz
Apr 18 '15 at 7:43
By the way a small correction, a bit pedantic maybe, but in the above formua, don't you mean $nin mathbb{N}_{0}$, since you used that in the earlier one.
– user1314
Apr 18 '15 at 7:50
@user1314 You mean the formula in my previous comment? No, I don't mean $n in mathbb{N}_0$ - $frac{1}{n}$ is not well-defined for $n=0$.
– saz
Apr 18 '15 at 7:52
oh of course, sorry, I kept thinking $mathbb{N}_{0}$ means $mathbb{N}$ without the 0, yes you are absolutely correct. Thanks for the help
– user1314
Apr 18 '15 at 7:54
|
show 1 more comment
Hint: $$(a,infty) = bigcup_{n in mathbb{N}_0} (a+n,a+n+1).$$
And yes: If you know that the Borel-$sigma$ algebra contains all open sets, then $(a,infty) in mathcal{B}(mathbb{R})$ follows also from the fact that $(a,infty)$ is an open set.
answered Apr 18 '15 at 7:31
saz
78.2k758123
Hint: $$(a,infty) = bigcup_{n in mathbb{N}_0} (a+n,a+n+1).$$
And yes: If you know that the Borel-$sigma$ algebra contains all open sets, then $(a,infty) in mathcal{B}(mathbb{R})$ follows also from the fact that $(a,infty)$ is an open set.
answered Apr 18 '15 at 7:31
saz
78.2k758123
edited Apr 18 '15 at 7:44
answered Apr 18 '15 at 7:31
saz
78.2k758123
answered Apr 18 '15 at 7:31
saz
78.2k758123
answered Apr 18 '15 at 7:31
saz
78.2k758123
78.2k758123
Ok, so even though the identity you gave is correct I think I will just use the fact $(a, infty)$ is an open interval, hence an open set. Just one last question, how would I show the above for $[a, infty)$. I can express $(a,b)$ as a union of intervals of the form $[a, infty)$ but can't show the other way around
– user1314
Apr 18 '15 at 7:36
@user1314 Hint: Use $$[a,infty) = bigcap_{n in mathbb{N}} (a- frac{1}{n},infty)$$
– saz
Apr 18 '15 at 7:43
By the way a small correction, a bit pedantic maybe, but in the above formua, don't you mean $nin mathbb{N}_{0}$, since you used that in the earlier one.
– user1314
Apr 18 '15 at 7:50
@user1314 You mean the formula in my previous comment? No, I don't mean $n in mathbb{N}_0$ - $frac{1}{n}$ is not well-defined for $n=0$.
– saz
Apr 18 '15 at 7:52
oh of course, sorry, I kept thinking $mathbb{N}_{0}$ means $mathbb{N}$ without the 0, yes you are absolutely correct. Thanks for the help
– user1314
Apr 18 '15 at 7:54
|
show 1 more comment
Ok, so even though the identity you gave is correct I think I will just use the fact $(a, infty)$ is an open interval, hence an open set. Just one last question, how would I show the above for $[a, infty)$. I can express $(a,b)$ as a union of intervals of the form $[a, infty)$ but can't show the other way around
– user1314
Apr 18 '15 at 7:36
@user1314 Hint: Use $$[a,infty) = bigcap_{n in mathbb{N}} (a- frac{1}{n},infty)$$
– saz
Apr 18 '15 at 7:43
By the way a small correction, a bit pedantic maybe, but in the above formua, don't you mean $nin mathbb{N}_{0}$, since you used that in the earlier one.
– user1314
Apr 18 '15 at 7:50
@user1314 You mean the formula in my previous comment? No, I don't mean $n in mathbb{N}_0$ - $frac{1}{n}$ is not well-defined for $n=0$.
– saz
Apr 18 '15 at 7:52
oh of course, sorry, I kept thinking $mathbb{N}_{0}$ means $mathbb{N}$ without the 0, yes you are absolutely correct. Thanks for the help
– user1314
Apr 18 '15 at 7:54
Ok, so even though the identity you gave is correct I think I will just use the fact $(a, infty)$ is an open interval, hence an open set. Just one last question, how would I show the above for $[a, infty)$. I can express $(a,b)$ as a union of intervals of the form $[a, infty)$ but can't show the other way around
– user1314
Apr 18 '15 at 7:36
Ok, so even though the identity you gave is correct I think I will just use the fact $(a, infty)$ is an open interval, hence an open set. Just one last question, how would I show the above for $[a, infty)$. I can express $(a,b)$ as a union of intervals of the form $[a, infty)$ but can't show the other way around
– user1314
Apr 18 '15 at 7:36
@user1314 Hint: Use $$[a,infty) = bigcap_{n in mathbb{N}} (a- frac{1}{n},infty)$$
– saz
Apr 18 '15 at 7:43
@user1314 Hint: Use $$[a,infty) = bigcap_{n in mathbb{N}} (a- frac{1}{n},infty)$$
– saz
Apr 18 '15 at 7:43
By the way a small correction, a bit pedantic maybe, but in the above formua, don't you mean $nin mathbb{N}_{0}$, since you used that in the earlier one.
– user1314
Apr 18 '15 at 7:50
By the way a small correction, a bit pedantic maybe, but in the above formua, don't you mean $nin mathbb{N}_{0}$, since you used that in the earlier one.
– user1314
Apr 18 '15 at 7:50
@user1314 You mean the formula in my previous comment? No, I don't mean $n in mathbb{N}_0$ - $frac{1}{n}$ is not well-defined for $n=0$.
– saz
Apr 18 '15 at 7:52
@user1314 You mean the formula in my previous comment? No, I don't mean $n in mathbb{N}_0$ - $frac{1}{n}$ is not well-defined for $n=0$.
– saz
Apr 18 '15 at 7:52
oh of course, sorry, I kept thinking $mathbb{N}_{0}$ means $mathbb{N}$ without the 0, yes you are absolutely correct. Thanks for the help
– user1314
Apr 18 '15 at 7:54
oh of course, sorry, I kept thinking $mathbb{N}_{0}$ means $mathbb{N}$ without the 0, yes you are absolutely correct. Thanks for the help
– user1314
Apr 18 '15 at 7:54
|
show 1 more comment
Denote $mathcal{B}=sigmaleft(tauright)=sigmaleft(left{ left(a,bright)mid a,binmathbb{R}right} right)$ (this equality was found out by you allready)
where $tau$ denotes the topology and $mathcal{B}_{1}=sigmaleft(left{ left(a,inftyright)mid ainmathbb{R}right} right)$.
Then $left{ left(a,inftyright)mid ainmathbb{R}right} subsettau$
so that $mathcal{B}_{1}subseteqmathcal{B}$.
To prove the converse
inclusion it is enough to show that $left{ left(a,bright)mid a,binmathbb{R}right} subsetmathcal{B}_{1}$.
We have $left(a,cright]=left(a,inftyright)-left(c,inftyright)inmathcal{B}_{1}$
for each $c$, and consequently $left(a,bright)=bigcup_{ninmathbb{N}}left(a,b-frac{1}{n}right]inmathcal{B}_{1}$.
answered Apr 18 '15 at 7:54
drhab
97.9k544129
I don't see any contradiction: $left(a,bright]=cap_{ninmathbb{N}}left(a,b+frac{1}{n}right]inmathcal{B}_{1}$. Maybe, $left(a,bright)=cup_{ninmathbb{N}}left(a,b-frac{1}{n}right]inmathcal{B}_{1}$.
– Aidas
Nov 20 '18 at 15:38
@Aidas You are right, and thank you for attending me. I repaired. I really don't understand (and remember) how I could write that.
– drhab
Nov 20 '18 at 16:12
add a comment |
Denote $mathcal{B}=sigmaleft(tauright)=sigmaleft(left{ left(a,bright)mid a,binmathbb{R}right} right)$ (this equality was found out by you allready)
where $tau$ denotes the topology and $mathcal{B}_{1}=sigmaleft(left{ left(a,inftyright)mid ainmathbb{R}right} right)$.
Then $left{ left(a,inftyright)mid ainmathbb{R}right} subsettau$
so that $mathcal{B}_{1}subseteqmathcal{B}$.
To prove the converse
inclusion it is enough to show that $left{ left(a,bright)mid a,binmathbb{R}right} subsetmathcal{B}_{1}$.
We have $left(a,cright]=left(a,inftyright)-left(c,inftyright)inmathcal{B}_{1}$
for each $c$, and consequently $left(a,bright)=bigcup_{ninmathbb{N}}left(a,b-frac{1}{n}right]inmathcal{B}_{1}$.
answered Apr 18 '15 at 7:54
drhab
97.9k544129
I don't see any contradiction: $left(a,bright]=cap_{ninmathbb{N}}left(a,b+frac{1}{n}right]inmathcal{B}_{1}$. Maybe, $left(a,bright)=cup_{ninmathbb{N}}left(a,b-frac{1}{n}right]inmathcal{B}_{1}$.
– Aidas
Nov 20 '18 at 15:38
@Aidas You are right, and thank you for attending me. I repaired. I really don't understand (and remember) how I could write that.
– drhab
Nov 20 '18 at 16:12
add a comment |
Denote $mathcal{B}=sigmaleft(tauright)=sigmaleft(left{ left(a,bright)mid a,binmathbb{R}right} right)$ (this equality was found out by you allready)
where $tau$ denotes the topology and $mathcal{B}_{1}=sigmaleft(left{ left(a,inftyright)mid ainmathbb{R}right} right)$.
Then $left{ left(a,inftyright)mid ainmathbb{R}right} subsettau$
so that $mathcal{B}_{1}subseteqmathcal{B}$.
To prove the converse
inclusion it is enough to show that $left{ left(a,bright)mid a,binmathbb{R}right} subsetmathcal{B}_{1}$.
We have $left(a,cright]=left(a,inftyright)-left(c,inftyright)inmathcal{B}_{1}$
for each $c$, and consequently $left(a,bright)=bigcup_{ninmathbb{N}}left(a,b-frac{1}{n}right]inmathcal{B}_{1}$.
answered Apr 18 '15 at 7:54
drhab
97.9k544129
Denote $mathcal{B}=sigmaleft(tauright)=sigmaleft(left{ left(a,bright)mid a,binmathbb{R}right} right)$ (this equality was found out by you allready)
where $tau$ denotes the topology and $mathcal{B}_{1}=sigmaleft(left{ left(a,inftyright)mid ainmathbb{R}right} right)$.
Then $left{ left(a,inftyright)mid ainmathbb{R}right} subsettau$
so that $mathcal{B}_{1}subseteqmathcal{B}$.
To prove the converse
inclusion it is enough to show that $left{ left(a,bright)mid a,binmathbb{R}right} subsetmathcal{B}_{1}$.
We have $left(a,cright]=left(a,inftyright)-left(c,inftyright)inmathcal{B}_{1}$
for each $c$, and consequently $left(a,bright)=bigcup_{ninmathbb{N}}left(a,b-frac{1}{n}right]inmathcal{B}_{1}$.
answered Apr 18 '15 at 7:54
drhab
97.9k544129
edited Nov 20 '18 at 16:10
answered Apr 18 '15 at 7:54
drhab
97.9k544129
answered Apr 18 '15 at 7:54
drhab
97.9k544129
answered Apr 18 '15 at 7:54
drhab
97.9k544129
97.9k544129
I don't see any contradiction: $left(a,bright]=cap_{ninmathbb{N}}left(a,b+frac{1}{n}right]inmathcal{B}_{1}$. Maybe, $left(a,bright)=cup_{ninmathbb{N}}left(a,b-frac{1}{n}right]inmathcal{B}_{1}$.
– Aidas
Nov 20 '18 at 15:38
@Aidas You are right, and thank you for attending me. I repaired. I really don't understand (and remember) how I could write that.
– drhab
Nov 20 '18 at 16:12
add a comment |
I don't see any contradiction: $left(a,bright]=cap_{ninmathbb{N}}left(a,b+frac{1}{n}right]inmathcal{B}_{1}$. Maybe, $left(a,bright)=cup_{ninmathbb{N}}left(a,b-frac{1}{n}right]inmathcal{B}_{1}$.
– Aidas
Nov 20 '18 at 15:38
@Aidas You are right, and thank you for attending me. I repaired. I really don't understand (and remember) how I could write that.
– drhab
Nov 20 '18 at 16:12
I don't see any contradiction: $left(a,bright]=cap_{ninmathbb{N}}left(a,b+frac{1}{n}right]inmathcal{B}_{1}$. Maybe, $left(a,bright)=cup_{ninmathbb{N}}left(a,b-frac{1}{n}right]inmathcal{B}_{1}$.
– Aidas
Nov 20 '18 at 15:38
I don't see any contradiction: $left(a,bright]=cap_{ninmathbb{N}}left(a,b+frac{1}{n}right]inmathcal{B}_{1}$. Maybe, $left(a,bright)=cup_{ninmathbb{N}}left(a,b-frac{1}{n}right]inmathcal{B}_{1}$.
– Aidas
Nov 20 '18 at 15:38
@Aidas You are right, and thank you for attending me. I repaired. I really don't understand (and remember) how I could write that.
– drhab
Nov 20 '18 at 16:12
@Aidas You are right, and thank you for attending me. I repaired. I really don't understand (and remember) how I could write that.
– drhab
Nov 20 '18 at 16:12
add a comment |
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