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How to check whether a set belongs to a $sigma$-algebra? Not understanding “a Borel function”
$begingroup$
Hello fine ladies and gentlemen,
Now, I have never been very good with measure theory and I am struggling to be able to do the exercise.
My definition of a sigma-algebra is the following.
(Definition) Let $X$ be a set. A collection of $Sigma$ subset of $X$ is called a sigma-algebra if
(i) $phi in Sigma$
(ii) $E in Sigma$ implies $X setminus E in Sigma$
(iii) $E_n in Sigma, ngeq 1$ implies $U_{n=1}^{infty}E_n in Sigma$.
I am not sure about what comes beneath the line "As a $sigma$-algebra $mathcal{T}( mathbb{R}_{geq 0}, mathbb{R}^d)$ is also generated by the following families...
And I am struggling with the exercise! Also, I would like a more intuitive understanding to explain the natural events part.
real-analysis measure-theory borel-sets
edited Feb 11 at 7:38
Alex Ravsky
43k32583
asked Feb 1 at 18:41
Catherine DrysdaleCatherine Drysdale
426311
$endgroup$
add a comment |
$begingroup$
Hello fine ladies and gentlemen,
Now, I have never been very good with measure theory and I am struggling to be able to do the exercise.
My definition of a sigma-algebra is the following.
(Definition) Let $X$ be a set. A collection of $Sigma$ subset of $X$ is called a sigma-algebra if
(i) $phi in Sigma$
(ii) $E in Sigma$ implies $X setminus E in Sigma$
(iii) $E_n in Sigma, ngeq 1$ implies $U_{n=1}^{infty}E_n in Sigma$.
I am not sure about what comes beneath the line "As a $sigma$-algebra $mathcal{T}( mathbb{R}_{geq 0}, mathbb{R}^d)$ is also generated by the following families...
And I am struggling with the exercise! Also, I would like a more intuitive understanding to explain the natural events part.
real-analysis measure-theory borel-sets
edited Feb 11 at 7:38
Alex Ravsky
43k32583
asked Feb 1 at 18:41
Catherine DrysdaleCatherine Drysdale
426311
$endgroup$
$begingroup$
Hi Catherine! Please make your title informative, and avoid greetings or statements about your problem that would make it less useful to other users (example: "very hard algebra problem", "struggling with normal operators", "confusing matrix multiplication", etc). Thank you!
$endgroup$
– Pedro Tamaroff♦
Feb 7 at 16:39
$begingroup$
Is it possible to change it now? I cannot see where to edit the question.
$endgroup$
– Catherine Drysdale
Feb 7 at 16:53
$begingroup$
There is a button at the bottom of the post, next to the tags that says 'edit'.
$endgroup$
– Pedro Tamaroff♦
Feb 7 at 16:55
add a comment |
$begingroup$
Hello fine ladies and gentlemen,
Now, I have never been very good with measure theory and I am struggling to be able to do the exercise.
My definition of a sigma-algebra is the following.
(Definition) Let $X$ be a set. A collection of $Sigma$ subset of $X$ is called a sigma-algebra if
(i) $phi in Sigma$
(ii) $E in Sigma$ implies $X setminus E in Sigma$
(iii) $E_n in Sigma, ngeq 1$ implies $U_{n=1}^{infty}E_n in Sigma$.
I am not sure about what comes beneath the line "As a $sigma$-algebra $mathcal{T}( mathbb{R}_{geq 0}, mathbb{R}^d)$ is also generated by the following families...
And I am struggling with the exercise! Also, I would like a more intuitive understanding to explain the natural events part.
real-analysis measure-theory borel-sets
edited Feb 11 at 7:38
Alex Ravsky
43k32583
asked Feb 1 at 18:41
Catherine DrysdaleCatherine Drysdale
426311
$endgroup$
Hello fine ladies and gentlemen,
Now, I have never been very good with measure theory and I am struggling to be able to do the exercise.
My definition of a sigma-algebra is the following.
(Definition) Let $X$ be a set. A collection of $Sigma$ subset of $X$ is called a sigma-algebra if
(i) $phi in Sigma$
(ii) $E in Sigma$ implies $X setminus E in Sigma$
(iii) $E_n in Sigma, ngeq 1$ implies $U_{n=1}^{infty}E_n in Sigma$.
I am not sure about what comes beneath the line "As a $sigma$-algebra $mathcal{T}( mathbb{R}_{geq 0}, mathbb{R}^d)$ is also generated by the following families...
And I am struggling with the exercise! Also, I would like a more intuitive understanding to explain the natural events part.
real-analysis measure-theory borel-sets
real-analysis measure-theory borel-sets
edited Feb 11 at 7:38
Alex Ravsky
43k32583
asked Feb 1 at 18:41
Catherine DrysdaleCatherine Drysdale
426311
edited Feb 11 at 7:38
Alex Ravsky
43k32583
asked Feb 1 at 18:41
Catherine DrysdaleCatherine Drysdale
426311
edited Feb 11 at 7:38
Alex Ravsky
43k32583
edited Feb 11 at 7:38
Alex Ravsky
43k32583
edited Feb 11 at 7:38
Alex Ravsky
43k32583
43k32583
asked Feb 1 at 18:41
Catherine DrysdaleCatherine Drysdale
426311
asked Feb 1 at 18:41
Catherine DrysdaleCatherine Drysdale
426311
asked Feb 1 at 18:41
Catherine DrysdaleCatherine Drysdale
426311
426311
$begingroup$
Hi Catherine! Please make your title informative, and avoid greetings or statements about your problem that would make it less useful to other users (example: "very hard algebra problem", "struggling with normal operators", "confusing matrix multiplication", etc). Thank you!
$endgroup$
– Pedro Tamaroff♦
Feb 7 at 16:39
$begingroup$
Is it possible to change it now? I cannot see where to edit the question.
$endgroup$
– Catherine Drysdale
Feb 7 at 16:53
$begingroup$
There is a button at the bottom of the post, next to the tags that says 'edit'.
$endgroup$
– Pedro Tamaroff♦
Feb 7 at 16:55
add a comment |
$begingroup$
Hi Catherine! Please make your title informative, and avoid greetings or statements about your problem that would make it less useful to other users (example: "very hard algebra problem", "struggling with normal operators", "confusing matrix multiplication", etc). Thank you!
$endgroup$
– Pedro Tamaroff♦
Feb 7 at 16:39
$begingroup$
Is it possible to change it now? I cannot see where to edit the question.
$endgroup$
– Catherine Drysdale
Feb 7 at 16:53
$begingroup$
There is a button at the bottom of the post, next to the tags that says 'edit'.
$endgroup$
– Pedro Tamaroff♦
Feb 7 at 16:55
$begingroup$
Hi Catherine! Please make your title informative, and avoid greetings or statements about your problem that would make it less useful to other users (example: "very hard algebra problem", "struggling with normal operators", "confusing matrix multiplication", etc). Thank you!
$endgroup$
– Pedro Tamaroff♦
Feb 7 at 16:39
$begingroup$
Hi Catherine! Please make your title informative, and avoid greetings or statements about your problem that would make it less useful to other users (example: "very hard algebra problem", "struggling with normal operators", "confusing matrix multiplication", etc). Thank you!
$endgroup$
– Pedro Tamaroff♦
Feb 7 at 16:39
$begingroup$
Is it possible to change it now? I cannot see where to edit the question.
$endgroup$
– Catherine Drysdale
Feb 7 at 16:53
$begingroup$
Is it possible to change it now? I cannot see where to edit the question.
$endgroup$
– Catherine Drysdale
Feb 7 at 16:53
$begingroup$
There is a button at the bottom of the post, next to the tags that says 'edit'.
$endgroup$
– Pedro Tamaroff♦
Feb 7 at 16:55
$begingroup$
There is a button at the bottom of the post, next to the tags that says 'edit'.
$endgroup$
– Pedro Tamaroff♦
Feb 7 at 16:55
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
By $X=mathcal A([0,1],Bbb R)$ and $sigma=mathcal T([0,1],Bbb R)$ I shall understand the natural counterparts of notions $mathcal A(Bbb R_{ge},Bbb R)$ and $mathcal T(Bbb R_{ge},Bbb R)$ defined above. That is $X$ is a set of all functions from $[0,1]$ to $Bbb R$ and $Sigma$ is a $sigma$-algebra on $X$ generated by respective cylindrical sets.
A key point to the solution is the following easy
Claim. Each set $EinSigma$ depends on countable many coordinates. Namely, there exists a countable subset $T$ of $[0,1]$ such that for each $fin X$ holds $fin E$ iff there exists a function $gin E$ such that $g|T=f|T$.
Proof. To show Implication $Rightarrow$ it suffices to take $g=f$. Denote by $Sigma’$ the family of all subsets of $X$ satisfying Implication $Leftarrow$. It is easy to check that $Sigma’$ is closed with respect to complements and countable unions, so conditions (ii) and (iii) from the definition of a $sigma$-algebra imply that $Sigma’$ is a $sigma$-algebra. It is easy to see that $Sigma’$ contains all cylindrical subsets of $X$, so it is a $sigma$-algebra containing $Sigma$. $square$
Let $E$ be the set form the first part of Exercise. Suppose to the contrary that $EinSigma$. Pick a countable set $Tsubset [0,1]$ satisfying Claim for the set $E$. Pick an arbitrary $t’in [0,1]setminus T$ and define a function $fin X$ by putting $f(t’)=1$ and $f(t)=0$ for all other $tin [0,1]$. Since $sup_{tin [0,1]} f(t)=f(t’)=1$, $fnotin E$. On the other hand, let $gequiv 0$ be the zero function on the set $[0,1]$. Since $gin E$ and $f(t)=g(t)$ for each $tin T$, $fin E$, a contradiction.
Let $E$ be the set form the second part of Exercise. Suppose to the contrary that $EinSigma$. Pick a countable set $Tsubset [0,1]$ satisfying Claim for the set $E$. Pick an arbitrary $t’in [0,1]setminus T$ and define a function $fin X$ by putting $f(t’)=0$ and $f(t)=1$ for all other $tin [0,1]$. Since $f(t’)=0$, $fin E$. On the other hand, let $gequiv 1$ be the unit function on the set $[0,1]$. Since $gnotin E$ and $f(t)=g(t)$ for each $tin T$, $fnotin E$, a contradiction.
answered Feb 10 at 20:24
Alex RavskyAlex Ravsky
43k32583
$endgroup$
$begingroup$
I am unsure what you mean by countably many coordinates and why that is relevant. Could you perhaps expand a little?
$endgroup$
– Catherine Drysdale
Feb 11 at 10:49
$begingroup$
@CatherineDrysdale A function $f:[0,1]toBbb R$ can be naturally considered as a point in the space $Bbb R^{[0,1]}$. Then for $tin [0,1]$, $t$-th coordinate of the function $f$ is a value $f(t)$. Similarly, a set $E$ of such functions is a subset of $Bbb R^{[0,1]}$. Given a subset $T$ of $[0,1]$ let $pi_T: Bbb R^{[0,1]}to Bbb R^{T}$ be the natural projection. Then $E$ depends on countable many coordinates means that there exists a countable subset $T$ of $[0,1]$ such that $E=pi_T^{-1}(pi^{T}(E))$.
$endgroup$
– Alex Ravsky
Feb 11 at 13:33
$begingroup$
The claim that the sets $EinSigma$ depend on countable many coordinates shows that they are not very complicated. This result about their structure is a fragment of a big topic called descriptive set theory.
$endgroup$
– Alex Ravsky
Feb 11 at 13:33
1
$begingroup$
Thank you! I will look into it. I am going over all of my old notes and definitions and it is finally coming together!
$endgroup$
– Catherine Drysdale
Feb 12 at 14:52
add a comment |
$begingroup$
For the first part of your question, the Borel sets are just the $sigma$-algebra generated by open sets (or some define as the ones generated by compact sets. These are the same in $mathbb{R}^n$). If you don't understand what is meant by the $sigma$-algebra generated by some set, you should prove as an exercise that the intersection of $sigma$-algebras is again a $sigma$-algebra and that the power set of a set is a $sigma$-algebra (the $sigma$-algebra generated by a set is then just the intersection of all $sigma$-algebras containing the set). Knowing this, I think you should be able to understand the first part. Note that it's quite difficult to determine whether or not a general set is actually in the Borel sets, so don't try to get some sort of nice form for them. Hopefully this is also enough to get you started on the exercise?
answered Feb 1 at 20:49
Jon HilleryJon Hillery
707
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Will try with your perspective and get back to it today!
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– Catherine Drysdale
Feb 2 at 11:25
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Hey, I have not looked at this problem for several days and I have come back to it, and I am still struggling slightly. I guess that I do not know what a sigma-algebra on space of functions means! Any help?
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– Catherine Drysdale
Feb 7 at 16:09
$begingroup$
In my limited experience in working with measures on spaces other than R^n or subsets of it, the only way I was able to think about the problem was to sort of pretend it was R^n and only remind myself what space I was in when necessary. This is probably extremely unhelpful, but I don't know what other advice to give. Hopefully someone with more experience will chime in?
$endgroup$
– Jon Hillery
Feb 7 at 19:20
add a comment |
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2 Answers
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2 Answers
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$begingroup$
By $X=mathcal A([0,1],Bbb R)$ and $sigma=mathcal T([0,1],Bbb R)$ I shall understand the natural counterparts of notions $mathcal A(Bbb R_{ge},Bbb R)$ and $mathcal T(Bbb R_{ge},Bbb R)$ defined above. That is $X$ is a set of all functions from $[0,1]$ to $Bbb R$ and $Sigma$ is a $sigma$-algebra on $X$ generated by respective cylindrical sets.
A key point to the solution is the following easy
Claim. Each set $EinSigma$ depends on countable many coordinates. Namely, there exists a countable subset $T$ of $[0,1]$ such that for each $fin X$ holds $fin E$ iff there exists a function $gin E$ such that $g|T=f|T$.
Proof. To show Implication $Rightarrow$ it suffices to take $g=f$. Denote by $Sigma’$ the family of all subsets of $X$ satisfying Implication $Leftarrow$. It is easy to check that $Sigma’$ is closed with respect to complements and countable unions, so conditions (ii) and (iii) from the definition of a $sigma$-algebra imply that $Sigma’$ is a $sigma$-algebra. It is easy to see that $Sigma’$ contains all cylindrical subsets of $X$, so it is a $sigma$-algebra containing $Sigma$. $square$
Let $E$ be the set form the first part of Exercise. Suppose to the contrary that $EinSigma$. Pick a countable set $Tsubset [0,1]$ satisfying Claim for the set $E$. Pick an arbitrary $t’in [0,1]setminus T$ and define a function $fin X$ by putting $f(t’)=1$ and $f(t)=0$ for all other $tin [0,1]$. Since $sup_{tin [0,1]} f(t)=f(t’)=1$, $fnotin E$. On the other hand, let $gequiv 0$ be the zero function on the set $[0,1]$. Since $gin E$ and $f(t)=g(t)$ for each $tin T$, $fin E$, a contradiction.
Let $E$ be the set form the second part of Exercise. Suppose to the contrary that $EinSigma$. Pick a countable set $Tsubset [0,1]$ satisfying Claim for the set $E$. Pick an arbitrary $t’in [0,1]setminus T$ and define a function $fin X$ by putting $f(t’)=0$ and $f(t)=1$ for all other $tin [0,1]$. Since $f(t’)=0$, $fin E$. On the other hand, let $gequiv 1$ be the unit function on the set $[0,1]$. Since $gnotin E$ and $f(t)=g(t)$ for each $tin T$, $fnotin E$, a contradiction.
answered Feb 10 at 20:24
Alex RavskyAlex Ravsky
43k32583
$endgroup$
$begingroup$
I am unsure what you mean by countably many coordinates and why that is relevant. Could you perhaps expand a little?
$endgroup$
– Catherine Drysdale
Feb 11 at 10:49
$begingroup$
@CatherineDrysdale A function $f:[0,1]toBbb R$ can be naturally considered as a point in the space $Bbb R^{[0,1]}$. Then for $tin [0,1]$, $t$-th coordinate of the function $f$ is a value $f(t)$. Similarly, a set $E$ of such functions is a subset of $Bbb R^{[0,1]}$. Given a subset $T$ of $[0,1]$ let $pi_T: Bbb R^{[0,1]}to Bbb R^{T}$ be the natural projection. Then $E$ depends on countable many coordinates means that there exists a countable subset $T$ of $[0,1]$ such that $E=pi_T^{-1}(pi^{T}(E))$.
$endgroup$
– Alex Ravsky
Feb 11 at 13:33
$begingroup$
The claim that the sets $EinSigma$ depend on countable many coordinates shows that they are not very complicated. This result about their structure is a fragment of a big topic called descriptive set theory.
$endgroup$
– Alex Ravsky
Feb 11 at 13:33
1
$begingroup$
Thank you! I will look into it. I am going over all of my old notes and definitions and it is finally coming together!
$endgroup$
– Catherine Drysdale
Feb 12 at 14:52
add a comment |
$begingroup$
By $X=mathcal A([0,1],Bbb R)$ and $sigma=mathcal T([0,1],Bbb R)$ I shall understand the natural counterparts of notions $mathcal A(Bbb R_{ge},Bbb R)$ and $mathcal T(Bbb R_{ge},Bbb R)$ defined above. That is $X$ is a set of all functions from $[0,1]$ to $Bbb R$ and $Sigma$ is a $sigma$-algebra on $X$ generated by respective cylindrical sets.
A key point to the solution is the following easy
Claim. Each set $EinSigma$ depends on countable many coordinates. Namely, there exists a countable subset $T$ of $[0,1]$ such that for each $fin X$ holds $fin E$ iff there exists a function $gin E$ such that $g|T=f|T$.
Proof. To show Implication $Rightarrow$ it suffices to take $g=f$. Denote by $Sigma’$ the family of all subsets of $X$ satisfying Implication $Leftarrow$. It is easy to check that $Sigma’$ is closed with respect to complements and countable unions, so conditions (ii) and (iii) from the definition of a $sigma$-algebra imply that $Sigma’$ is a $sigma$-algebra. It is easy to see that $Sigma’$ contains all cylindrical subsets of $X$, so it is a $sigma$-algebra containing $Sigma$. $square$
Let $E$ be the set form the first part of Exercise. Suppose to the contrary that $EinSigma$. Pick a countable set $Tsubset [0,1]$ satisfying Claim for the set $E$. Pick an arbitrary $t’in [0,1]setminus T$ and define a function $fin X$ by putting $f(t’)=1$ and $f(t)=0$ for all other $tin [0,1]$. Since $sup_{tin [0,1]} f(t)=f(t’)=1$, $fnotin E$. On the other hand, let $gequiv 0$ be the zero function on the set $[0,1]$. Since $gin E$ and $f(t)=g(t)$ for each $tin T$, $fin E$, a contradiction.
Let $E$ be the set form the second part of Exercise. Suppose to the contrary that $EinSigma$. Pick a countable set $Tsubset [0,1]$ satisfying Claim for the set $E$. Pick an arbitrary $t’in [0,1]setminus T$ and define a function $fin X$ by putting $f(t’)=0$ and $f(t)=1$ for all other $tin [0,1]$. Since $f(t’)=0$, $fin E$. On the other hand, let $gequiv 1$ be the unit function on the set $[0,1]$. Since $gnotin E$ and $f(t)=g(t)$ for each $tin T$, $fnotin E$, a contradiction.
answered Feb 10 at 20:24
Alex RavskyAlex Ravsky
43k32583
$endgroup$
$begingroup$
I am unsure what you mean by countably many coordinates and why that is relevant. Could you perhaps expand a little?
$endgroup$
– Catherine Drysdale
Feb 11 at 10:49
$begingroup$
@CatherineDrysdale A function $f:[0,1]toBbb R$ can be naturally considered as a point in the space $Bbb R^{[0,1]}$. Then for $tin [0,1]$, $t$-th coordinate of the function $f$ is a value $f(t)$. Similarly, a set $E$ of such functions is a subset of $Bbb R^{[0,1]}$. Given a subset $T$ of $[0,1]$ let $pi_T: Bbb R^{[0,1]}to Bbb R^{T}$ be the natural projection. Then $E$ depends on countable many coordinates means that there exists a countable subset $T$ of $[0,1]$ such that $E=pi_T^{-1}(pi^{T}(E))$.
$endgroup$
– Alex Ravsky
Feb 11 at 13:33
$begingroup$
The claim that the sets $EinSigma$ depend on countable many coordinates shows that they are not very complicated. This result about their structure is a fragment of a big topic called descriptive set theory.
$endgroup$
– Alex Ravsky
Feb 11 at 13:33
1
$begingroup$
Thank you! I will look into it. I am going over all of my old notes and definitions and it is finally coming together!
$endgroup$
– Catherine Drysdale
Feb 12 at 14:52
add a comment |
$begingroup$
By $X=mathcal A([0,1],Bbb R)$ and $sigma=mathcal T([0,1],Bbb R)$ I shall understand the natural counterparts of notions $mathcal A(Bbb R_{ge},Bbb R)$ and $mathcal T(Bbb R_{ge},Bbb R)$ defined above. That is $X$ is a set of all functions from $[0,1]$ to $Bbb R$ and $Sigma$ is a $sigma$-algebra on $X$ generated by respective cylindrical sets.
A key point to the solution is the following easy
Claim. Each set $EinSigma$ depends on countable many coordinates. Namely, there exists a countable subset $T$ of $[0,1]$ such that for each $fin X$ holds $fin E$ iff there exists a function $gin E$ such that $g|T=f|T$.
Proof. To show Implication $Rightarrow$ it suffices to take $g=f$. Denote by $Sigma’$ the family of all subsets of $X$ satisfying Implication $Leftarrow$. It is easy to check that $Sigma’$ is closed with respect to complements and countable unions, so conditions (ii) and (iii) from the definition of a $sigma$-algebra imply that $Sigma’$ is a $sigma$-algebra. It is easy to see that $Sigma’$ contains all cylindrical subsets of $X$, so it is a $sigma$-algebra containing $Sigma$. $square$
Let $E$ be the set form the first part of Exercise. Suppose to the contrary that $EinSigma$. Pick a countable set $Tsubset [0,1]$ satisfying Claim for the set $E$. Pick an arbitrary $t’in [0,1]setminus T$ and define a function $fin X$ by putting $f(t’)=1$ and $f(t)=0$ for all other $tin [0,1]$. Since $sup_{tin [0,1]} f(t)=f(t’)=1$, $fnotin E$. On the other hand, let $gequiv 0$ be the zero function on the set $[0,1]$. Since $gin E$ and $f(t)=g(t)$ for each $tin T$, $fin E$, a contradiction.
Let $E$ be the set form the second part of Exercise. Suppose to the contrary that $EinSigma$. Pick a countable set $Tsubset [0,1]$ satisfying Claim for the set $E$. Pick an arbitrary $t’in [0,1]setminus T$ and define a function $fin X$ by putting $f(t’)=0$ and $f(t)=1$ for all other $tin [0,1]$. Since $f(t’)=0$, $fin E$. On the other hand, let $gequiv 1$ be the unit function on the set $[0,1]$. Since $gnotin E$ and $f(t)=g(t)$ for each $tin T$, $fnotin E$, a contradiction.
answered Feb 10 at 20:24
Alex RavskyAlex Ravsky
43k32583
$endgroup$
By $X=mathcal A([0,1],Bbb R)$ and $sigma=mathcal T([0,1],Bbb R)$ I shall understand the natural counterparts of notions $mathcal A(Bbb R_{ge},Bbb R)$ and $mathcal T(Bbb R_{ge},Bbb R)$ defined above. That is $X$ is a set of all functions from $[0,1]$ to $Bbb R$ and $Sigma$ is a $sigma$-algebra on $X$ generated by respective cylindrical sets.
A key point to the solution is the following easy
Claim. Each set $EinSigma$ depends on countable many coordinates. Namely, there exists a countable subset $T$ of $[0,1]$ such that for each $fin X$ holds $fin E$ iff there exists a function $gin E$ such that $g|T=f|T$.
Proof. To show Implication $Rightarrow$ it suffices to take $g=f$. Denote by $Sigma’$ the family of all subsets of $X$ satisfying Implication $Leftarrow$. It is easy to check that $Sigma’$ is closed with respect to complements and countable unions, so conditions (ii) and (iii) from the definition of a $sigma$-algebra imply that $Sigma’$ is a $sigma$-algebra. It is easy to see that $Sigma’$ contains all cylindrical subsets of $X$, so it is a $sigma$-algebra containing $Sigma$. $square$
Let $E$ be the set form the first part of Exercise. Suppose to the contrary that $EinSigma$. Pick a countable set $Tsubset [0,1]$ satisfying Claim for the set $E$. Pick an arbitrary $t’in [0,1]setminus T$ and define a function $fin X$ by putting $f(t’)=1$ and $f(t)=0$ for all other $tin [0,1]$. Since $sup_{tin [0,1]} f(t)=f(t’)=1$, $fnotin E$. On the other hand, let $gequiv 0$ be the zero function on the set $[0,1]$. Since $gin E$ and $f(t)=g(t)$ for each $tin T$, $fin E$, a contradiction.
Let $E$ be the set form the second part of Exercise. Suppose to the contrary that $EinSigma$. Pick a countable set $Tsubset [0,1]$ satisfying Claim for the set $E$. Pick an arbitrary $t’in [0,1]setminus T$ and define a function $fin X$ by putting $f(t’)=0$ and $f(t)=1$ for all other $tin [0,1]$. Since $f(t’)=0$, $fin E$. On the other hand, let $gequiv 1$ be the unit function on the set $[0,1]$. Since $gnotin E$ and $f(t)=g(t)$ for each $tin T$, $fnotin E$, a contradiction.
answered Feb 10 at 20:24
Alex RavskyAlex Ravsky
43k32583
answered Feb 10 at 20:24
Alex RavskyAlex Ravsky
43k32583
answered Feb 10 at 20:24
Alex RavskyAlex Ravsky
43k32583
answered Feb 10 at 20:24
Alex RavskyAlex Ravsky
43k32583
43k32583
$begingroup$
I am unsure what you mean by countably many coordinates and why that is relevant. Could you perhaps expand a little?
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– Catherine Drysdale
Feb 11 at 10:49
$begingroup$
@CatherineDrysdale A function $f:[0,1]toBbb R$ can be naturally considered as a point in the space $Bbb R^{[0,1]}$. Then for $tin [0,1]$, $t$-th coordinate of the function $f$ is a value $f(t)$. Similarly, a set $E$ of such functions is a subset of $Bbb R^{[0,1]}$. Given a subset $T$ of $[0,1]$ let $pi_T: Bbb R^{[0,1]}to Bbb R^{T}$ be the natural projection. Then $E$ depends on countable many coordinates means that there exists a countable subset $T$ of $[0,1]$ such that $E=pi_T^{-1}(pi^{T}(E))$.
$endgroup$
– Alex Ravsky
Feb 11 at 13:33
$begingroup$
The claim that the sets $EinSigma$ depend on countable many coordinates shows that they are not very complicated. This result about their structure is a fragment of a big topic called descriptive set theory.
$endgroup$
– Alex Ravsky
Feb 11 at 13:33
1
$begingroup$
Thank you! I will look into it. I am going over all of my old notes and definitions and it is finally coming together!
$endgroup$
– Catherine Drysdale
Feb 12 at 14:52
add a comment |
$begingroup$
I am unsure what you mean by countably many coordinates and why that is relevant. Could you perhaps expand a little?
$endgroup$
– Catherine Drysdale
Feb 11 at 10:49
$begingroup$
@CatherineDrysdale A function $f:[0,1]toBbb R$ can be naturally considered as a point in the space $Bbb R^{[0,1]}$. Then for $tin [0,1]$, $t$-th coordinate of the function $f$ is a value $f(t)$. Similarly, a set $E$ of such functions is a subset of $Bbb R^{[0,1]}$. Given a subset $T$ of $[0,1]$ let $pi_T: Bbb R^{[0,1]}to Bbb R^{T}$ be the natural projection. Then $E$ depends on countable many coordinates means that there exists a countable subset $T$ of $[0,1]$ such that $E=pi_T^{-1}(pi^{T}(E))$.
$endgroup$
– Alex Ravsky
Feb 11 at 13:33
$begingroup$
The claim that the sets $EinSigma$ depend on countable many coordinates shows that they are not very complicated. This result about their structure is a fragment of a big topic called descriptive set theory.
$endgroup$
– Alex Ravsky
Feb 11 at 13:33
1
$begingroup$
Thank you! I will look into it. I am going over all of my old notes and definitions and it is finally coming together!
$endgroup$
– Catherine Drysdale
Feb 12 at 14:52
$begingroup$
I am unsure what you mean by countably many coordinates and why that is relevant. Could you perhaps expand a little?
$endgroup$
– Catherine Drysdale
Feb 11 at 10:49
$begingroup$
I am unsure what you mean by countably many coordinates and why that is relevant. Could you perhaps expand a little?
$endgroup$
– Catherine Drysdale
Feb 11 at 10:49
$begingroup$
@CatherineDrysdale A function $f:[0,1]toBbb R$ can be naturally considered as a point in the space $Bbb R^{[0,1]}$. Then for $tin [0,1]$, $t$-th coordinate of the function $f$ is a value $f(t)$. Similarly, a set $E$ of such functions is a subset of $Bbb R^{[0,1]}$. Given a subset $T$ of $[0,1]$ let $pi_T: Bbb R^{[0,1]}to Bbb R^{T}$ be the natural projection. Then $E$ depends on countable many coordinates means that there exists a countable subset $T$ of $[0,1]$ such that $E=pi_T^{-1}(pi^{T}(E))$.
$endgroup$
– Alex Ravsky
Feb 11 at 13:33
$begingroup$
@CatherineDrysdale A function $f:[0,1]toBbb R$ can be naturally considered as a point in the space $Bbb R^{[0,1]}$. Then for $tin [0,1]$, $t$-th coordinate of the function $f$ is a value $f(t)$. Similarly, a set $E$ of such functions is a subset of $Bbb R^{[0,1]}$. Given a subset $T$ of $[0,1]$ let $pi_T: Bbb R^{[0,1]}to Bbb R^{T}$ be the natural projection. Then $E$ depends on countable many coordinates means that there exists a countable subset $T$ of $[0,1]$ such that $E=pi_T^{-1}(pi^{T}(E))$.
$endgroup$
– Alex Ravsky
Feb 11 at 13:33
$begingroup$
The claim that the sets $EinSigma$ depend on countable many coordinates shows that they are not very complicated. This result about their structure is a fragment of a big topic called descriptive set theory.
$endgroup$
– Alex Ravsky
Feb 11 at 13:33
$begingroup$
The claim that the sets $EinSigma$ depend on countable many coordinates shows that they are not very complicated. This result about their structure is a fragment of a big topic called descriptive set theory.
$endgroup$
– Alex Ravsky
Feb 11 at 13:33
1
1
$begingroup$
Thank you! I will look into it. I am going over all of my old notes and definitions and it is finally coming together!
$endgroup$
– Catherine Drysdale
Feb 12 at 14:52
$begingroup$
Thank you! I will look into it. I am going over all of my old notes and definitions and it is finally coming together!
$endgroup$
– Catherine Drysdale
Feb 12 at 14:52
add a comment |
$begingroup$
For the first part of your question, the Borel sets are just the $sigma$-algebra generated by open sets (or some define as the ones generated by compact sets. These are the same in $mathbb{R}^n$). If you don't understand what is meant by the $sigma$-algebra generated by some set, you should prove as an exercise that the intersection of $sigma$-algebras is again a $sigma$-algebra and that the power set of a set is a $sigma$-algebra (the $sigma$-algebra generated by a set is then just the intersection of all $sigma$-algebras containing the set). Knowing this, I think you should be able to understand the first part. Note that it's quite difficult to determine whether or not a general set is actually in the Borel sets, so don't try to get some sort of nice form for them. Hopefully this is also enough to get you started on the exercise?
answered Feb 1 at 20:49
Jon HilleryJon Hillery
707
$endgroup$
$begingroup$
Will try with your perspective and get back to it today!
$endgroup$
– Catherine Drysdale
Feb 2 at 11:25
$begingroup$
Hey, I have not looked at this problem for several days and I have come back to it, and I am still struggling slightly. I guess that I do not know what a sigma-algebra on space of functions means! Any help?
$endgroup$
– Catherine Drysdale
Feb 7 at 16:09
$begingroup$
In my limited experience in working with measures on spaces other than R^n or subsets of it, the only way I was able to think about the problem was to sort of pretend it was R^n and only remind myself what space I was in when necessary. This is probably extremely unhelpful, but I don't know what other advice to give. Hopefully someone with more experience will chime in?
$endgroup$
– Jon Hillery
Feb 7 at 19:20
add a comment |
$begingroup$
For the first part of your question, the Borel sets are just the $sigma$-algebra generated by open sets (or some define as the ones generated by compact sets. These are the same in $mathbb{R}^n$). If you don't understand what is meant by the $sigma$-algebra generated by some set, you should prove as an exercise that the intersection of $sigma$-algebras is again a $sigma$-algebra and that the power set of a set is a $sigma$-algebra (the $sigma$-algebra generated by a set is then just the intersection of all $sigma$-algebras containing the set). Knowing this, I think you should be able to understand the first part. Note that it's quite difficult to determine whether or not a general set is actually in the Borel sets, so don't try to get some sort of nice form for them. Hopefully this is also enough to get you started on the exercise?
answered Feb 1 at 20:49
Jon HilleryJon Hillery
707
$endgroup$
$begingroup$
Will try with your perspective and get back to it today!
$endgroup$
– Catherine Drysdale
Feb 2 at 11:25
$begingroup$
Hey, I have not looked at this problem for several days and I have come back to it, and I am still struggling slightly. I guess that I do not know what a sigma-algebra on space of functions means! Any help?
$endgroup$
– Catherine Drysdale
Feb 7 at 16:09
$begingroup$
In my limited experience in working with measures on spaces other than R^n or subsets of it, the only way I was able to think about the problem was to sort of pretend it was R^n and only remind myself what space I was in when necessary. This is probably extremely unhelpful, but I don't know what other advice to give. Hopefully someone with more experience will chime in?
$endgroup$
– Jon Hillery
Feb 7 at 19:20
add a comment |
$begingroup$
For the first part of your question, the Borel sets are just the $sigma$-algebra generated by open sets (or some define as the ones generated by compact sets. These are the same in $mathbb{R}^n$). If you don't understand what is meant by the $sigma$-algebra generated by some set, you should prove as an exercise that the intersection of $sigma$-algebras is again a $sigma$-algebra and that the power set of a set is a $sigma$-algebra (the $sigma$-algebra generated by a set is then just the intersection of all $sigma$-algebras containing the set). Knowing this, I think you should be able to understand the first part. Note that it's quite difficult to determine whether or not a general set is actually in the Borel sets, so don't try to get some sort of nice form for them. Hopefully this is also enough to get you started on the exercise?
answered Feb 1 at 20:49
Jon HilleryJon Hillery
707
$endgroup$
For the first part of your question, the Borel sets are just the $sigma$-algebra generated by open sets (or some define as the ones generated by compact sets. These are the same in $mathbb{R}^n$). If you don't understand what is meant by the $sigma$-algebra generated by some set, you should prove as an exercise that the intersection of $sigma$-algebras is again a $sigma$-algebra and that the power set of a set is a $sigma$-algebra (the $sigma$-algebra generated by a set is then just the intersection of all $sigma$-algebras containing the set). Knowing this, I think you should be able to understand the first part. Note that it's quite difficult to determine whether or not a general set is actually in the Borel sets, so don't try to get some sort of nice form for them. Hopefully this is also enough to get you started on the exercise?
answered Feb 1 at 20:49
Jon HilleryJon Hillery
707
answered Feb 1 at 20:49
Jon HilleryJon Hillery
707
answered Feb 1 at 20:49
Jon HilleryJon Hillery
707
answered Feb 1 at 20:49
Jon HilleryJon Hillery
707
707
$begingroup$
Will try with your perspective and get back to it today!
$endgroup$
– Catherine Drysdale
Feb 2 at 11:25
$begingroup$
Hey, I have not looked at this problem for several days and I have come back to it, and I am still struggling slightly. I guess that I do not know what a sigma-algebra on space of functions means! Any help?
$endgroup$
– Catherine Drysdale
Feb 7 at 16:09
$begingroup$
In my limited experience in working with measures on spaces other than R^n or subsets of it, the only way I was able to think about the problem was to sort of pretend it was R^n and only remind myself what space I was in when necessary. This is probably extremely unhelpful, but I don't know what other advice to give. Hopefully someone with more experience will chime in?
$endgroup$
– Jon Hillery
Feb 7 at 19:20
add a comment |
$begingroup$
Will try with your perspective and get back to it today!
$endgroup$
– Catherine Drysdale
Feb 2 at 11:25
$begingroup$
Hey, I have not looked at this problem for several days and I have come back to it, and I am still struggling slightly. I guess that I do not know what a sigma-algebra on space of functions means! Any help?
$endgroup$
– Catherine Drysdale
Feb 7 at 16:09
$begingroup$
In my limited experience in working with measures on spaces other than R^n or subsets of it, the only way I was able to think about the problem was to sort of pretend it was R^n and only remind myself what space I was in when necessary. This is probably extremely unhelpful, but I don't know what other advice to give. Hopefully someone with more experience will chime in?
$endgroup$
– Jon Hillery
Feb 7 at 19:20
$begingroup$
Will try with your perspective and get back to it today!
$endgroup$
– Catherine Drysdale
Feb 2 at 11:25
$begingroup$
Will try with your perspective and get back to it today!
$endgroup$
– Catherine Drysdale
Feb 2 at 11:25
$begingroup$
Hey, I have not looked at this problem for several days and I have come back to it, and I am still struggling slightly. I guess that I do not know what a sigma-algebra on space of functions means! Any help?
$endgroup$
– Catherine Drysdale
Feb 7 at 16:09
$begingroup$
Hey, I have not looked at this problem for several days and I have come back to it, and I am still struggling slightly. I guess that I do not know what a sigma-algebra on space of functions means! Any help?
$endgroup$
– Catherine Drysdale
Feb 7 at 16:09
$begingroup$
In my limited experience in working with measures on spaces other than R^n or subsets of it, the only way I was able to think about the problem was to sort of pretend it was R^n and only remind myself what space I was in when necessary. This is probably extremely unhelpful, but I don't know what other advice to give. Hopefully someone with more experience will chime in?
$endgroup$
– Jon Hillery
Feb 7 at 19:20
$begingroup$
In my limited experience in working with measures on spaces other than R^n or subsets of it, the only way I was able to think about the problem was to sort of pretend it was R^n and only remind myself what space I was in when necessary. This is probably extremely unhelpful, but I don't know what other advice to give. Hopefully someone with more experience will chime in?
$endgroup$
– Jon Hillery
Feb 7 at 19:20
add a comment |
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Hi Catherine! Please make your title informative, and avoid greetings or statements about your problem that would make it less useful to other users (example: "very hard algebra problem", "struggling with normal operators", "confusing matrix multiplication", etc). Thank you!
$endgroup$
– Pedro Tamaroff♦
Feb 7 at 16:39
$begingroup$
Is it possible to change it now? I cannot see where to edit the question.
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– Catherine Drysdale
Feb 7 at 16:53
$begingroup$
There is a button at the bottom of the post, next to the tags that says 'edit'.
$endgroup$
– Pedro Tamaroff♦
Feb 7 at 16:55