Excellent video, as usual. Just a note: from what I understand from books, the difference between an A and a sigma-A is that regarding the union requirement, in the former the union set is COUNTABLE, while in the latter it's COUNTABLE INFINITE. COUNTABLE means that there is a correspondence between each set and A natural number (injective function). COUNTABLE INFINITE means that there is a one-to-one correspondence between each set and THE natural numbers (bijective function).
I have the same doubt, Viji told that in algebra union is for two sets but i think we can do that union for any finite number of set and it will still belong in the algebra
@@DeepakSingh-le6di yes for sure. I think Viji brought just an example talking about two sets. It'd make no sense that such property holds for two sets and not for three or n
Viji, thank you so much. I have been trying to understand these math language heavy probability primers all day and was so stressed by how confusing everything was. It seems like everyone either talks about the intuitive ideas + pictures WITHOUT using the math language, or just sticks with the language. You have done both and you walked me through it so easily. I wish I would have found your channel 8 hours ago. Thank you!
Thank you for a very clear exposition. At the point where you mention hospitals, I am thinking (for a brief moment) "How does a video know, and adjust, to where I live?"
why is it necessary to define an algebra in accordance with the three conditions? What is the intuition? If we are interested for example of the function value in the range of a subset (a,b) as a new element from the universal set a,b,c,d,e,f, why should I include the element (c,d,e,f) as a complement in the algebra on the original set? Can you give an example? I have a second question: would ((a,b), (c),(d),(e),(f), (a,b,c,d,e,f), omega) , all brackets should be curly brackets, be an algebra? I mean, can I form the complement as a sum of elements in the new set or should I include the subset (c,d,e,f) instead as a complement? Thank you for the answer!
Very few of those good in maths can actually teach...If only the average of maths genius could teach maths in the way you have just demonstrated, then it would not be so dreadful as a subject! Absolutely brillant!
Thank you very much for these videos. I think I have understood the definition of the algebra A, but it took me I while because i found the statement of the second condition somehow ambiguous: "if x belongs to A then the complement of x must also belong to A". I mean, in this definition by "the complement of x" I can understand the subset of S composed by all the elements of S removing the elements x, but I could also (wrongly) understand it to be the subset of A composed by all the elements of A removing the element x. Shouldn't the definition some how include the clarification that the complement operation is meant on S, not on A? Could you please comment on this?
Hi Enrico! Thank you for your comment. These videos are part of a playlist on measure theory that can be found on my channel. The first video is on sets and can be found here: ruclips.net/video/NMAK10PyRDE/видео.html ... this video will clarify what I mean by the complement. I think at the beginning of the videos I typically refer to the previous video in the series so hopefully that will prompt people to watch those videos to get up to speed. Thank you!
I don't understand the practical difference between the algebra and the Sigma algebra. Eg, if you have 3 elements a, b, and c, per algebra rules there will also be a+b. Call that union d. Then d+c = a+b+c is also in the algebra, ie it's also a sigma algebra. Is this incorrect?
I like the video. But I'm confused, in slide 6 (6th minute), you are saying that Sigma Algebra is different from an algebra, since it has got extra requirement as follows, (a) "Any countable union of elements in sigma-algebra A must also belongs to A" My question is, is the above condition (a) for Sigma-Algebra been already implied by following condition (b) for Algebra? (b) "Union of any 2 elements in an Algebra must belongs to the same Algebra"
This is the most elegant presentation of measure theory I have watched. A kid watching this will understand Measure theory.
Excellent video, as usual.
Just a note: from what I understand from books, the difference between an A and a sigma-A is that regarding the union requirement, in the former the union set is COUNTABLE, while in the latter it's COUNTABLE INFINITE.
COUNTABLE means that there is a correspondence between each set and A natural number (injective function).
COUNTABLE INFINITE means that there is a one-to-one correspondence between each set and THE natural numbers (bijective function).
I have the same doubt, Viji told that in algebra union is for two sets but i think we can do that union for any finite number of set and it will still belong in the algebra
@@DeepakSingh-le6di yes for sure. I think Viji brought just an example talking about two sets. It'd make no sense that such property holds for two sets and not for three or n
This is the best video on sigma-algebras for non-mathematicians I could find. Thank you so much.
Viji, thank you so much. I have been trying to understand these math language heavy probability primers all day and was so stressed by how confusing everything was. It seems like everyone either talks about the intuitive ideas + pictures WITHOUT using the math language, or just sticks with the language. You have done both and you walked me through it so easily. I wish I would have found your channel 8 hours ago. Thank you!
Thank you !!!
hi sara can you plz tell me about borel field ?
This is the clearest explanation of the fundamentals of measure theory I've ever seen.
This is truly well explained.
Wow....such a beautiful progression of concepts
Best video on sigma algebra’s.
Thank you for a very clear exposition. At the point where you mention hospitals, I am thinking (for a brief moment) "How does a video know, and adjust, to where I live?"
dlfiftyone you don't live in a hospital, do you?
What a crystal clear presentation!
Excellent video lecture. Please consider adding historical background of this algebra.
Mind blowing series about measure Theory!!!! Thank u so much.And plz upload more...
Great presentations. Thank-you so much!
Thank you Viji, this cleared much of this stuff up for me!
Your explanation is wonderful!
Great clear easy to understand explanation. You should make more videos on math or any other topics. This channel deserves more views and subscribers
Thank you for your best.
Awesome!!!plz upload more....
why is it necessary to define an algebra in accordance with the three conditions? What is the intuition? If we are interested for example of the function value in the range of a subset (a,b) as a new element from the universal set a,b,c,d,e,f, why should I include the element (c,d,e,f) as a complement in the algebra on the original set? Can you give an example? I have a second question: would ((a,b), (c),(d),(e),(f), (a,b,c,d,e,f), omega) , all brackets should be curly brackets, be an algebra? I mean, can I form the complement as a sum of elements in the new set or should I include the subset (c,d,e,f) instead as a complement? Thank you for the answer!
great teacher
Very few of those good in maths can actually teach...If only the average of maths genius could teach maths in the way you have just demonstrated, then it would not be so dreadful as a subject! Absolutely brillant!
great!! makes life easy!!!
Thank you very much for these videos. I think I have understood the definition of the algebra A, but it took me I while because i found the statement of the second condition somehow ambiguous: "if x belongs to A then the complement of x must also belong to A". I mean, in this definition by "the complement of x" I can understand the subset of S composed by all the elements of S removing the elements x, but I could also (wrongly) understand it to be the subset of A composed by all the elements of A removing the element x.
Shouldn't the definition some how include the clarification that the complement operation is meant on S, not on A? Could you please comment on this?
Hi Enrico!
Thank you for your comment. These videos are part of a playlist on measure theory that can be found on my channel. The first video is on sets and can be found here: ruclips.net/video/NMAK10PyRDE/видео.html ... this video will clarify what I mean by the complement. I think at the beginning of the videos I typically refer to the previous video in the series so hopefully that will prompt people to watch those videos to get up to speed. Thank you!
excellent !
Thank you so much, this has helped a lot with my degree!
very nice thanks!
I don't understand the practical difference between the algebra and the Sigma algebra. Eg, if you have 3 elements a, b, and c, per algebra rules there will also be a+b. Call that union d. Then d+c = a+b+c is also in the algebra, ie it's also a sigma algebra. Is this incorrect?
I am confused about this too.
I like the video. But I'm confused, in slide 6 (6th minute), you are saying that Sigma Algebra is different from an algebra, since it has got extra requirement as follows,
(a) "Any countable union of elements in sigma-algebra A must also belongs to A"
My question is, is the above condition (a) for Sigma-Algebra been already implied by following condition (b) for Algebra?
(b) "Union of any 2 elements in an Algebra must belongs to the same Algebra"
so, for the Algebra it is any 2 elements ... for a sigma algebra it can be the union of 3 elements or the union of 4 elements and so on.
10:05 each unit set is an element of the generated set
Perfect!
I dont understand borel measure
that is one sexy voice