in the example at time stamp (15:13) you constructed intervals of the form (0, 1/n] but these intervals do not satisfy a_n > b_(n-1) condition you state above. it would be kind of you to clarify if i misunderstood what you are trying to state.
The orange example at 12:51 is just an example to show that script A on the top of the page is not a sigma-algebra. If your question was about why it is relevant to script A: just one interval of the form (. , .] is in A (take n=1 in the union). Let's change the index of the example from n to k (I should have done that in the video to avoid confusion), and consider a sequence of intervals (0, 1-1/k] as k=2...infinity. For each k, this is in script A (with n=1 in the definition of script A each time). Hence, the union of these things for all k's should also be in script A, if script A would be a sigma-algebra. But it is not, since the union is not in script A: it is not of the form (. , .], nor a finite combination of these. For the next bit starting at 14:44 you could forget about this example in orange: it continues the discussion of the script A algebra on the top of the page. I hope this helps.
In my research, I need a sequence of probability spaces. All the spaces have a uniform probability measure. Therefore, I was wondering if I could simplify my notation by omitting the probability measure from the notation?
If you are using the measure for anything (e.g., considering probabilities of events, taking expectations) then I suppose you will need to have some notation for it anyway?
@@martonbalazsuob4363 Well, the measure exists. But if it is always a uniform probability, then I suggest it isn't necessary to include in the notation. It can be implicit that the probability measure is a uniform probability. This would simplify the notation greatly. Is this a reasonable way of writing a paper?
Yes, but the countability of subsets of the sample space gives us the right foundations to build a useful theory of probability. That's the important bit, we don't need the actual sigma-algebra to be countable.
you are the best teacher in this field. really thank you for your helping
in the example at time stamp (15:13) you constructed intervals of the form (0, 1/n] but these intervals do not satisfy a_n > b_(n-1) condition you state above. it would be kind of you to clarify if i misunderstood what you are trying to state.
The orange example at 12:51 is just an example to show that script A on the top of the page is not a sigma-algebra. If your question was about why it is relevant to script A: just one interval of the form (. , .] is in A (take n=1 in the union). Let's change the index of the example from n to k (I should have done that in the video to avoid confusion), and consider a sequence of intervals (0, 1-1/k] as k=2...infinity. For each k, this is in script A (with n=1 in the definition of script A each time). Hence, the union of these things for all k's should also be in script A, if script A would be a sigma-algebra. But it is not, since the union is not in script A: it is not of the form (. , .], nor a finite combination of these.
For the next bit starting at 14:44 you could forget about this example in orange: it continues the discussion of the script A algebra on the top of the page. I hope this helps.
In my research, I need a sequence of probability spaces. All the spaces have a uniform probability measure. Therefore, I was wondering if I could simplify my notation by omitting the probability measure from the notation?
If you are using the measure for anything (e.g., considering probabilities of events, taking expectations) then I suppose you will need to have some notation for it anyway?
@@martonbalazsuob4363 Well, the measure exists. But if it is always a uniform probability, then I suggest it isn't necessary to include in the notation. It can be implicit that the probability measure is a uniform probability. This would simplify the notation greatly. Is this a reasonable way of writing a paper?
@@PikesCore24 Sorry I don't think I can help more without seeing the details.
@4:15 but the power set of countably infinite set is uncountable
Yes, but the countability of subsets of the sample space gives us the right foundations to build a useful theory of probability. That's the important bit, we don't need the actual sigma-algebra to be countable.