Nice informative video. As per my understanding, at 6:40, Omega={1,2,3} and {Omega}={{1,2,3}} are two different thing. {Omega} is a member of T not the Omega. Kindly correct me if I am wrong.
your first condition should be that omega is an element of sigma, not a subset. {omega} != omega, {omega} = {{1,2,3}}. There shouldn't be brackets around omega inside the examples. {null_set} != null_set. There shouldn't be brackets around the null_set in the examples.
At about 6:15, you define the trivial set as T = { {Ø}, {Ω} }, but I think you mean T = {Ø, Ω}, without the extraneous braces. Ø denotes the empty set, and {Ø} is a set with one element (which is the empty set), so they are different. For T to be a sigma-algebra, Ø and Ω themselves must be elements of T.
There is a mistake in the notation I think. When you want to indicate that, in your first example, T = {emptyset, omegaset}, you should write emptyset without parentheses, otherwise {emptyset} and {omega} mean a set containing another set... Also, the first condition at the beginning, you should not use operator "contains" but operator "in".
when you say "F is closed under countable unions", you shouldnt just mention the finite Union, but also the countable infite union of a series Ai, where it then would say "If A1,A2,A3,...Ai,... is an element of F, then so is the union UAi from i=1 to infinity"
As far as I understood, because of condition 1, it is not necessary to say infinite set. All you need is to change A1,A2, ... to A1 union (A1)^c (complement of A1). By this you can change infinite union to finite union. Indeed, you will have a finite union of sets.
@@VahidOnTheMove It might be late but I would like to give my thoughts. suppose omega is [0,2]. and the building sets for F are of the form [0,1-1/n] and let's include their complements and all the finite unions. the set [0,1) is not in F but is in the union on all the sets we started with. F does contain omega simply because it contains the sets and their compliments,it does obey the closure under compliments and finite union by definition. F doesn't have [0,1) because if you think about the group from topology thin this is an open set in the induced topology from R onto omega but doesn't include any set of the compliments of the sets [0,1/n]. so it isn't an open set in the topology from F
I have a question about the examples: Why is the sample set inside brackets {Ω}? shouldn't have to be without brackets like Ω ? because we already know that represents {1,2,3} so if that is inside brackets we get: {{1,2,3}} which is not at the level of the other subsets of the collection of each example, precluding to be measurable. I talking that instead of Z = {{{}},{Ω}} we should write Z = {{},Ω} and instead of Z = {{{}},{Ω},{1,2}, {3}, {2,3}} we should write : Z = {{},Ω,{1,2}, {3}, {2,3}}
thank you so much! I have paid for books which do not explain half as clear as you did, with the examples and all. I also thank scadqwqw for additional clarification.
I have another question, in the first property of the sigma-field it says that Ω ⊂ F. But as I understand the containment symbol (⊂) is used for subsets, but in this context Ω is not refered to a subset but an element of F, so shouldn't be written as Ω ∈ F the first property? Also in the second property It has A ∈ F which I consider it is correct.
Great, but at 12:53 you should add that is the cardinal of any sigma-algebra that lies between the cardinals of the trivial and power set instead of use "
Mind the notation of the first condition: you use the symbol of subset-of instead of the symbol of member-of. Also the definition you give later on corresponds to 'finite union', not to 'countable union'.
+Kevin Chen You can think of the operation union as "all the elements that belong to both A and B". Likewise, you can define the intersection as "all the elements unique to both A and B". So when you have Omega U 0, you are essentially asking "if I combine all the elements of the set and the empty set, what will I get". Obviously, you get the elements of the set because the null set contains nothing. Similarly, when asking "Omega intesect 0", you are looking for all the elements you can find in both Omega and the empty set. Well, Omega has elements, but the empty set has nothing. Therefore, they have no common element, nothing between them. So the result is the empty set.
![σ-algebras | [generated; partition; Borel]-sigma-algebras & much more](http://i.ytimg.com/vi/Co04Ah1rdwo/mqdefault.jpg)
![σ-algebras | [generated; partition; Borel]-sigma-algebras & much more](/img/tr.png)
Well this is certainly clearer than anything else on RUclips. Nice job coloring all those Venn diagrams in one go with no mistakes!
I watched 2-3 sigma-algebra before this and this video had the better explanation for me, thank you.
Nice informative video.
As per my understanding, at 6:40, Omega={1,2,3} and {Omega}={{1,2,3}} are two different thing. {Omega} is a member of T not the Omega. Kindly correct me if I am wrong.
your first condition should be that omega is an element of sigma, not a subset.
{omega} != omega, {omega} = {{1,2,3}}. There shouldn't be brackets around omega inside the examples.
{null_set} != null_set. There shouldn't be brackets around the null_set in the examples.
Nathaniel Gregg exactly
I observed the same
U are the best!!!!!!!!!!!! Even in my native language I cound´t find someone with this great and clear explanation.
I just realized that my college professor used the exact same notes for explaining Sigma algebra! thankyou
Besides the notation thing, great material! Really makes me wanna rip through my probability problems lol.
Hey ya, I find your video very clear and comprehensive.
Can you provide a sequence of watching?
Also can you make more videos like these.
Thank you!
The examples made the explanation lucid!
At about 6:15, you define the trivial set as T = { {Ø}, {Ω} }, but I think you mean T = {Ø, Ω}, without the extraneous braces. Ø denotes the empty set, and {Ø} is a set with one element (which is the empty set), so they are different. For T to be a sigma-algebra, Ø and Ω themselves must be elements of T.
Thank you for you video! You've made it very easy to understand.
thank you so much, I noticed that you put "phi" inside curly brackets "{ }"; in set theory it states that curly brackets "{ }" is equivalent to "phi";
There is a mistake in the notation I think. When you want to indicate that, in your first example, T = {emptyset, omegaset}, you should write emptyset without parentheses, otherwise {emptyset} and {omega} mean a set containing another set...
Also, the first condition at the beginning, you should not use operator "contains" but operator "in".
Totally agree with u
Great description and examples! This cleared things up for me
Thanks good i finally find an example about what isn't a sigma-álgebra, thanks man.
thank you sir, you are far better than my teacher, Danke
its similar to properties of discrete probability distribution.p(X)=1 and 0
when you say "F is closed under countable unions", you shouldnt just mention the finite Union, but also the countable infite union of a series Ai, where it then would say "If A1,A2,A3,...Ai,... is an element of F, then so is the union UAi from i=1 to infinity"
As far as I understood, because of condition 1, it is not necessary to say infinite set. All you need is to change A1,A2, ... to A1 union (A1)^c (complement of A1). By this you can change infinite union to finite union. Indeed, you will have a finite union of sets.
@@VahidOnTheMove It might be late but I would like to give my thoughts. suppose omega is [0,2]. and the building sets for F are of the form [0,1-1/n] and let's include their complements and all the finite unions. the set [0,1) is not in F but is in the union on all the sets we started with. F does contain omega simply because it contains the sets and their compliments,it does obey the closure under compliments and finite union by definition. F doesn't have [0,1) because if you think about the group from topology thin this is an open set in the induced topology from R onto omega but doesn't include any set of the compliments of the sets [0,1/n]. so it isn't an open set in the topology from F
Helped me with my homework. Thanks
I have a question about the examples: Why is the sample set inside brackets {Ω}? shouldn't have to be without brackets like Ω ? because we already know that represents {1,2,3} so if that is inside brackets we get: {{1,2,3}} which is not at the level of the other subsets of the collection of each example, precluding to be measurable. I talking that instead of Z = {{{}},{Ω}} we should write Z = {{},Ω} and instead of Z = {{{}},{Ω},{1,2}, {3}, {2,3}} we should write : Z = {{},Ω,{1,2}, {3}, {2,3}}
There seems to be something wrong with property 1. Omega is an element of F rather than a subset of F.
This is true. He says that Omega is a member of F, which is correct, but he have used the wrong symbol.
A nice and a short video. 0:49 the sign of inclusion is inaccurate. Sigma belongs to F as an element, but not as a subset.
Thanks for the video. It is clear, and answered all my questions :D
Very helpful. Thank you so much
Thank you for taking the time to make this.
Thanks, great video, friend.
This leacture is really awesome
thank you so much! I have paid for books which do not explain half as clear as you did, with the examples and all. I also thank scadqwqw for additional clarification.
well explained
Simple and very clear explanation od sigma algebra.
Not totally sure, but think that empty set and Omega may not be in bracket, because they are allredy a set, so is like put a set into a set
Are uncountable sigma algebras of computational interest, given they run smack dab into AoC+CH buzz saw?
you deserve heaven sir
Awesome. Easy to understand.
I have another question, in the first property of the sigma-field it says that Ω ⊂ F. But as I understand the containment symbol (⊂) is used for subsets, but in this context Ω is not refered to a subset but an element of F, so shouldn't be written as Ω ∈ F the first property? Also in the second property It has A ∈ F which I consider it is correct.
Great, but at 12:53 you should add that is the cardinal of any sigma-algebra that lies between the cardinals of the trivial and power set instead of use "
Thank you. The video is fantastic.
thanks for clear explanation.
This is great please make more videos on probability measure!
wrong, sigma algrebra uses infinity union. a finity union define only a algebra.
I guess you are right. if F is theta-field and if A belongs to F than the infinity union of An belongs to F.
Thanks, very clear. Why not build on this video and explain Measure Theory?
you do very nice, make more video on measure theory
wonderful
the example in 7:30 is actually a set with an empty set inside of it. So, the condition for a sigma-algebra is NOT satisfied I would argue
Thank you very much! Very good explanation!
Thanks, very very helpful!!!!
thank u so much it made me understand this topic....
thanks, that was actually pretty helpful. Keep it up!
this exemples works, because you uses a finity set.
Mind the notation of the first condition: you use the symbol of subset-of instead of the symbol of member-of.
Also the definition you give later on corresponds to 'finite union', not to 'countable union'.
really really helpful, thanks!
Thanks
Very clear! Thanks!
Muy bueno!! Very good
Very clear video, thanks a lot ;-)
really helpful~
How about the the Borel sigma field?
Thankyou thankyou thankyou!!
Thank you!
Dzięki !!!
Thanks a lot. Very clear. ;-)
Thanks!
THANK YOU!!
think the same
I'm lost in this example; T = { {Ø}, {Ω} }, why ØUΩ=Ω and ØnΩ=Ø?
+Kevin Chen
You can think of the operation union as "all the elements that belong to both A and B". Likewise, you can define the intersection as "all the elements unique to both A and B". So when you have Omega U 0, you are essentially asking "if I combine all the elements of the set and the empty set, what will I get". Obviously, you get the elements of the set because the null set contains nothing. Similarly, when asking "Omega intesect 0", you are looking for all the elements you can find in both Omega and the empty set. Well, Omega has elements, but the empty set has nothing. Therefore, they have no common element, nothing between them. So the result is the empty set.
THANK YOUU
Thanks, I've got it
Danke;-)
is {empty set} = empty set ?
Absolutely not
No. He made a mistake in the video
:D
Superficial stuff, nothing useful.
Thanks a lot ra Dhootha